DTH Compressor Sizing and Duty Confirmation

Q: Can borehole depth alone determine compressor size?

No. Depth changes drill-string losses, water exposure and operating conditions. Sizing also needs the hammer data, hole and pipe geometry, air demand and a supported pressure-flow operating point.

Q: Should hammer air and cleaning air be added together?

First identify the flow paths and put both requirements on one reference basis. Hammer exhaust can serve hole cleaning. Add only distinct simultaneous consumers and justified losses; confirm that the hammer or approved bypass can pass the required flow.

Q: How is water head used in a pressure calculation?

For an illustrative static freshwater column, pressure is approximately 0.0981 bar per metre. Use the effective column rather than total hole depth. Do not add it again if the stated hammer inlet requirement already includes exhaust backpressure.

Q: Does a booster increase available primary airflow?

A booster raises pressure within its documented operating envelope. It does not remedy a shortage of upstream mass flow. Verify inlet conditions, throughput, losses and final delivered duty for the combined package.

Q: Can maximum CFM and maximum pressure be used together?

Only when manufacturer data confirms that simultaneous operating point for the offered configuration and site conditions. Separate catalogue maxima do not establish the required duty.

Q: What information should accompany a sizing enquiry?

Send hole diameter and target depth, pipe outside and inside diameters, hammer and bit identity, available consumption data, expected water conditions, site temperature and altitude, hose layout, operating hours and the intended duty stages.

Choose a DTH compressor by the pressure and airflow it can deliver together at the site. Begin with the hole, drill string, hammer and return conditions; establish a common flow basis; then compare the complete duty with documented compressor and booster capability. This page gives the calculations, worked examples and decision records needed to explain that selection.

How to use the calculations

The numerical cases are illustrative assumptions, not Reddy fleet ratings, quotations, guaranteed depths or universal settings. All CFM comparisons within an example assume the same declared reference pressure, temperature and moisture basis. Values labelled actual flow apply only at the stated local thermodynamic condition. If a real case lacks that information, record it as unknown.

The pressure-budget examples use hammer differential pressure and an explicitly assumed exhaust backpressure. Sections 7 and 21 explain the alternative when the supplier gives a hammer inlet pressure already corrected for exhaust conditions. Freshwater head calculations use density 1000 kg/m³ and gravitational acceleration 9.81 m/s². Circulating air-water-cuttings mixtures require a justified return-flow assessment; static head alone is not a complete model.

A shared hammer/exhaust path does not guarantee that any calculated excess air can pass through the selected hammer. A manufacturer-approved flushing or bypass configuration may be necessary. The maximum of hammer and cleaning demands is a preliminary lower bound after conversion to one basis, not proof that the flow route can deliver it.

Define the duty before requesting equipment

A defensible sizing decision connects the application’s air demand with the offered compressor’s supported duty. It should be possible to explain why the machine was selected, which assumptions matter and what changes would require a new review. A CFM label alone is not that explanation.

Start with the exact hammer and its documented demand at the intended operating condition. Add the drilling specialist’s assessment of the hole, drill string and return conditions. Identify separate operating stages, including any different requirement during flushing or another approved activity.

Do not automatically add a hammer-flow figure to a hole-cleaning-flow figure as though they were unrelated consumers. The reviewer must establish the actual flow paths and whether each figure describes the same air serving successive functions. Genuine simultaneous additional consumers should be identified separately.

Record airflow units, reference conditions and the pressure boundary for both sides of the comparison. State whether a supplied value is free-air delivery, another reference-volume quantity or a local actual volume. Refer to the Annular Velocity guide when translating a geometric calculation into a defined flow basis.

Request model-specific compressor evidence for the proposed duty and site conditions. Do not combine an airflow maximum and a pressure maximum from different operating points. The offered configuration and its applicable performance information belong in the selection record.

Any allowance for distribution, site conditions, uncertainty or future duty should be named and justified by the responsible reviewer. Avoid an unexplained percentage added repeatedly at different stages. A transparent selection shows what has been allowed for and what remains unresolved.

The final duty statement should identify equipment, application, boundaries and limitations in plain language. It should also name changes that invalidate the selection, such as different tooling or an altered air arrangement. A quotation can then refer to a common technical brief instead of relying on separate verbal descriptions.

Request a duty-based proposal: submit the project enquiry brief with manufacturer data and the contractor’s inputs. The earlier DTH Compressor Selection guide remains the introduction; this page explains the decision record used to close the selection.

Calculation and selection reference

The following 225 numbered sections cover geometry, demand, pressure, equipment matching, economics, changing conditions, units and an integrated worked example.

1 The Two Fundamental Compressor Requirements

Every DTH compressor-selection calculation must independently establish:

Airflow Requirement

$Q_{r e q u i r e d}$

and:

Pressure Requirement

$P_{r e q u i r e d}$

The compressor must satisfy both simultaneously at the required operating condition.

Therefore:

$Q_{a v a i l a b l e} \geq Q_{r e q u i r e d}$

and:

$P_{a v a i l a b l e} \geq P_{r e q u i r e d}$

Meeting only one condition is insufficient.

2 Why Compressor Horsepower Alone Is Not a Selection Method

Engine power does not directly tell us whether a compressor is suitable for a particular DTH application.

Two machines with similar engine ratings may have different:

airflow
maximum pressure
compressor architecture
control characteristics
ambient derating
operating envelopes

Therefore:

$Engine Power \neq DTH Compressor Selection$

The drilling system requires verified pressure-flow performance.

3 Required Input Data

Before calculation, establish:

Borehole

target hole diameter $D_{h}$ ,

target depth $L$ ,

expected final depth,

casing programme.

Drill String

drill-pipe outside diameter $D_{p}$ ,

drill-pipe inside diameter $D_{i}$ ,

joint geometry,

expected total length.

Hammer

model,

nominal size,

recommended bit range,

manufacturer airflow curve,

manufacturer operating-pressure range.

Formation

hardness,

abrasiveness,

expected cuttings characteristics,

fractured/competent condition.

Water

expected groundwater,

water-column/head estimate,

likely water inflow.

Site

altitude,

ambient temperature,

access/logistics.

Without these inputs, compressor sizing is preliminary rather than final.

4 Step 1 — Select Required Hole Diameter

Let:

$D_{h} = Borehole Diameter$

This parameter influences:

bit selection
hammer selection
annular area
rock volume removed
required hole-cleaning airflow
A larger hole generally requires greater flushing capability

5 Step 2 — Select Compatible Bit

The bit must satisfy:

$D_{b i t} \approx D_{h}$

subject to:

hammer compatibility
formation
casing requirement
manufacturer recommendations

Bit diameter establishes the effective drilled-hole diameter used in the annular-flow calculation.

6 Step 3 — Select Compatible DTH Hammer

Select a hammer compatible with:

bit shank
hole diameter
formation
rig capability

Then obtain manufacturer data for:

$Q_{h} (P)$

Where:

$Q_{h}$ = hammer air consumption,

$P$ = hammer operating pressure.

Do not estimate hammer consumption from diameter alone when verified manufacturer data is available.

7 Define the hammer pressure boundary

Record whether the manufacturer value is hammer inlet gauge pressure, inlet absolute pressure, or inlet-to-exhaust differential pressure. These quantities cannot be substituted for each other.

For the pressure-budget examples on this page, $Δ P_{h}$ means the assumed pressure difference required across the hammer. The inlet requirement is then:

$P_{h, i n, g} = P_{h, o u t, g} + Δ P_{h}$ .

If the supplier instead gives an inlet pressure already corrected for the actual exhaust backpressure, use that inlet requirement directly and do not add the same backpressure a second time. A catalogue curve measured with atmospheric exhaust does not, by itself, establish performance under water backpressure. Obtain the selected hammer manufacturer’s interpretation and limits.

8 Step 5 — Establish Hammer Air Demand

From the manufacturer airflow curve determine:

$Q_{h}$

at the intended:

$Δ P_{h}$

This is the first major airflow requirement.

But it may not be the final compressor airflow requirement.

Hole cleaning must also be checked.

9 Step 6 — Calculate Annular Area

For a simplified circular bore and drill pipe:

$A_{a} = \frac{π}{4} (D_{h}^{2} - D_{p}^{2})$

Where:

$A_{a}$ = annular area,

$D_{h}$ = borehole diameter,

$D_{p}$ = drill-pipe outside diameter.

Use consistent units.

10 Worked Annular-Area Example

Assume:

$D_{h} = 0.1524 m$

and:

$D_{p} = 0.089 m$

Then:

$A_{a} = \frac{π}{4} ({0.1524}^{2} - {0.089}^{2})$

Calculate:

${0.1524}^{2} = 0.023226$

${0.089}^{2} = 0.007921$

Therefore:

$A_{a} = \frac{π}{4} (0.015305)$

$A_{a} \approx 0.0120 m^{2}$

This is the approximate annular return-flow area.

11 Step 7 — Establish Required Annular Velocity

However, the appropriate velocity is application dependent.

It depends on:

cuttings size
density
formation
bore geometry
water conditions
air expansion

12 Critical Flow-Basis Warning

The airflow used in:

$Q_{a} = V_{a} A_{a}$

must represent volumetric flow at the conditions where annular velocity is being evaluated.

Do not insert compressor FAD directly into an in-hole actual-volume equation without converting conditions appropriately.

This is one of the most common sources of incorrect drilling-air calculations.

13 Step 8 — Compare Hammer Demand and Cleaning Demand

The compressor must satisfy both functions.

Conceptually:

$Q_{r e q u i r e d} \geq \max (Q_{h}, Q_{c l e a n i n g})$

plus allowances for:

legitimate leakage
operating variation
appropriate engineering reserve

Do not simply add two complete airflow values if the hammer exhaust itself is performing the hole-cleaning function.

Avoid double-counting the same air.

14 Airflow Balance

A more useful system relationship is:

$Q_{c o m p r e s s o r} = Q_{h a m m e r p a t h} + Q_{o t h e r l e g i t i m a t e d e m a n d} + Q_{l o s s e s}$

The hammer exhaust normally becomes part of the return airflow.

Therefore hammer air consumption and annular cleaning are interconnected requirements.

15 Step 9 — Calculate Surface Pressure Losses

Pressure can be lost through:

compressor outlet
aftercooling/separation components where fitted
hose
valves
manifold
couplings
drill head/swivel

Represent:

$Δ P_{s} = \sum Δ P_{s u r f a c e}$

These losses should preferably come from:

manufacturer data
pressure-drop calculations
measured field data

16 Step 10 — Calculate Drill-String Pressure Loss

Conceptually:

$Δ P_{d} = f (L, D_{i}, Q, ρ, ϵ, fittings)$

Where:

$L$ = drill-string length,

$D_{i}$ = internal diameter,

$Q$ = airflow,

$ρ$ = air density,

$ϵ$ = roughness.

Because compressed air is compressible, rigorous high-pressure calculations require an appropriate compressible-flow method.

A simple incompressible-water pressure-drop equation should not be blindly applied.

17 Why Drill-Pipe Internal Diameter Matters

For a given flow:

$D_{i} ↓ \Rightarrow V_{p i p e} ↑$

which can substantially increase frictional pressure loss.

Therefore a drill pipe selected only for mechanical strength can unintentionally create a pneumatic restriction.

The drill string must satisfy both:

$Mechanical Requirement$

and:

$Airflow Requirement$

18 Step 11 — Calculate Hydrostatic Backpressure

For a water column:

$P_{w} = ρ_{w} g h$

For freshwater, approximately:

$P_{w} \approx 0.0981 h bar$

when $h$ is in metres.

19 Worked Water-Head Example

Suppose the effective water head opposing the drilling air is:

$h = 150 m$

Then:

$P_{w} \approx 0.0981 \times 150$

$P_{w} \approx 14.7 b a r$

This is already a substantial pressure requirement before adding:

hammer differential pressure
drill-string loss
surface losses
controlled margin

20 Water Head Is Not Necessarily Equal to Hole Depth

A 300 m borehole does not automatically create:

300m
of hydrostatic water head
The relevant fluid column depends on actual borehole and groundwater conditions

Therefore:

$Hole Depth \neq Water Head$

unless the actual conditions make them equivalent.

21 Build the source pressure budget

Using a hammer differential-pressure requirement, the planning relationship is:

$P_{s o u r c e, g} = Δ P_{h} + P_{h, o u t, g} + Δ P_{s} + Δ P_{d} + Δ P_{o} + P_{m}$ .

Here $P_{h, o u t, g}$ is the gauge pressure at the hammer exhaust; $Δ P_{s}$ is the surface supply loss; $Δ P_{d}$ is drill-string supply loss; $Δ P_{o}$ contains additional losses not already counted; and $P_{m}$ is a named operating allowance.

For the illustrative static freshwater cases, $P_{h, o u t, g}$ is approximated by $P_{w} = ρ_{w} g h / 10^{5}$ bar. This approximation does not predict the moving air-water-cuttings return. Do not add full static water head to an exhaust pressure that already includes it. State the return-flow model and avoid counting annular losses twice.

An alternative route, where the supplier supplies the required hammer inlet gauge pressure for the actual exhaust condition, is $P_{s o u r c e, g} = P_{h, i n, g} + Δ P_{s} + Δ P_{d} + Δ P_{o} + P_{m}$ . Both routes must refer to the same operating case.

22 Do Not Add Arbitrary Pressure Margin

The engineering margin should account for legitimate uncertainty such as:

depth variation
measured loss variation
water conditions
operating fluctuations

It should not be used to conceal:

unknown pressure losses
poor calculation
leakage
restrictions
The objective is controlled reserve—not arbitrary oversizing

23 Step 13 — Establish Required Compressor Flow

At the relevant reference condition:

$Q_{c} \geq Q_{r e q u i r e d}$

with appropriate allowance for actual site conditions.

The selected compressor must deliver the required airflow at the required pressure, not merely at a lower catalogue pressure.

24 Pressure–Flow Operating Point

The true selection point is:

$(Q_{r e q u i r e d}, P_{r e q u i r e d})$

Therefore compressor comparison should ask:

Can this machine deliver the required flow at this pressure under these site conditions?

Not merely:

What is its maximum CFM?

or:

What is its maximum pressure?

25 Maximum CFM and Maximum Pressure May Not Be Simultaneous

Depending on compressor design, a machine’s:

$Q_{m a x}$

and:

$P_{m a x}$

may not represent one simultaneous operating point.

Therefore use verified pressure-flow performance information rather than combining unrelated maximum specifications.

26 Step 14 — Correct for Altitude

At increased altitude, atmospheric pressure and air density decrease.

This can affect:

engine power
compressor mass-flow capability
cooling

Therefore:

$Q_{a v a i l a b l e, s i t e}$

may differ from rated reference-condition performance.

Use manufacturer altitude correction or derating data.

27 Step 15 — Correct for Ambient Temperature

High ambient temperature can influence:

air density
cooling
engine capability
compressor operating temperature
Use equipment-specific correction data where required
Do not invent a universal percentage derating per degree

28 Step 16 — Verify Engine Capability

A compressor package must have sufficient engine capability to produce the required pressure-flow condition under site conditions.

Therefore verify:

altitude derating
ambient conditions
compressor loading
cooling capacity

The compressor air end cannot continuously deliver a condition that the prime mover cannot support.

29 Step 17 — Check Compressor Operating Reserve

After site correction:

$Q_{r e s e r v e} = Q_{a v a i l a b l e, s i t e} - Q_{r e q u i r e d}$

and:

$P_{r e s e r v e} = P_{a v a i l a b l e} - P_{r e q u i r e d}$

Reserve should be positive and technically justified.

A system operating continuously with virtually zero reserve can become vulnerable to:

depth increase
water entry
leakage
wear
environmental changes

30 Airflow Reserve Percentage

A project-specific indicator is:

$R_{Q} (%) = \frac{Q_{a v a i l a b l e} - Q_{r e q u i r e d}}{Q_{r e q u i r e d}} \times 100$

This is useful for comparison but should not be treated as a universal required reserve percentage.

31 Pressure Reserve Percentage

Similarly:

$R_{P} (%) = \frac{P_{a v a i l a b l e} - P_{r e q u i r e d}}{P_{r e q u i r e d}} \times 100$

Again, acceptable reserve depends on:

application
equipment
uncertainty
manufacturer operating envelope

32 Step 18 — Determine Whether a Booster Is Required

A booster becomes relevant when:

$P_{r e q u i r e d} > P_{p r a c t i c a l, p r i m a r y}$

while the primary system can still provide suitable upstream mass flow for the booster.

Therefore:

$Booster Decision = Pressure Deficit with Adequate Upstream Flow$

not simply:

$Deep Hole \Rightarrow Booster$

33 Booster Is Not an Airflow Substitute

If:

$Q_{p r i m a r y} < Q_{r e q u i r e d}$

a booster does not solve the fundamental upstream mass-flow deficit.

The correct response may require:

greater primary compressor capacity
different compressor architecture
reduced avoidable losses
revised hammer/bit configuration

34 Booster Inlet Requirement

Verify:

$Q_{p r i m a r y} \geq Q_{b o o s t e r, i n l e t}$

at the required inlet pressure and reference conditions.

Also verify:

booster allowable inlet pressure
maximum discharge pressure
pressure ratio
cooling
temperature
duty cycle

35 Booster Pressure Ratio

Using absolute pressures:

$P R = \frac{P_{o u t, a b s}}{P_{i n, a b s}}$

Example:

Primary compressor:

$P_{i n} = 20 b a r (g)$

Required booster outlet:

$P_{o u t} = 35 b a r (g)$

Approximately:

$P_{i n, a b s} = 21 b a r$

$P_{o u t, a b s} = 36 b a r$

Therefore:

$P R = \frac{36}{21}$

$P R \approx 1.71$

Actual allowable operating conditions must come from booster manufacturer data.

36 Step 19 — Verify Complete Pressure Rating

Every downstream component must be rated appropriately for the maximum credible pressure.

Verify:

booster discharge piping
hose
manifold
valves
couplings
drill head
swivel
drill string
associated components
The lowest-rated component can establish the safe system limit

37 Step 20 — Check Compressor Fuel/Productivity Implication

Once several technically acceptable compressor configurations are identified, compare:

L/h
L/m
ROP
₹/m
Availability

The smallest technically suitable machine is not automatically the lowest-cost drilling solution.

38 Compressor Sizing Calculation Sheet

A controlled selection worksheet should contain:

Parameter	Symbol	Value
Hole diameter	$D_{h}$	—
Pipe OD	$D_{p}$	—
Pipe ID	$D_{i}$	—
Depth	$L$	—
Hammer	—	—
Hammer pressure	$Δ P_{h}$	—
Hammer airflow	$Q_{h}$	—
Annular area	$A_{a}$	—
Cleaning airflow	$Q_{c l e a n}$	—
Surface loss	$Δ P_{s}$	—
Drill-string loss	$Δ P_{d}$	—
Water head	$h$	—
Water pressure	$P_{w}$	—
Required source pressure	$P_{r e q}$	—
Required airflow	$Q_{r e q}$	—
Altitude	—	—
Temperature	—	—
Site-corrected compressor flow	$Q_{s i t e}$	—
Compressor max working pressure	—	—
Booster required	—	Yes/No

This worksheet creates traceability between drilling requirement and equipment selection.

39 Worked Example A — Shallow Dry DTH Hole

Assume, for illustration:

hole diameter = 115 mm
drill pipe OD = 76 mm
depth = 80 m
dry borehole

selected hammer requires a manufacturer-verified airflow $Q_{h}$ ,

assumed hammer differential pressure = 18 bar,

calculated/verified surface + drill-string losses = 2 bar,

controlled margin = 2 bar.

With:

$P_{w} = 0$

required source pressure becomes:

$P_{r e q} = 18 + 2 + 2$

$P_{r e q} = 22 b a r (g)$

The compressor must then be checked for adequate airflow at approximately this operating pressure.

The example intentionally does not invent hammer CFM; that value must come from the selected hammer’s verified data.

40 Example A — Annular Area

Using:

$D_{h} = 0.115 m$

and:

$D_{p} = 0.076 m$

the annular area is:

$A_{a} = \frac{π}{4} ({0.115}^{2} - {0.076}^{2})$

${0.115}^{2} = 0.013225$

${0.076}^{2} = 0.005776$

Therefore:

$A_{a} = \frac{π}{4} (0.007449)$

$A_{a} \approx 0.00585 m^{2}$

This area is then used to evaluate whether the selected hammer airflow provides adequate return-flow velocity under the relevant in-hole conditions.

41 Example A — Compressor Selection Logic

The selection sequence becomes:

$Q_{h a m m e r} \to Q_{c l e a n i n g} \to Q_{r e q u i r e d}$

and:

$P_{h a m m e r} + Δ P_{s y s t e m} + P_{m a r g i n} \to P_{r e q u i r e d}$

For Example A:

$P_{r e q u i r e d} \approx 22 b a r (g)$

The candidate compressor must therefore provide:

$Q_{a v a i l a b l e, s i t e} \geq Q_{r e q u i r e d}$

at:

$P \approx 22 b a r (g)$

after any applicable site correction.

If the primary compressor can satisfy both requirements with appropriate reserve, a booster is unnecessary.

42 Example A — Why Maximum Compressor Pressure Is Irrelevant

Suppose a candidate compressor has a maximum rating substantially above 22 bar.

That alone does not make it the preferred machine.

The important question remains:

What airflow can the compressor continuously deliver at the required operating pressure, under actual site conditions?

Therefore:

$P_{m a x} > P_{r e q u i r e d}$

is necessary only to the extent required for a suitable operating envelope.

Excess maximum pressure does not automatically create drilling value.

43 Worked Example B — Deeper Dry Hole

Assume:

hole diameter = 152 mm
drill-pipe OD = 89 mm
target depth = 300 m
dry borehole
hammer differential pressure assumption = 20 bar
surface-system loss = 1.5 bar
calculated/verified drill-string loss = 4.0 bar
other justified loss = 0.5 bar
engineering margin = 2.0 bar

Because the hole is dry:

$P_{w} = 0$

Therefore:

$P_{r e q u i r e d} = 20 + 1.5 + 4.0 + 0.5 + 2.0$

$P_{r e q u i r e d} = 28 b a r (g)$

The pressure requirement has increased substantially even though no hydrostatic head exists.

44 Example B — Annular Area

Using:

$D_{h} = 0.152 m$

and:

$D_{p} = 0.089 m$

$A_{a} = \frac{π}{4} ({0.152}^{2} - {0.089}^{2})$

${0.152}^{2} = 0.023104$

${0.089}^{2} = 0.007921$

Therefore:

$A_{a} = \frac{π}{4} (0.015183)$

$A_{a} \approx 0.011925 m^{2}$

This is more than twice the annular area in the smaller-hole Example A.

Therefore the same actual return airflow would produce a substantially lower annular velocity.

45 Example B — Annular Velocity Sensitivity

Suppose actual in-hole volumetric flow at the evaluated location is:

$Q_{a}$

Then:

$V_{a} = \frac{Q_{a}}{0.011925}$

If the bore diameter increases while $Q_{a}$ remains unchanged:

$A_{a} ↑$

therefore:

$V_{a} ↓$

This illustrates why compressor selection for a larger bore cannot be based only on hammer consumption.

Hole cleaning must also be verified.

46 Example B — Primary Compressor Decision

Suppose the candidate primary compressor is rated for the required airflow but only at:

25bar

while the calculated source requirement is:

28bar

Then:

$P_{a v a i l a b l e} < P_{r e q u i r e d}$

The configuration fails the pressure gate.

Possible engineering responses include:

select a higher-pressure compressor
evaluate a properly matched booster
reduce verified avoidable pressure losses
reconsider the complete drilling architecture

Do not simply assume the 25-bar machine will be adequate because the difference appears small.

47 Worked Example C — Deep Water-Bearing Hole

Assume:

hole diameter = 165 mm
drill-pipe OD = 102 mm
target depth = 400 m
hammer differential pressure assumption = 20 bar
surface loss = 1.5 bar
drill-string loss = 4.5 bar
other losses = 1.0 bar
effective freshwater head = 120 m
engineering margin = 2 bar

First calculate hydrostatic pressure:

$P_{w} = 0.0981 \times 120$

$P_{w} \approx 11.77 b a r$

Then:

$P_{r e q u i r e d} = 20 + 1.5 + 4.5 + 1.0 + 11.77 + 2$

$P_{r e q u i r e d} \approx 40.77 b a r (g)$

This illustrates how groundwater can fundamentally change compressor architecture.

48 Example C — Why Hole Depth Alone Would Have Failed

The hole depth is:

400m

but the water head used in the calculation is only:

120m

Therefore using:

$400 \times 0.0981$

would incorrectly assume a 400 m effective water column.

The correct pressure calculation must use the actual relevant fluid head.

This reinforces:

$Drilling Depth \neq Hydrostatic Head$

49 Example C — Booster Decision

Suppose the primary compressor can supply the required mass flow at approximately:

20bar(g)

but cannot reach:

40.77bar(g)
Then the system has principally a pressure deficit

A high-pressure booster may therefore be evaluated, provided:

$Q_{p r i m a r y} \geq Q_{b o o s t e r, i n l e t}$

and the complete booster operating envelope is suitable.

50 Example C — Booster Pressure Ratio

Assume:

$P_{i n} = 20 b a r (g)$

and:

$P_{o u t} = 41 b a r (g)$

approximately.

Convert to absolute pressure:

$P_{i n, a b s} \approx 21 b a r$

$P_{o u t, a b s} \approx 42 b a r$

Then:

$P R = \frac{42}{21}$

$P R \approx 2.0$

Whether this pressure ratio is permissible must be established from the actual booster manufacturer’s performance data.

51 Example C — Complete High-Pressure Rating Check

If boosted discharge pressure is approximately:

41bar(g)
do not check only the booster

Verify the ratings of:

discharge hose
manifold
valves
couplings
drill head
swivel
drill string
other downstream components

The safe operating limit is constrained by the weakest appropriately identified component in the pressure circuit.

52 Worked Example D — Large-Diameter Hole

Assume:

hole diameter = 203 mm
drill-pipe OD = 114 mm

Then:

$A_{a} = \frac{π}{4} ({0.203}^{2} - {0.114}^{2})$

${0.203}^{2} = 0.041209$

${0.114}^{2} = 0.012996$

Difference:

0.028213

Therefore:

$A_{a} = \frac{π}{4} (0.028213)$

$A_{a} \approx 0.02216 m^{2}$

This is a large annular area.

Hole-cleaning airflow can therefore become a major selection driver.

53 Example D — Required Actual Flow for Selected Velocity

If an engineering assessment establishes a target actual annular velocity:

$V_{t a r g e t}$

then:

$Q_{a c t u a l} = A_{a} V_{t a r g e t}$

For illustration only, if:

$V_{t a r g e t} = 20 m / s$

then:

$Q_{a c t u a l} = 0.02216 \times 20$

$Q_{a c t u a l} = 0.4432 m^{3} / s$

or:

$Q_{a c t u a l} \approx 26.6 m^{3} / m i n$

Important: this is actual volumetric flow at the location/conditions being evaluated—not automatically compressor FAD.

The chosen target velocity itself must be justified for the actual drilling application.

54 Example D — Convert Actual Flow Carefully

To relate an in-hole actual flow to compressor reference flow, use an appropriate gas relationship.

For an idealized mass balance:

$\frac{P_{1} Q_{1}}{T_{1}} \approx \frac{P_{2} Q_{2}}{T_{2}}$

subject to the actual system and reference definitions.

Therefore:

$Q_{F A D}$

cannot be assumed numerically equal to:

$Q_{a n n u l u s, a c t u a l}$

at high pressure.

55 Standard vs Actual Flow Conversion Concept

For the same mass flow:

$\dot{m} = ρ Q$

Therefore:

$ρ_{1} Q_{1} = ρ_{2} Q_{2}$

and, using the ideal-gas approximation:

$Q_{2} = Q_{1} \frac{P_{1}}{P_{2}} \frac{T_{2}}{T_{1}}$

provided pressures are absolute and the selected reference basis is consistent.

Real drilling systems require appropriate allowance for:

pressure variation
temperature
leakage
hammer exhaust behaviour

56 Do Not Double Count Air Expansion

A common error is to:

use compressor free-air flow directly as annular actual flow, and then
separately add an air-expansion factor
This can produce an internally inconsistent calculation
Choose one defined mass-flow/reference basis and convert systematically

57 Worked Example E — Altitude Correction Concept

Assume a compressor satisfies the requirement at its reference rating.

The drilling site is at significant altitude.

Do not simply use:

$Q_{r a t e d}$

as:

$Q_{a v a i l a b l e, s i t e}$

Instead obtain manufacturer correction data and calculate:

$Q_{s i t e} = Q_{r a t e d} \times K_{a l t i t u d e}$

where:

$K_{a l t i t u d e}$

is the verified equipment-specific correction factor.

The same principle applies to engine capability.

58 Example E — Selection Gate After Derating

Suppose:

$Q_{r e q u i r e d} = 1000 C F M$

and a machine is nominally rated:

$Q_{r a t e d} = 1050 C F M$

If site correction produces:

$Q_{s i t e} = 960 C F M$

then:

$Q_{s i t e} < Q_{r e q u i r e d}$

The machine fails the airflow gate despite apparently exceeding the requirement on the catalogue rating.

59 Temperature Correction Concept

Similarly:

$Q_{s i t e} = f (Q_{r a t e d}, A l t i t u d e, T e m p e r a t u r e, E q u i p m e n t D e s i g n)$

Do not invent a universal correction factor.

Use verified manufacturer information for the candidate compressor.

60 Combined Site Derating

Where manufacturer methodology permits:

$Q_{s i t e} = Q_{r a t e d} K_{A} K_{T} K_{o t h e r}$

Where:

$K_{A}$ = altitude factor,

$K_{T}$ = temperature factor,

$K_{o t h e r}$ = other specified correction.

Only use multiplicative factors if the manufacturer specifies that method.

61 Worked Example F — Pressure-Loss Reduction

Suppose:

$Δ P_{h} = 20 b a r$

water head:

$P_{w} = 10 b a r$

and original system losses are:

$\sum Δ P = 8 b a r$

with:

$P_{m} = 2 b a r$

Then:

$P_{r e q u i r e d} = 20 + 10 + 8 + 2$

$P_{r e q u i r e d} = 40 b a r$

Now suppose engineering improvements reduce avoidable losses by:

3bar

New losses:

5bar

Then:

$P_{r e q u i r e d, n e w} = 20 + 10 + 5 + 2$

$P_{r e q u i r e d, n e w} = 37 b a r$

A distribution-system improvement has reduced source pressure requirement by 3 bar without changing the hammer or borehole.

62 Example F — Why This Matters

That 3-bar reduction may:

reduce booster duty
reduce compression energy
improve operating reserve
lower thermal loading
It may even change which compressor/booster combination is technically viable

Therefore:

$Reduce Avoidable Losses Before Buying More Pressure$

63 Worked Example G — Airflow Deficit

Suppose:

$Q_{r e q u i r e d} = 1200 C F M$

but the selected primary compressor can provide only:

$Q_{a v a i l a b l e} = 1000 C F M$

at the required operating pressure.

Then:

$Q_{d e f i c i t} = 1200 - 1000$

$Q_{d e f i c i t} = 200 C F M$

A booster does not solve this fundamental airflow deficit.

The primary compressor architecture must be reconsidered.

64 Worked Example H — Pressure Deficit

Suppose:

$Q_{r e q u i r e d} = 1000 C F M$

and the primary compressor provides:

$Q_{a v a i l a b l e} = 1100 C F M$

but:

$P_{r e q u i r e d} = 35 b a r$

while primary practical pressure is:

25bar

Then airflow is adequate:

$1100 > 1000$

but pressure is not:

$25 < 35$

This is the type of condition in which booster assistance may be evaluated.

65 Worked Example I — Both Flow and Pressure Deficient

Suppose:

$Q_{r e q u i r e d} = 1200 C F M$

$P_{r e q u i r e d} = 35 b a r$

but the candidate primary system provides:

$Q_{a v a i l a b l e} = 950 C F M$

and:

$P_{a v a i l a b l e} = 25 b a r$

Then:

$Q_{a v a i l a b l e} < Q_{r e q u i r e d}$

and:

$P_{a v a i l a b l e} < P_{r e q u i r e d}$

The system fails both selection gates.

A booster attached to the same undersized upstream system does not constitute a complete engineering solution.

66 Four Compressor Selection Conditions

Every candidate falls into one of four broad conditions:

Condition	Airflow	Pressure	Engineering Direction
A	Adequate	Adequate	Primary compressor potentially suitable
B	Adequate	Inadequate	Evaluate higher-pressure machine/booster
C	Inadequate	Adequate	Increase airflow capability
D	Inadequate	Inadequate	Redesign compressor architecture

This simple matrix prevents pressure and airflow from being confused.

67 Hammer Consumption Curve

Hammer airflow should ideally be evaluated at several pressures:

$Q_{h} (P_{1})$

$Q_{h} (P_{2})$

$Q_{h} (P_{3})$

because air consumption can change as hammer operating pressure changes.

Therefore a single CFM value without associated pressure may be incomplete.

68 Compressor Performance Curve

Similarly, candidate compressor data should establish:

$Q_{c} (P)$

The selection condition is where:

$Q_{c} (P_{r e q u i r e d}) \geq Q_{r e q u i r e d}$

subject to site corrections.

This is much stronger than comparing nameplate maximums.

69 Graphical Selection Method

Plot:

Hammer/System Requirement

$Q_{r e q u i r e d} (P)$

and:

Compressor Capability

$Q_{c} (P)$

The candidate compressor must maintain adequate capability around the required operating region.

This provides a visual engineering method for comparing alternatives.

70 Operating Point

The actual operating point occurs where the compressor/system controls and downstream demand establish a stable combination of:

(Q,P)

The selected compressor should operate within its approved and efficient range at this point.

71 Avoid Selecting Exactly at the Limit

A compressor whose corrected capability is exactly:

$Q_{r e q u i r e d}$

and:

$P_{r e q u i r e d}$

has essentially no reserve for:

uncertainty
wear
leakage
deeper drilling
water changes
Therefore practical selection should include justified reserve

72 Avoid Excessive Reserve

Conversely, excessive reserve can increase:

purchase/rental cost
fuel consumption
logistics

The goal is:

$Requirement + Justified Reserve$

not arbitrary oversizing.

73 Selection Reserve vs Safety Factor

These terms should not be confused.

Operating Reserve

Capacity available for normal variation.

Safety Factor

A design concept applied where relevant to mechanical/engineering limits.

Do not treat additional compressor pressure as a substitute for properly rated pressure-system components.

74 Pressure Equipment Safety

If the calculated system requires high pressure, component ratings must be independently verified.

Never use:

$P_{m a r g i n}$

to justify operating an underrated hose or coupling.

Safe working pressure remains an equipment constraint.

75 Worked Example J — Compressor Comparison

Assume the calculated drilling requirement is:

$Q_{r e q u i r e d} = 1050 C F M$

$P_{r e q u i r e d} = 28 b a r$

after site correction.

Three candidate systems provide:

Compressor A
1000CFM@28bar
Compressor B
1150CFM@28bar
Compressor C
1400CFM@28bar

Then:

A fails the airflow requirement
B satisfies the calculated requirement with moderate airflow reserve

C also satisfies the technical requirement but provides substantially greater reserve.

Airflow reserve for B:

$R_{Q, B} = \frac{1150 - 1050}{1050} \times 100$

$R_{Q, B} \approx 9.5 %$

For C:

$R_{Q, C} = \frac{1400 - 1050}{1050} \times 100$

$R_{Q, C} \approx 33.3 %$

This calculation does not automatically establish B or C as the preferred compressor.

The technically acceptable candidates must next be compared for:

actual pressure-flow capability
expected future depth
water conditions
fuel consumption
availability
rental/ownership cost
operating reserve
cost per accepted metre

76 Technical Gate Before Economic Comparison

The selection process must occur in the correct order:

TECHNICAL QUALIFICATION → ECONOMIC COMPARISON.

Do not economically compare a compressor that fails:

$Q_{a v a i l a b l e} \geq Q_{r e q u i r e d}$

or:

$P_{a v a i l a b l e} \geq P_{r e q u i r e d}$

as though it were an equivalent alternative.

A cheaper machine that cannot perform the required drilling duty is not a valid economic option.

77 Worked Example K — Fuel/Productivity Comparison

Suppose Compressors B and C both satisfy the technical requirement.

Compressor B

Fuel:

38L/h

Effective ROP:

18m/h

Then:

$F_{m, B} = \frac{38}{18}$

$F_{m, B} \approx 2.11 L / m$

Compressor C

Fuel:

46L/h

Effective ROP:

21m/h

Then:

$F_{m, C} = \frac{46}{21}$

$F_{m, C} \approx 2.19 L / m$

In this hypothetical case, C drills faster but B consumes slightly less fuel per metre.

Neither result alone determines the final selection.

78 Example K — Cost per Metre

Suppose total attributable hourly operating cost is:

$C_{B, h} = ₹ 6, 500 / h$

$C_{C, h} = ₹ 7, 200 / h$

Then:

$C_{B, m} = \frac{6500}{18}$

$C_{B, m} \approx ₹ 361.11 / m$

and:

$C_{C, m} = \frac{7200}{21}$

$C_{C, m} \approx ₹ 342.86 / m$

Therefore C has higher:

L/m

but lower:

₹/m
under these assumptions because of higher productivity

79 Productivity Requirement Can Change Selection

Suppose the project must complete:

420m/day
within the available productive operating window

If B cannot reliably achieve the required daily output but C can, then the larger machine’s productivity capability becomes commercially important.

Compressor sizing should therefore consider:

$Technical Air Requirement + Required Project Production$

80 Required Daily Production

If project requirement is:

$M_{d}$

and available productive drilling time is:

$T_{p}$

then required effective ROP is:

$R O P_{r e q u i r e d} = \frac{M_{d}}{T_{p}}$

Example:

$M_{d} = 160 m$

$T_{p} = 8 h$

Then:

$R O P_{r e q u i r e d} = \frac{160}{8}$

$R O P_{r e q u i r e d} = 20 m / h$

A technically adequate compressor-hammer system producing only 15 m/h may therefore fail the project productivity requirement.

81 Technical Capacity vs Project Capacity

These must be distinguished.

Technical Capacity

Can the compressor operate the hammer and clean the hole?

Project Capacity

Can the complete drilling system produce the required accepted metres within the available schedule?

A machine can pass the first test and fail the second.

82 Worked Example L — Productivity Gate

Suppose:

System B

$R O P = 18 m / h$

System C

$R O P = 22 m / h$

Available productive time:

$T_{p} = 8 h / d a y$

Then:

$M_{B} = 18 \times 8 = 144 m / d a y$

$M_{C} = 22 \times 8 = 176 m / d a y$

If contractual/project requirement is:

160m/day

then:

$144 < 160$

while:

$176 > 160$

System C satisfies the simplified productivity gate; B does not under these assumptions.

83 Availability Must Be Included

Nominal daily production assumes equipment remains available.

If:

$A = 90 %$

then simplified effective available drilling time becomes:

$T_{a v a i l a b l e} = T_{s} A$

If utilization is also:

then:

$T_{p r o d u c t i v e} = T_{s} A U$

Therefore:

$M_{d a y} = T_{s} \times A \times U \times R O P$

before any accepted-yield adjustment.

84 Worked Example M — Availability Effect

Assume:

$T_{s} = 10 h$

$A = 0.90$

$U = 0.80$

$R O P = 20 m / h$

Then:

$T_{p} = 10 \times 0.90 \times 0.80$

$T_{p} = 7.2 h$

Therefore:

$M_{d a y} = 7.2 \times 20$

$M_{d a y} = 144 m$

A 20 m/h hammer-on-bottom penetration rate therefore does not automatically mean 200 m/day from a 10-hour shift.

85 Accepted Yield

If only:

$Y = 95 %$

of gross output becomes accepted output:

$M_{a c c e p t e d} = M_{g r o s s} Y$

For:

$M_{g r o s s} = 144 m$

$M_{a c c e p t e d} = 144 \times 0.95$

$M_{a c c e p t e d} = 136.8 m / d a y$

This is the more meaningful production number for project economics.

86 Complete Productivity Equation

Therefore:

$M_{a c c e p t e d} = T_{s} \times A \times U \times R O P \times Y$

This equation should accompany compressor sizing whenever project schedule matters.

87 Worked Example N — Compressor Too Small Economically

Suppose a smaller compressor technically operates the hammer but produces:

$R O P = 12 m / h$

while a properly matched larger system produces:

$R O P = 20 m / h$

If both systems are required for 1,000 accepted metres, simplified productive drilling hours are:

Small system:

$T_{1} = \frac{1000}{12} \approx 83.3 h$

Larger system:

$T_{2} = \frac{1000}{20} = 50 h$

Difference:

$Δ T \approx 33.3 h$

The smaller machine’s lower hourly cost may be overwhelmed by additional operating time.

88 Worked Example O — Compressor Excessively Large

Suppose required airflow is:

900CFM

at:

20bar

A candidate compressor can supply:

1600CFM
at the required pressure

If the hammer and hole-cleaning system cannot use the additional capacity productively, the larger machine may add:

fuel
rental cost
transport burden
without equivalent ROP improvement

Therefore:

$Q_{a v a i l a b l e} ≫ Q_{r e q u i r e d}$

requires economic justification.

89 Compressor Utilization Window

A suitable compressor should operate effectively across the expected demand range:

$Q_{m i n} \to Q_{m a x}$

and:

$P_{m i n} \to P_{m a x}$

The project should avoid selecting a machine that is:

inadequate at the upper requirement
grossly inefficient throughout most of the lower requirement

90 Variable pressure operation

Where the package permits it, the operating setting may follow the approved stage pressure target. On this page $P_{r e q u i r e d}$ already includes the explicitly chosen pressure allowance. Do not add that allowance again when translating the selection target into a setting. Apply the manufacturer’s control procedure, tolerances and maximum equipment limits; a worksheet result is not an instruction to change a regulator.

91 Variable Demand by Depth

The drilling requirement may evolve approximately as:

$D ↑$

leading to changes in:

$Δ P_{s t r i n g}$

$P_{w}$

$Q_{c l e a n i n g}$

and potentially:

$P_{r e q u i r e d}$

Therefore compressor selection should be based on the expected complete drilling envelope, not only surface conditions.

92 Depth-Stage Selection

A project can model:

Stage 1

$0 - D_{1}$

Stage 2

$D_{1} - D_{2}$

Stage 3

$D_{2} - D_{f i n a l}$

For each stage calculate:

$Q_{r e q u i r e d, i}$

and:

$P_{r e q u i r e d, i}$

This reveals whether one compressor can efficiently cover the complete hole.

93 Staged Compressor Architecture

Where project scale justifies it, possible strategies include:

Strategy A
One compressor for complete depth
Strategy B
Lower-pressure compressor initially, then change to high-pressure equipment
Strategy C
Primary compressor throughout, booster introduced at deeper stage

The technical and economic optimum depends on:

changeover cost
project metres
depth profile
equipment availability

94 Booster Introduction Depth

Define:

$D_{B}$

as the depth at which:

$P_{r e q u i r e d} (D_{B})$

approaches or exceeds practical primary-compressor capability.

The booster should not be introduced solely because a predetermined depth number has been reached.

Actual system pressure requirement controls the decision.

95 Worked Example P — Booster Start Depth

Suppose calculated pressure requirement is:

Depth	Required Pressure
100 m	22 bar
200 m	24 bar
300 m	27 bar
350 m	31 bar
400 m	36 bar

Assume primary compressor practical operating capability for the required airflow is:

30bar

Then the pressure requirement first exceeds this capability between approximately:

300m

and:

350m

The actual booster introduction decision should consider measured field conditions and suitable operating reserve.

96 Interpolation Concept

If a simplified linear approximation between 300 and 350 m is considered for planning:

At 300 m:

$P = 27 b a r$

At 350 m:

$P = 31 b a r$

Increase:

4bar

over:

50m

To reach 30 bar requires:

3bar
above the 300 m condition

Approximate depth increment:

$50 \times \frac{3}{4} = 37.5 m$

Therefore:

$D \approx 337.5 m$

This is only an illustrative interpolation.

Real pressure requirement may not vary linearly with depth.

97 Field Verification Overrides Planning Approximation

The calculated:

$D_{B}$

is a planning value.

During drilling, verify:

actual source pressure
drill-head pressure
hammer performance
water conditions
ROP

A sudden water-bearing zone can change booster requirement earlier than a dry-hole calculation predicts.

98 Worked Example Q — Water Entry Changes Selection

Suppose at 280 m:

$P_{r e q u i r e d} = 27 b a r$

and the primary compressor remains adequate.

At 300 m, a significant water zone introduces an additional:

8bar
of effective hydrostatic backpressure

New requirement becomes approximately:

27+8
35bar

The compressor architecture may therefore change abruptly because of water—not simply because another 20 m was drilled.

99 Dynamic Recalculation

DTH compressor sizing should therefore be treated as:

$Initial Design + Field Verification + Recalculation When Conditions Change$

Important triggers include:

significant water entry
bore diameter change
hammer change
bit change
drill-string change
unexpected pressure loss

100 Change-Control Rule

When a major drilling variable changes, do not assume the original compressor calculation remains valid.

Recalculate at least the affected portions.

Examples:

Larger Bit
Recalculate annular area and cleaning requirement
Different Hammer
Recalculate airflow and pressure demand
Deeper Target
Recalculate drill-string losses and water exposure
Different Drill Pipe
Recalculate internal pressure loss and annular geometry

101 Worked Example R — Larger Bit Installed

Original:

$D_{h} = 152 m m$

New:

$D_{h} = 203 m m$

with the same:

$D_{p} = 102 m m$

Original annular area:

$A_{a 1} = \frac{π}{4} ({0.152}^{2} - {0.102}^{2})$

${0.152}^{2} = 0.023104$

${0.102}^{2} = 0.010404$

$A_{a 1} \approx 0.00997 m^{2}$

New:

$A_{a 2} = \frac{π}{4} ({0.203}^{2} - {0.102}^{2})$

$A_{a 2} \approx 0.02419 m^{2}$

The annular area increases by more than two times.

Therefore the original cleaning-air calculation is no longer valid.

102 Example R — Area Ratio

$R_{A} = \frac{A_{a 2}}{A_{a 1}}$

$R_{A} = \frac{0.02419}{0.00997}$

$R_{A} \approx 2.43$

At the same actual annular volumetric flow:

$V_{a 2} = \frac{V_{a 1}}{2.43}$

approximately.

This demonstrates how dramatically larger bore diameter can alter hole-cleaning performance.

103 Worked Example S — Larger Drill Pipe

Suppose hole diameter remains:

203mm

but drill-pipe OD increases from:

89mm

to:

114mm
The larger pipe reduces annular area
This can increase annular velocity for the same actual return flow

However, drill-pipe selection must also consider:

internal diameter
pressure loss
mechanical strength
rig compatibility

Changing pipe size can improve one part of the pneumatic system while worsening another.

104 Annular Area vs Pipe Internal Area

Two different areas matter:

Internal Flow Area

$A_{i} = \frac{π D_{i}^{2}}{4}$

controls compressed-air velocity inside the drill pipe.

Annular Return Area

$A_{a} = \frac{π}{4} (D_{h}^{2} - D_{p}^{2})$

controls return-flow velocity around the pipe.

Therefore drill-pipe sizing creates a two-sided optimization problem.

105 Larger Pipe — Potential Benefit

A larger internal diameter can reduce:

$Δ P_{s t r i n g}$

for a given flow.

A larger external diameter can also reduce annular area and potentially increase:

$V_{a}$

for the same actual return flow.

However, the larger pipe also affects:

weight
rig load
handling
hole clearance
Selection must therefore remain system-based

106 Drill-Pipe Sizing Principle

Do not select drill pipe solely from mechanical strength or outside diameter. Its internal diameter affects pressure delivery, while its outside diameter affects annular return geometry.

Thus:

$Drill Pipe = Structural Component + Pressure Conduit + Annular Geometry Component$

107 Worked Example T — Internal Pipe Velocity

Suppose actual volumetric flow inside a drill pipe at the evaluated condition is:

$Q = 0.20 m^{3} / s$

and internal diameter:

$D_{i} = 0.075 m$

Internal area:

$A_{i} = \frac{π}{4} {(0.075)}^{2}$

$A_{i} \approx 0.00442 m^{2}$

Then:

$V_{i} = \frac{0.20}{0.00442}$

$V_{i} \approx 45.2 m / s$

This illustrates why high-flow drill strings can experience significant frictional pressure loss.

108 Example T — Larger Internal Diameter

If internal diameter increases to:

$D_{i} = 0.090 m$

then:

$A_{i} = \frac{π}{4} {(0.090)}^{2}$

$A_{i} \approx 0.00636 m^{2}$

For the same actual volumetric flow:

$Q = 0.20 m^{3} / s$

the internal velocity becomes:

$V_{i} = \frac{0.20}{0.00636}$

$V_{i} \approx 31.4 m / s$

Compared with the previous 75 mm ID pipe:

$45.2 m / s \to 31.4 m / s$

This illustrates why increasing internal flow area can materially reduce air velocity and potentially reduce frictional pressure loss.

The actual pressure-loss calculation must still account for compressible flow and the complete drill-string geometry.

109 Internal Area Ratio

For the two pipe sizes:

$A_{75} \approx 0.00442 m^{2}$

$A_{90} \approx 0.00636 m^{2}$

Therefore:

$\frac{A_{90}}{A_{75}} = \frac{0.00636}{0.00442}$

$\approx 1.44$

The 90 mm internal diameter provides approximately 44% more internal flow area.

This demonstrates why relatively modest diameter changes can materially affect pneumatic performance.

110 Diameter Has a Strong Pressure-Loss Influence

For internal flow systems, pressure loss is highly sensitive to pipe diameter.

The exact relationship for high-pressure compressed air depends on:

flow regime
density variation
pressure
temperature
pipe roughness
fittings
compressibility

Therefore:

Do not estimate deep DTH drill-string pressure loss solely from pipe length. Internal diameter can be equally or more important.

111 Drill-String Length Effect

For otherwise comparable conditions:

$L ↑ \Rightarrow Δ P_{s t r i n g} ↑$

Therefore the pressure requirement should be recalculated as drilling progresses.

A compressor selected for:

100m

may not necessarily remain adequate at:

400m
even when hammer and bit remain unchanged

112 Joint Losses

The drill string is not a perfectly smooth continuous pipe.

It contains:

threaded connections
internal transitions
tool joints
possible diameter changes

Therefore:

$Δ P_{s t r i n g} = Δ P_{p i p e} + \sum Δ P_{j o i n t}$

where applicable.

For high-flow deep drilling, accumulated local losses can become significant.

113 Leakage Across Drill-Pipe Joints

With increasing depth:

$N_{j o i n t s} ↑$

Therefore potential leakage locations also increase.

Conceptually:

$Q_{h a m m e r} = Q_{c o m p r e s s o r} - Q_{s u r f a c e l e a k a g e} - \sum Q_{j o i n t l e a k a g e}$

This is why drill-string integrity must be considered in deep-hole compressor sizing.

114 Worked Example U — Leakage Allowance

Suppose calculated useful airflow requirement is:

$Q_{u s e f u l} = 1000 C F M$

and a controlled design allowance of:

$5 %$

is justified for the specific system.

Then:

$Q_{d e s i g n} = 1000 (1.05)$

$Q_{d e s i g n} = 1050 C F M$

However, this does not mean known leaks should be accepted.

Known significant leakage should be repaired.

115 Reserve vs Leakage Allowance

These concepts should remain separate.

Leakage Allowance

Accounts for realistic minor system losses where justified.

Operating Reserve

Provides capacity for:

changing conditions
depth
wear
uncertainty
Do not hide both inside one arbitrary percentage

116 Worked Example V — Combined Airflow Allowances

Suppose:

$Q_{u s e f u l} = 1000 C F M$

justified minor-loss allowance:

$Q_{L} = 50 C F M$

and operating reserve:

$Q_{R} = 100 C F M$

Then:

$Q_{s e l e c t i o n} = 1000 + 50 + 100$

$Q_{s e l e c t i o n} = 1150 C F M$

The calculation should show each component separately.

This makes the sizing decision auditable.

117 Airflow Selection Equation

A transparent relationship is:

$Q_{s e l e c t i o n} = Q_{u s e f u l} + Q_{l e g i t i m a t e l o s s e s} + Q_{r e s e r v e}$

where:

$Q_{u s e f u l}$

must already satisfy the governing hammer and hole-cleaning requirement without double-counting the same air.

118 Pressure selection equation

Use the boundary-defined budget in Section 21. In the static freshwater examples only:

$P_{s e l e c t i o n, g} = Δ P_{h} + P_{w} + Δ P_{s} + Δ P_{d} + Δ P_{o} + P_{R}$ .

$Δ P_{h}$ is hammer differential pressure, $P_{w}$ is the assumed exhaust backpressure represented by static head, and $P_{R}$ is the allowance counted once. For a supplier-defined hammer inlet requirement that already includes exhaust conditions, replace the first two terms with that inlet requirement. Confirm the actual manufacturer’s pressure and consumption basis before selecting a machine.

119 Required Pressure Is Dynamic

For depth $D$ :

$P_{s e l e c t i o n} = P_{s e l e c t i o n} (D)$

because:

$Δ P_{d} = f (D)$

and:

$P_{w} = f (water conditions)$

Therefore the final compressor selection should normally consider the maximum credible operating requirement, not merely the initial drilling condition.

120 Required Airflow Can Also Change

Although hammer consumption may dominate, required effective airflow can change with:

bit diameter
hammer pressure
annular geometry
leakage
drilling method

Therefore:

$Q_{s e l e c t i o n} = Q_{s e l e c t i o n} (D, geometry, P, tooling)$

conceptually.

121 Maximum Requirement Envelope

Define:

$Q_{m a x, r e q} = \max (Q_{r e q u i r e d, 1}, Q_{r e q u i r e d, 2}, . . ., Q_{r e q u i r e d, n})$

and:

$P_{m a x, r e q} = \max (P_{r e q u i r e d, 1}, P_{r e q u i r e d, 2}, . . ., P_{r e q u i r e d, n})$

But do not automatically combine these two independent maxima into one fictitious operating point.

The maximum airflow and maximum pressure may occur at different drilling stages.

122 Simultaneous Requirement Is What Matters

The actual compressor duty at stage $i$ is:

$(Q_{i}, P_{i})$

Therefore equipment must satisfy every required operating point:

$(Q_{1}, P_{1})$

$(Q_{2}, P_{2})$

$\dots$

$(Q_{n}, P_{n})$

The compressor selection envelope should contain all relevant points.

123 Worked Example W — Multi-Stage Requirement

Suppose:

Stage	Airflow	Pressure
Shallow	1100 CFM	18 bar
Intermediate	1100 CFM	24 bar
Deep	1050 CFM	30 bar
Deep + Water	1000 CFM	38 bar

The selected architecture must accommodate the relevant combinations.

A compressor capable of:

1200CFM@20bar

does not automatically satisfy:

1000CFM@38bar

124 Multi-Stage Architecture Decision

The table above may lead to evaluation of:

One High-Pressure Compressor

or:

Primary Compressor + Booster at Depth

or:

Equipment Change by Stage

The preferred architecture depends on:

pressure-flow capability
project metres
changeover cost
fuel
availability
logistics

125 Worked Example X — One Compressor vs Booster

Assume the project requires:

First 300 m
1100CFM@24bar
Final 150 m
1000CFM@38bar

Two possible architectures:

Option A: High-pressure compressor for all 450 m
Option B: Primary compressor for complete hole + booster for final 150 m
Both may be technically feasible
reliability

126 Do Not Select Architecture From Pressure Alone

Even if Option B reaches 38 bar, verify that the booster can process the required:

$\dot{m}$

at the actual inlet condition.

Similarly, Option A must provide adequate:

Q
at 38 bar
Pressure capability alone does not establish suitability

127 Compressor Capacity at Working Pressure

For every candidate, obtain:

$Q (P_{w o r k i n g})$

not simply:

$Q_{n a m e p l a t e, m a x}$

If verified data is unavailable, the selection remains provisional.

Do not invent intermediate pressure-flow performance.

128 Manufacturer Data Gate

Before final equipment approval, obtain where applicable:

compressor performance curve/data
site derating information
hammer airflow curve
booster inlet/outlet envelope
maximum temperature limits
component pressure ratings
Manufacturer data confirms whether a specific machine can supply it

129 Field Verification Gate

After mobilization, verify actual:

$P_{c o m p r e s s o r}$

$P_{d r i l l h e a d}$

ROP

$Cuttings Return$

and, where available:

Q
Calculated suitability must ultimately be confirmed by safe field performance

130 Compare losses across matching measurement boundaries

Read source and drill-head pressures simultaneously during the same stable duty. Their difference estimates only the losses between those two taps:

$Δ P_{s u r f a c e, m e a s u r e d} = P_{s o u r c e, g} - P_{d r i l l h e a d, g}$ .

Compare that value with the calculated surface route between the same taps. It does not measure the pressure loss down the drill string or establish pressure at the down-hole hammer. Drill-string loss requires a separately justified calculation or appropriate down-hole measurement. Record tap locations, instrument accuracy, flow estimate and operating state with every comparison.

131 Investigate higher surface loss

If the measured source-to-drill-head loss exceeds the calculated loss for that route, inspect the hose, surface valves, manifolds, fittings, treatment devices where fitted and measurement assumptions. Identify leaking connections and confirm simultaneous readings at the same duty. Assess drill-string restrictions separately because they lie beyond the drill-head pressure tap. Resolve the cause before considering any pressure-setting change.

132 Lower-Than-Expected Pressure Loss

If measured loss is lower than predicted, the system may have more pressure reserve than expected.

However, do not immediately reduce design requirements until:

measurement accuracy
drilling condition
depth
airflow
are confirmed

133 Pressure Instrument Accuracy

A 2-bar apparent pressure difference may be meaningless if gauges are inaccurate.

Therefore important sizing validation should use suitable instruments with known condition/calibration appropriate to the task.

Measurement quality controls engineering quality.

134 Flow Measurement Challenge

Airflow is often more difficult to measure in field drilling than pressure.

Where direct flow measurement is unavailable, use:

verified compressor performance information
machine operating data
hammer manufacturer consumption data
while clearly identifying assumptions
Do not convert uncertain flow estimates into false precision

135 Compressor Sizing Confidence Levels

A useful classification is:

Preliminary
Based on estimated geology/water and nominal equipment data
Engineered
Based on defined hole geometry, hammer data and calculated losses
Verified
Supported by actual site measurements and equipment performance
This prevents preliminary calculations from being presented as final certainty

136 Worked Example Y — Preliminary Selection

Suppose only these values are known:

target depth = 300 m
hole diameter = 152 mm
approximate hammer class

But the following are unknown:

hammer airflow curve
drill-pipe ID
groundwater head
altitude
A final compressor size cannot yet be responsibly calculated

The correct output is a preliminary compressor class, with the missing inputs explicitly listed.

137 Missing-Input Rule

Do not manufacture numbers merely to complete a compressor-sizing worksheet.

If a required input is unknown:

$MARK IT UNKNOWN$

Then determine:

whether it can be measured
obtained from manufacturer data
reasonably bounded
This is superior to false numerical precision

138 Bounding Unknown Inputs

Sometimes an unknown can be represented as a range.

Example:

$P_{w} = 5 to 12 b a r$

Then calculate:

$P_{r e q u i r e d, m i n}$

and:

$P_{r e q u i r e d, m a x}$

This produces a selection envelope rather than a single unsupported value.

139 Worked Example Z — Water-Head Range

Assume:

$Δ P_{h} = 20 b a r$

system losses:

$\sum Δ P = 6 b a r$

margin:

2bar

and expected water head:

$50 m \leq h \leq 120 m$

Minimum hydrostatic pressure:

$P_{w, m i n} = 0.0981 \times 50$

$P_{w, m i n} \approx 4.91 b a r$

Maximum:

$P_{w, m a x} = 0.0981 \times 120$

$P_{w, m a x} \approx 11.77 b a r$

140 Example Z — Pressure Envelope

Minimum required source pressure:

$P_{m i n} = 20 + 6 + 2 + 4.91$

$P_{m i n} \approx 32.91 b a r$

Maximum:

$P_{m a x} = 20 + 6 + 2 + 11.77$

$P_{m a x} \approx 39.77 b a r$

The selection problem is therefore better represented as:

$P_{r e q u i r e d} \approx 33 - 40 b a r$

until groundwater conditions are better defined.

141 Sensitivity Analysis

For uncertain variables, determine which one most strongly affects compressor selection.

Potential sensitivity variables include:

water head
hammer pressure
drill-string pressure loss
airflow
altitude

Focus data-gathering effort on the variables capable of changing the equipment decision.

142 Water-Head Sensitivity

For freshwater:

$\frac{d P_{w}}{d h} \approx 0.0981 b a r / m$

Therefore every additional:

10m

of effective water head adds approximately:

0.981bar
of hydrostatic pressure
This makes water head a highly transparent pressure sensitivity

143 Pressure-Loss Sensitivity

Every avoidable:

1bar

of additional system pressure loss adds approximately:

1bar
to required source pressure

This makes pressure-drop control particularly valuable when compressor or booster capability is near its limit.

144 Airflow Sensitivity to Hole Diameter

Annular area is:

$A_{a} = \frac{π}{4} (D_{h}^{2} - D_{p}^{2})$

Therefore hole diameter enters as a squared term.

A modest increase in bit diameter can produce a much larger increase in annular area.

This can materially change the cleaning-air requirement.

145 Worked Diameter Sensitivity

Assume pipe OD:

$D_{p} = 0.102 m$

152 mm Hole

$A_{152} = \frac{π}{4} ({0.152}^{2} - {0.102}^{2})$

$A_{152} \approx 0.00997 m^{2}$

178 mm Hole

$A_{178} = \frac{π}{4} ({0.178}^{2} - {0.102}^{2})$

$A_{178} \approx 0.016713 m^{2}$

203 mm Hole

$A_{203} \approx 0.02419 m^{2}$

The increase is substantial.

146 Relative Annular Area

Relative to the 152 mm hole:

$\frac{A_{178}}{A_{152}} \approx 1.6756$

and:

$\frac{A_{203}}{A_{152}} \approx 2.43$

Thus, for the same actual annular flow:

$V_{a}$

would fall inversely with these area increases.

This is why “just use a larger bit” can fundamentally alter compressor requirements.

147 Hammer Change Sensitivity

Changing the hammer can change:

$Q_{h} (P)$

and:

$Δ P_{h}$

Therefore a compressor calculation tied to one hammer should not automatically be reused for another hammer of similar nominal diameter.

Use the actual hammer manufacturer’s performance data.

148 Bit Change Sensitivity

Changing only bit diameter can alter:

annular area
cuttings volume
cleaning demand

Therefore the compressor may need reassessment even when the hammer remains unchanged.

149 Drill-Pipe Change Sensitivity

Changing pipe geometry alters:

Internal Side

$D_{i} \to Δ P_{s t r i n g}$

External Side

$D_{p} \to A_{a} \to V_{a}$

Therefore drill-pipe change requires both supply-path and return-path recalculation.

150 Depth Change Sensitivity

Increasing target depth changes:

drill-string length
potential pressure loss
joint count
water exposure
handling time

Therefore a compressor selected for 200 m should not automatically be approved for 500 m merely because the hammer remains the same.

151 Compressor Selection Failure Modes

Incorrect sizing can fail in four principal ways:

Flow-Undersized

$Q_{a v a i l a b l e} < Q_{r e q u i r e d}$

Pressure-Undersized

$P_{a v a i l a b l e} < P_{r e q u i r e d}$

Both Undersized

Both conditions fail.

Excessively Oversized

Technically capable but potentially inefficient or uneconomic.

The objective is balanced sizing.

152 Symptoms of Flow Undersizing

Possible symptoms include:

compressor unable to maintain desired pressure under hammer demand
weak hole cleaning
cuttings accumulation
poor water lifting
low ROP

These symptoms require diagnosis because leakage or restriction can produce similar effects.

153 Symptoms of Pressure Undersizing

Possible symptoms include:

hammer performs satisfactorily at shallow depth but weakens as depth increases
compressor operates continuously near its pressure limit
penetration deteriorates after significant water entry
drill-head pressure falls below the hammer’s required operating range
hammer becomes intermittent under increasing backpressure
adequate cuttings airflow appears available but impact performance remains poor
These symptoms do not independently prove compressor pressure undersizing

First verify:

$P_{c o m p r e s s o r}$

$P_{d r i l l h e a d}$

$Δ P_{s y s t e m}$

and:

$P_{h y d r o s t a t i c}$

A restriction or excessive pressure loss can imitate an undersized compressor.

154 Symptoms of Combined Flow and Pressure Undersizing

When both:

$Q_{a v a i l a b l e} < Q_{r e q u i r e d}$

and:

$P_{a v a i l a b l e} < P_{r e q u i r e d}$

the drilling system may exhibit:

weak hammer action
inability to maintain working pressure
poor cuttings return
low penetration
difficulty clearing water
compressor operating continuously at maximum load

In this condition, simply adding a booster to an inadequate primary compressor is unlikely to provide a properly balanced solution.

155 Symptoms of Excessive Compressor Capacity

Oversizing may be indicated when:

required hammer airflow is substantially below installed capacity
the machine spends significant time in inefficient low-demand operation
additional CFM provides no measurable ROP improvement
rental/ownership cost is disproportionately high
fuel/metre is poorer than a correctly sized alternative

However, apparent excess capacity may be intentional reserve for:

deeper drilling
future water entry
larger bits
changing hammer requirements
Therefore oversizing should be judged against the complete project envelope

156 Compressor Selection Is a Constraint Problem

A candidate compressor must satisfy multiple constraints simultaneously:

$Q_{c} \geq Q_{r e q u i r e d}$

$P_{c} \geq P_{r e q u i r e d}$

$T_{o p e r a t i n g} \leq T_{a l l o w a b l e}$

$P_{c o m p o n e n t s} \geq P_{s y s t e m, m a x}$

and must remain compatible with:

hammer
booster where applicable
rig
hoses
drill string
site conditions
Selection therefore cannot be reduced to one specification

157 Minimum Technically Acceptable Compressor

Define the minimum technically acceptable configuration as one that satisfies every required operating point after site correction with justified reserve.

Conceptually:

$C_{m i n} = \min (C) subject to all technical constraints$

This does not necessarily mean the smallest physical compressor.

158 Economic Selection After Technical Qualification

Once several systems pass the technical gate, compare:

$C_{a c c e p t e d, m}$

$F_{a c c e p t e d, m}$

$M_{d a y}$

Availability

Mobilization

Reliability

The preferred project configuration is determined by complete drilling economics, not compressor size alone.

159 Compressor Selection Matrix

Parameter	Candidate A	Candidate B	Candidate C
Site-corrected airflow	—	—	—
Airflow requirement	—	—	—
Working pressure capability	—	—	—
Pressure requirement	—	—	—
Airflow reserve	—	—	—
Pressure reserve	—	—	—
Booster required	—	—	—
Fuel L/h	—	—	—
Expected L/m	—	—	—
Expected ROP	—	—	—
Availability	—	—	—
₹/accepted metre	—	—	—

This provides a transparent technical-commercial comparison.

160 PASS / FAIL Technical Gate

Each candidate should first receive individual technical dispositions:

AIRFLOW
PASS/FAIL
PRESSURE
PASS/FAIL
SITE DERATING
PASS/FAIL
HAMMER COMPATIBILITY
PASS/FAIL
HIGH-PRESSURE CIRCUIT
PASS/FAIL
BOOSTER COMPATIBILITY — IF APPLICABLE
PASS/FAIL

A candidate failing a mandatory technical gate should not proceed as an equivalent economic option.

161 Selection Margin Must Be Transparent

Instead of writing:

“Add 20% safety factor.”

show the actual basis:

$Q_{u s e f u l} = 1000 C F M$

$Q_{m i n o r l o s s e s} = 50 C F M$

$Q_{o p e r a t i n g r e s e r v e} = 100 C F M$

Therefore:

$Q_{s e l e c t i o n} = 1150 C F M$

Likewise for pressure.

Transparent engineering is superior to unexplained percentages.

162 Do Not Stack Margins Repeatedly

A common sizing error is adding:

hammer manufacturer’s allowance
consultant allowance
compressor reserve
contractor safety factor
without checking whether they cover the same uncertainty
Repeated margins can create severe oversizing

Therefore:

$Identify Each Margin Once$

and state what it protects against.

163 Worst-Case vs Credible-Case Design

Designing for every imaginable extreme can make the compressor unnecessarily large.

Instead distinguish:

Expected Condition
Normal project operation
Credible Maximum Condition
Realistically foreseeable demanding operation
Abnormal/Emergency Condition
Managed separately where appropriate

The compressor should be selected for the defined engineering envelope, not unlimited hypothetical extremes.

164 Scenario-Based Compressor Sizing

Create scenarios such as:

Scenario A — Dry Hole

$P_{w} = 0$

Scenario B — Expected Water

$P_{w} = P_{e x p e c t e d}$

Scenario C — Credible High Water

$P_{w} = P_{h i g h}$

Calculate:

(Q,P)
for each scenario
This makes the effect of uncertainty visible

165 Scenario Example

Assume:

$Δ P_{h} = 20 b a r$

$\sum Δ P = 6 b a r$

$P_{R} = 2 b a r$

Dry

$P_{r e q} = 28 b a r$

50 m Water Head

$P_{w} = 4.91 b a r$

$P_{r e q} \approx 32.91 b a r$

120 m Water Head

$P_{w} = 11.77 b a r$

$P_{r e q} \approx 39.77 b a r$

The equipment architecture can change materially across these scenarios.

166 Architecture Decision from Scenario Analysis

The previous example could support evaluation of:

Dry Project
High-pressure primary compressor alone
Moderate Water
Higher-pressure compressor or booster evaluation
High Water
Engineered high-pressure/booster package

The correct architecture depends on how probable and consequential each scenario is.

167 Probability Should Not Be Invented

If groundwater information is insufficient, do not claim:

“There is a 70% chance of requiring a booster.”
unless defensible project data supports that probability

Instead state:

known conditions
uncertainty
pressure range
contingency plan

168 Contingency Planning

When booster need is uncertain, a project may prepare:

compatible connection points
appropriately rated hose/manifold plan
booster availability
mobilization procedure

This can reduce emergency delay without necessarily operating the booster from the beginning.

169 Compressor Selection and Mobilization Risk

An undersized compressor can require emergency replacement.

An excessively large compressor can impose unnecessary cost from day one.

Therefore selection balances:

$Undersizing Risk$

against:

$Oversizing Cost$

using actual project uncertainty.

170 Selection Confidence Matrix

Input	Confidence	Impact on Selection
Hole diameter	High/Medium/Low	—
Final depth	High/Medium/Low	—
Hammer data	High/Medium/Low	—
Drill-pipe ID	High/Medium/Low	—
Water head	High/Medium/Low	—
Formation	High/Medium/Low	—
Altitude	High/Medium/Low	—
Compressor curve	High/Medium/Low	—

High-impact, low-confidence variables deserve further investigation before equipment commitment.

171 Data Quality Gate

If hammer airflow is unknown:

$Q_{r e q u i r e d}$

cannot be finalized.

If water head is unknown:

$P_{r e q u i r e d}$

may need to remain a range.

If compressor pressure-flow performance is unavailable:

$Q_{a v a i l a b l e} (P)$

cannot be verified.

172 Preliminary vs Final Selection

Preliminary Selection

Suitable for:

budgeting
equipment planning
early project discussions
Final Engineered Selection

Requires:

verified hammer data
defined drill-string geometry
pressure-loss analysis
groundwater assumptions
site corrections
candidate compressor performance
Field-Verified Selection
Adds actual drilling measurements
This distinction protects engineering integrity

173 Compressor Sizing Calculation Record

Every final selection should retain:

input data
source of hammer data
pressure-loss assumptions
groundwater assumption
site conditions
compressor performance data
correction factors
reserve basis
booster calculation
final disposition
This allows the calculation to be reviewed later

174 Calculation Revision Control

If the drilling plan changes, issue a revised calculation.

Example:

then:

Do not overwrite the original assumptions without traceability.

175 Unit Control

DTH compressor calculations frequently mix:

CFM
m³/min
psi
bar
mm
inch
ft
Unit errors can produce major sizing mistakes
Therefore every calculation should explicitly state units

176 Basic Flow Conversion

Approximately:

$1 m^{3} / m i n \approx 35.315 C F M$

Therefore:

$Q_{C F M} \approx 35.315 Q_{m^{3} / m i n}$

and:

$Q_{m^{3} / m i n} \approx \frac{Q_{C F M}}{35.315}$

provided both values refer to equivalent reference conditions.

177 Worked Flow Conversion

Suppose:

$Q = 30 m^{3} / m i n$

Then:

$Q_{C F M} = 30 \times 35.315$

$Q \approx 1059 C F M$

Again, the conversion changes units—not reference conditions.

178 Pressure Conversion

Approximately:

$1 b a r \approx 14.504 p s i$

Therefore:

$P_{p s i} \approx 14.504 P_{b a r}$

and:

$P_{b a r} \approx \frac{P_{p s i}}{14.504}$

179 Worked pressure conversion

For 35 bar, $P_{p s i} = 35 \times 14.504 \approx 507.6$ psi. Preserve gauge or absolute basis during the conversion. Pressure alone does not establish stored energy; contained volume, gas state and the system boundary also matter.

180 Diameter Conversion

$1 i n = 25.4 m m$

Therefore:

$D_{m m} = 25.4 D_{i n}$

and:

$D_{i n} = \frac{D_{m m}}{25.4}$

Use exact dimensional data wherever possible rather than nominal names.

181 Area Calculation Requires Consistent Units

If:

$D_{h}$

is entered in metres, then:

$A_{a}$

will be in:

$m^{2}$

Do not mix millimetres and metres inside:

$A_{a} = \frac{π}{4} (D_{h}^{2} - D_{p}^{2})$

without conversion.

182 Gauge vs Absolute Pressure

Most field pressure gauges display:

$P_{g}$

But gas-law and pressure-ratio calculations may require:

$P_{a b s}$

Approximately:

$P_{a b s} = P_{g} + P_{a t m}$

Use local atmospheric pressure where higher accuracy is required.

183 Worked Gauge-to-Absolute Example

At approximately sea-level atmospheric pressure:

$P_{g} = 20 b a r$

Then:

$P_{a b s} \approx 21 b a r$

This difference becomes particularly important in booster pressure-ratio calculations.

184 Temperature Must Be Absolute in Gas Equations

When using:

$\frac{P_{1} Q_{1}}{T_{1}} = \frac{P_{2} Q_{2}}{T_{2}}$

temperature must be expressed on an absolute scale such as kelvin.

Do not insert degrees Celsius directly into gas-law ratios.

185 Celsius to Kelvin

$T_{K} = T_{C} + 273.15$

Example:

$30^{\circ} C$

becomes:

303.15K

186 Absolute-Pressure and Temperature Discipline

Many incorrect compressor calculations arise from mixing:

gauge pressure with absolute pressure
Celsius with Kelvin
FAD with actual flow

Therefore every calculation sheet should state:

$Pressure Basis + Temperature Basis + Flow Basis$

before equations are applied.

187 Simplified Gas Conversion Example

Suppose a defined free-air reference flow is:

$Q_{1} = 1000 C F M$

at:

$P_{1} = 1 b a r (a b s)$

$T_{1} = 293 K$

At a hypothetical location where:

$P_{2} = 10 b a r (a b s)$

and:

$T_{2} = 323 K$

idealized actual volume is:

$Q_{2} = Q_{1} \frac{P_{1}}{P_{2}} \frac{T_{2}}{T_{1}}$

$Q_{2} = 1000 \times \frac{1}{10} \times \frac{323}{293}$

$Q_{2} \approx 110 C F M$

This demonstrates why FAD and high-pressure actual volume are numerically very different.

188 Simplified Conversion Is Not Full Borehole Modelling

The previous calculation assumes a defined thermodynamic state.

Actual DTH drilling involves:

pressure changes through hammer
exhaust expansion
heat transfer
leakage
water
two-phase cuttings transport

Therefore simple gas-law conversion is useful for understanding but does not replace rigorous borehole-flow modelling where required.

189 Calculation Complexity Should Match Project Risk

For routine shallow drilling, a relatively simple sizing method may be adequate.

For:

deep drilling
large diameter
severe water head
very high pressure
critical projects
more detailed engineering is justified

Do not use either:

unnecessary complexity for simple work
oversimplified calculations for high-risk work

190 Engineering Escalation Criteria

Escalate to more detailed analysis when:

required pressure approaches equipment limits
booster operation is required
water head is substantial
annular cleaning is uncertain
drill-string pressure loss is large
multiple compressor stages are considered
site conditions are severe

191 Compressor Selection Workflow — Master Sequence

The complete workflow is:

$DEFINE HOLE$

↓

$SELECT BIT$

↓

$SELECT HAMMER$

↓

$OBTAIN HAMMER Q (P)$

↓

$CALCULATE ANNULAR AREA$

↓

$VERIFY CLEANING AIR$

↓

$ESTABLISH AIRFLOW REQUIREMENT$

↓

$CALCULATE SURFACE LOSSES$

↓

$CALCULATE DRILL-STRING LOSSES$

↓

$CALCULATE WATER HEAD$

↓

$ESTABLISH PRESSURE REQUIREMENT$

↓

$CORRECT FOR SITE CONDITIONS$

↓

$ADD JUSTIFIED RESERVE$

↓

$CHECK PRIMARY COMPRESSOR$

↓

$CHECK BOOSTER NEED$

↓

$VERIFY COMPONENT RATINGS$

↓

$COMPARE ENERGY + COST$

↓

$FIELD VERIFY$

192 Selection Rule — Airflow

Never select compressor airflow solely from hammer nominal diameter. Use the hammer’s verified air-consumption data and independently verify that the resulting exhaust airflow can provide adequate borehole cleaning for the actual annular geometry.

193 Pressure selection rule

Define the hammer requirement and exhaust condition first. Add only losses and allowances outside that stated boundary. Use Section 21 to distinguish a differential-pressure requirement from an inlet-pressure requirement, and record water/backpressure once.

194 Selection Rule — Booster

Use a booster to solve a demonstrated high-pressure requirement when adequate upstream mass flow exists. Do not use a booster as a substitute for fundamentally insufficient primary airflow.

195 Selection Rule — Depth

Never select a drilling compressor from borehole depth alone. Depth influences the system through drill-string pressure loss, water exposure, leakage opportunity, handling and circulation—not through a universal CFM-or-pressure-per-metre rule.

196 Selection Rule — Water

Use actual or reasonably bounded hydrostatic head, not total hole depth, when calculating groundwater backpressure.

For freshwater:

$P_{w} \approx 0.0981 h bar$

with $h$ in metres.

Thus approximately:

$50 m \to 4.9 b a r$

$100 m \to 9.8 b a r$

$150 m \to 14.7 b a r$

$200 m \to 19.6 b a r$

These values represent hydrostatic pressure only.

They do not include:

hammer pressure requirement
drill-string loss
hose/manifold loss
other restrictions
operating reserve

197 Selection Rule — Annular Velocity

Never assume that sufficient air to operate the hammer automatically means sufficient air to clean the borehole.

Verify:

$A_{a} = \frac{π}{4} (D_{h}^{2} - D_{p}^{2})$

and:

$V_{a} = \frac{Q_{a}}{A_{a}}$

using an airflow value appropriate to the thermodynamic conditions at the point being evaluated.

198 Selection Rule — Flow Reference

Never directly compare FAD, SCFM, ACFM and Nm³/min unless their reference conditions are understood and, where necessary, converted to a common basis.

The underlying relationship is mass flow:

$\dot{m} = ρ Q$

The same mass of air occupies very different volumes at atmospheric and high-pressure conditions.

199 Selection Rule — Pressure Basis

Use gauge pressure for clearly defined field operating values where appropriate, but use absolute pressure for gas-law and compression-ratio calculations.

Approximately:

$P_{a b s} = P_{g} + P_{a t m}$

and:

$P R = \frac{P_{o u t, a b s}}{P_{i n, a b s}}$

200 Selection Rule — Site Conditions

A compressor that satisfies the requirement on a reference-condition datasheet does not automatically satisfy the requirement at a hot or high-altitude drilling site.

Verify:

$Q_{a v a i l a b l e, s i t e}$

and prime-mover capability using equipment-specific correction information.

201 Selection Rule — Reserve

Provide justified operating reserve for credible changes in drilling conditions, but do not repeatedly stack arbitrary safety margins until the compressor becomes grossly oversized.

Every allowance should have a defined purpose.

202 Selection Rule — Pressure Loss

Before purchasing additional pressure capability, identify how much existing compressor pressure is being consumed by avoidable restrictions.

Measure:

$Δ P = P_{c o m p r e s s o r} - P_{d r i l l h e a d}$

where practical.

Then determine whether the loss is:

necessary
expected
excessive
correctable

203 Selection Rule — Drill Pipe

Treat the drill string as both a structural system and a compressed-air distribution system.

Internal diameter affects:

$Δ P_{s t r i n g}$

while outside diameter affects:

$A_{a}$

and therefore:

$V_{a}$

The two effects must be considered together.

204 Selection Rule — Compressor Catalogue

Do not select a compressor by matching one catalogue CFM number with one hammer CFM number.

The correct selection point is:

$(Q_{r e q u i r e d}, P_{r e q u i r e d})$

after accounting for:

site conditions
pressure losses
water
hole cleaning
reserve

205 Selection Rule — Maximum Ratings

Maximum airflow and maximum pressure printed for a compressor should not be assumed to occur simultaneously unless verified by manufacturer performance data.

Use:

$Q_{c} (P)$

rather than unrelated maximum specifications.

206 Selection Rule — Field Validation

A compressor calculation is not fully closed until actual drilling confirms that the selected system can maintain required hammer performance, hole cleaning, pressure stability and productivity.

Therefore:

CALCULATE → SELECT → COMMISSION → MEASURE → VERIFY.

207 Final Worked Example — Integrated Deep-Hole Calculation

Consider a hypothetical deep DTH project.

Borehole

$D_{h} = 165 m m$

Drill Pipe

Outside diameter:

$D_{p} = 102 m m$

Target Depth

$L = 350 m$

Hammer Requirement

For this hypothetical example, assume the hammer differential-pressure requirement is:

$Δ P_{h} = 20 bar differential$

and:

$Q_{h} = Q_{m a n u f a c t u r e r}$

The airflow value is deliberately left manufacturer-dependent rather than invented.

Effective Water Head

$h = 100 m$

Surface Loss

$Δ P_{s} = 1.5 b a r$

Drill-String Loss

$Δ P_{d} = 4 b a r$

Other Justified Loss

$Δ P_{o} = 0.5 b a r$

Operating Margin

$P_{R} = 2 b a r$

208 Integrated Example — Annular Area

Convert:

$D_{h} = 0.165 m$

$D_{p} = 0.102 m$

Then:

$A_{a} = \frac{π}{4} ({0.165}^{2} - {0.102}^{2})$

${0.165}^{2} = 0.027225$

${0.102}^{2} = 0.010404$

Therefore:

$A_{a} = \frac{π}{4} (0.016821)$

$A_{a} \approx 0.01321 m^{2}$

This annular area is then used with the appropriate actual airflow to verify hole-cleaning velocity.

209 Integrated Example — Hydrostatic Pressure

For:

$h = 100 m$

$P_{w} = 0.0981 \times 100$

$P_{w} \approx 9.81 b a r$

210 Integrated Example — Required Source Pressure

Calculate:

$P_{r e q u i r e d} = Δ P_{h} + P_{w} + Δ P_{s} + Δ P_{d} + Δ P_{o} + P_{R}$

Substitute:

$P_{r e q u i r e d} = 20 + 9.81 + 1.5 + 4 + 0.5 + 2$

Therefore:

$P_{r e q u i r e d} \approx 37.81 b a r (g)$

This is the approximate source-pressure target under the stated hypothetical assumptions.

211 Integrated Example — Airflow Gate

Assume for this numerical demonstration:

$Q_{h} = 950 C F M$

at the selected operating condition for illustration only.

After the annular-cleaning calculation and flow-reference conversion, suppose the engineering assessment establishes:

$Q_{u s e f u l} = 1000 C F M$

Then justified minor losses:

$Q_{L} = 50 C F M$

and operating reserve:

$Q_{R} = 100 C F M$

give:

$Q_{s e l e c t i o n} = 1000 + 50 + 100$

$Q_{s e l e c t i o n} = 1150 C F M$

These airflow figures are illustrative; actual hammer and cleaning requirements must use verified project data.

212 Integrated Example — Candidate Primary Compressor

Suppose the site-corrected primary compressor can deliver:

1200CFM

but only up to the required flow at approximately:

25bar(g)

Then:

Airflow

$1200 > 1150$

PASS

Pressure

$25 < 37.81$

FAIL

The primary compressor is therefore not independently sufficient.

213 Integrated Example — Booster Evaluation

Because upstream airflow is adequate but final pressure is insufficient, a booster can be evaluated.

Required approximate outlet:

$P_{o u t} \approx 37.81 b a r (g)$

Primary inlet:

$P_{i n} \approx 25 b a r (g)$

Convert approximately to absolute:

$P_{o u t, a b s} \approx 38.81 b a r$

$P_{i n, a b s} \approx 26 b a r$

Therefore:

$P R = \frac{38.81}{26}$

$P R \approx 1.49$

The candidate booster must then be verified for:

required inlet mass flow
inlet pressure
discharge pressure
pressure ratio
discharge temperature
cooling
duty cycle

214 Integrated Example — High-Pressure Circuit

Before operation at approximately:

38bar(g)
verify all downstream equipment for the actual maximum credible system pressure

This includes:

hose assemblies
couplings
manifold
valves
swivel
drill head
drill-string components
No calculation can override a component’s approved pressure limit

215 Integrated Example — Site Derating

Suppose the compressor’s catalogue airflow is:

1250CFM

but manufacturer site correction gives:

$K_{s i t e} = 0.96$

under the specified conditions.

Then:

$Q_{s i t e} = 1250 \times 0.96$

$Q_{s i t e} = 1200 C F M$

This is the figure that should be compared with:

$Q_{s e l e c t i o n} = 1150 C F M$

—not the uncorrected 1250 CFM rating.

216 Separate included allowance from extra capacity

The example selects 1150 CFM, comprising 1000 CFM useful duty, 50 CFM minor-loss allowance and 100 CFM planned operating allowance. The site-corrected candidate supplies 1200 CFM at its stated primary operating point.

Additional capacity above the selection target is $1200 - 1150 = 50$ CFM, or $50 / 1150 \times 100 = 4.35 %$ . This is additional headroom after the 100 CFM allowance already included. Relative to useful duty plus minor losses, total headroom is $1200 - 1050 = 150$ CFM, or $150 / 1050 \times 100 = 14.29 %$ .

These are different denominators, not competing answers. Neither percentage establishes booster-delivered flow or overall package suitability. Verify the coupled compressor and booster duty with common reference conditions and all package losses.

217 Integrated Example — What Could Invalidate the Selection?

The calculation should be revisited if:

water head exceeds 100 m
target depth increases
bit diameter changes
hammer changes
drill-pipe geometry changes
measured pressure loss exceeds 4 bar
site conditions differ from the correction basis
leakage becomes significant
Compressor selection is therefore tied to a defined operating envelope

218 Integrated Example — Higher Water Head

Suppose effective water head later increases from:

100m

to:

150m

Additional hydrostatic pressure:

$Δ P_{w} = 0.0981 (150 - 100)$

$Δ P_{w} = 4.905 b a r$

New approximate requirement:

$P_{r e q u i r e d, n e w} = 37.81 + 4.905$

$P_{r e q u i r e d, n e w} \approx 42.72 b a r (g)$

This may materially alter booster selection and downstream pressure-rating requirements.

219 Integrated Example — Pressure-Loss Improvement

Suppose engineering changes reduce drill-string/system losses by:

2bar

Then:

$P_{r e q u i r e d, n e w} = 42.72 - 2$

$P_{r e q u i r e d, n e w} \approx 40.72 b a r (g)$

This illustrates why pressure-loss control can reduce the high-pressure burden even after water conditions worsen.

220 Integrated Example — Final Selection Gate

The configuration is not approved merely because:

$Q_{s i t e} \geq Q_{s e l e c t i o n}$

and:

$P_{b o o s t e r} \geq P_{r e q u i r e d}$

Also verify:

annular cleaning
hammer compatibility
compressor/booster thermal performance
high-pressure component ratings
rig capability
expected ROP
fuel/metre
cost/metre
field performance
This is the complete engineering gate

221 Master Calculation Sheet

For every DTH compressor-selection case, record:

A — Hole

$D_{h}, L$

B — Drill String

$D_{p}, D_{i}, L_{d}$

C — Hammer

$Δ P_{h}, Q_{h} (P)$

D — Annulus

$A_{a} = \frac{π}{4} (D_{h}^{2} - D_{p}^{2})$

E — Hole Cleaning

$V_{a} = \frac{Q_{a}}{A_{a}}$

F — Hydrostatic Head

$P_{w} = ρ g h$

G — System Losses

$Δ P_{s} + Δ P_{d} + Δ P_{o}$

H — Required Source Pressure

$P_{r e q u i r e d} = Δ P_{h} + P_{w} + \sum Δ P + P_{R}$

I — Required Airflow

$Q_{s e l e c t i o n} = Q_{u s e f u l} + Q_{L} + Q_{R}$

J — Site Correction

$Q_{s i t e} = f (Q_{r a t e d}, A l t i t u d e, T e m p e r a t u r e)$

K — Primary Compressor Gate

$Q_{s i t e} \geq Q_{s e l e c t i o n} ?$

$P_{p r i m a r y} \geq P_{r e q u i r e d} ?$

L — Booster Gate

If required:

$P R = \frac{P_{o u t, a b s}}{P_{i n, a b s}}$

M — Final Verification

safety,

performance,

energy,

economics,

field validation.

222 Compressor Selection Checklist

Before approving a compressor, confirm:

hole diameter defined,

final depth defined,

bit selected,

hammer selected,

hammer airflow curve available,

hammer pressure range available,

drill-pipe OD known,

drill-pipe ID known,

annular area calculated,

hole-cleaning requirement evaluated,

airflow reference basis defined,

surface pressure loss evaluated,

drill-string pressure loss evaluated,

water head evaluated,

hydrostatic pressure calculated,

operating reserve justified,

altitude checked,

temperature checked,

site-corrected compressor capacity established,

pressure-flow operating point verified,

booster requirement evaluated,

booster inlet flow verified,

booster pressure ratio calculated using absolute pressure,

downstream pressure ratings verified,

expected fuel/metre considered,

expected cost/metre considered,

field-verification plan established.

223 Calculation Integrity Checklist

Before releasing the calculation, check:

all dimensions use consistent units
pressure basis is identified
absolute pressure used where required
temperature converted to absolute scale where required
FAD and actual flow are not confused
hammer air is not double-counted as separate cleaning air
hole depth is not automatically used as water head
catalogue maximum CFM and maximum pressure are not incorrectly combined
altitude correction is equipment-specific
margins are not duplicated
unknown values are identified
illustrative assumptions are labelled
manufacturer limits remain controlling

224 Final engineering decision

At every required drilling stage, compare site-corrected equipment capability with the simultaneous pressure and flow target. Use one common airflow reference and the pressure boundaries in Section 21. Verify hammer operation and cuttings transport as separate requirements served partly by the same air stream. A numerical capacity comparison is one part of the decision; equipment compatibility, return conditions, ratings and commissioning evidence must also be satisfied.

225 Final Selection Principle

The correct compressor is not the machine with the largest CFM or highest pressure. It is the technically verified compressor—or compressor/booster architecture—that delivers the required mass flow and working pressure at actual site conditions, preserves adequate annular cleaning, operates within all equipment ratings, provides justified reserve, and produces acceptable metres reliably and economically.

In compact form:

HOLE → BIT → HAMMER → AIRFLOW → ANNULAR CLEANING → PRESSURE LOSSES → WATER HEAD → SITE CORRECTION → COMPRESSOR → BOOSTER → FIELD VERIFICATION.

Questions about DTH compressor sizing

Can borehole depth alone determine compressor size?

No. Depth changes drill-string losses, water exposure and operating conditions. Sizing also needs the hammer data, hole and pipe geometry, air demand and a supported pressure-flow operating point.

Should hammer air and cleaning air be added together?

First identify the flow paths and put both requirements on one reference basis. Hammer exhaust can serve hole cleaning. Add only distinct simultaneous consumers and justified losses; confirm that the hammer or approved bypass can pass the required flow.

How is water head used in a pressure calculation?

For an illustrative static freshwater column, pressure is approximately 0.0981 bar per metre. Use the effective column rather than total hole depth. Do not add it again if the stated hammer inlet requirement already includes exhaust backpressure.

Does a booster increase available primary airflow?

A booster raises pressure within its documented operating envelope. It does not remedy a shortage of upstream mass flow. Verify inlet conditions, throughput, losses and final delivered duty for the combined package.

Can maximum CFM and maximum pressure be used together?

Only when manufacturer data confirms that simultaneous operating point for the offered configuration and site conditions. Separate catalogue maxima do not establish the required duty.

What information should accompany a sizing enquiry?

Send hole diameter and target depth, pipe outside and inside diameters, hammer and bit identity, available consumption data, expected water conditions, site temperature and altitude, hose layout, operating hours and the intended duty stages.

Request a duty based compressor proposal

Send the drilling brief, hammer data and available nameplate or layout photographs to reddycompressors@gmail.com. Identify whether the requirement is purchase, rental, booster evaluation or reassessment of an existing package. Reddy Compressors can use the available duty information to guide the next enquiry step. Equipment availability and project suitability require confirmation.