Hydraulics · nozzles · field units
Bit hydraulics: nozzle TFA, jet velocity, bit pressure drop and HSI
Bit hydraulics describes how much of the pump's energy is spent at the bit nozzles, where it cleans cuttings off the bit face and the bottom of the hole. Five numbers do the work: the total flow area (TFA) of the nozzles, the jet velocity, the pressure drop across the bit, the bit hydraulic horsepower and the horsepower per square inch of hole (HSI). In field units: TFA = Σ π ÷ 4 × (nozzle ÷ 32)², Vn = 0.3208 × Q ÷ TFA, ΔPbit = MW × Q² ÷ (10,858 × TFA²), HHPb = ΔPbit × Q ÷ 1714 and HSI = 1.27 × HHPb ÷ bit diameter². This page explains each formula, works one example end to end, and gives a TFA table for common nozzle sizes.
Reference material only. Nozzle selection, flow rate and pressure limits come from the drilling programme, the bit supplier and the rig's pump and surface equipment ratings. Worked examples are illustrative arithmetic, not a design.
1. The bit hydraulics formulas
Total flow area
TFA (in²) = Σ 0.7854 × (n ÷ 32)²
Jet (nozzle) velocity
Vn (ft/s) = 0.3208 × Q ÷ TFA
Pressure drop across the bit
ΔPbit (psi) = MW × Q² ÷ (10,858 × TFA²)
Bit hydraulic horsepower
HHPb = ΔPbit × Q ÷ 1714
Horsepower per square inch
HSI = 1.27 × HHPb ÷ D²
Jet impact force
IF (lbf) = MW × Vn × Q ÷ 1930
Q is the flow rate in gpm, MW the mud weight in ppg, n the nozzle size in 32nds of an inch and D the bit diameter in inches. The 10,858 in the bit pressure drop formula is 12,032 × 0.95²: it already includes a nozzle discharge coefficient of 0.95, which is the value usually assumed for jet nozzles. Without that assumption the textbook form is ΔPbit = 8.311 × 10⁻⁵ × MW × Q² ÷ (Cd² × TFA²). The 1714 converts psi × gpm into hydraulic horsepower, and 1.27 is 4 ÷ π, which turns the bit diameter into the area of the hole bottom.
2. Total flow area (TFA) of bit nozzles
Bit nozzles are sized in 32nds of an inch, so a "14" is a 14/32 in diameter jet. Each nozzle's area is π ÷ 4 × (size ÷ 32)², and the TFA is the sum of all of them. Mixed nozzles are added individually: two 12s and one 13 give 0.1104 + 0.1104 + 0.1296 = 0.350 in². Bits with fixed ports have a TFA quoted by the manufacturer; use that figure directly.
| Nozzle | 1 nozzle | 3 nozzles | 4 nozzles | 5 nozzles | 6 nozzles |
|---|---|---|---|---|---|
| 8/32 | 0.0491 | 0.147 | 0.196 | 0.245 | 0.295 |
| 9/32 | 0.0621 | 0.186 | 0.249 | 0.311 | 0.373 |
| 10/32 | 0.0767 | 0.230 | 0.307 | 0.383 | 0.460 |
| 11/32 | 0.0928 | 0.278 | 0.371 | 0.464 | 0.557 |
| 12/32 | 0.1104 | 0.331 | 0.442 | 0.552 | 0.663 |
| 13/32 | 0.1296 | 0.389 | 0.518 | 0.648 | 0.778 |
| 14/32 | 0.1503 | 0.451 | 0.601 | 0.752 | 0.902 |
| 15/32 | 0.1726 | 0.518 | 0.690 | 0.863 | 1.035 |
| 16/32 | 0.1963 | 0.589 | 0.785 | 0.982 | 1.178 |
| 17/32 | 0.2217 | 0.665 | 0.887 | 1.108 | 1.330 |
| 18/32 | 0.2485 | 0.746 | 0.994 | 1.243 | 1.491 |
| 19/32 | 0.2769 | 0.831 | 1.108 | 1.384 | 1.661 |
| 20/32 | 0.3068 | 0.920 | 1.227 | 1.534 | 1.841 |
TFA in square inches, calculated from π ÷ 4 × (n ÷ 32)². The values match published bit-supplier TFA tables to the third decimal.
3. Worked example: 12¼ in hole section
A 12¼ in bit with three 14/32 nozzles, circulating 650 gpm of 10.5 ppg mud. The standpipe pressure while drilling is 3,200 psi. All figures are illustrative.
| Step | Working | Answer |
|---|---|---|
| TFA | 3 × 0.7854 × (14 ÷ 32)² | 0.451 in² |
| Jet velocity | 0.3208 × 650 ÷ 0.451 | ≈ 462 ft/s |
| Bit pressure drop | 10.5 × 650² ÷ (10,858 × 0.451²) | ≈ 2,010 psi |
| Share of standpipe | 2,010 ÷ 3,200 | ≈ 63% |
| Bit hydraulic horsepower | 2,010 × 650 ÷ 1714 | ≈ 762 hhp |
| HSI | 1.27 × 762 ÷ 12.25² | ≈ 6.4 hp/in² |
| Jet impact force | 10.5 × 462 × 650 ÷ 1930 | ≈ 1,635 lbf |
The rest of the standpipe pressure, about 1,190 psi here, is the parasitic loss: friction inside the surface lines, drill pipe, drill collars and the annulus. The annular part of that loss is what raises ECD. The pump output that sets Q comes from the triplex pump output formula, and the jet velocity is a different number from the annular velocity, which is what lifts cuttings up the hole.
4. Choosing nozzles for a target pressure drop
Turn the bit pressure drop formula round to find the TFA that gives a chosen ΔPbit, then split it between the nozzles:
Required TFA
TFA = √(MW × Q² ÷ (10,858 × ΔPbit))
Nozzle size from area
n (32nds) = 32 × √(4 × A ÷ π)
Classical hydraulics optimisation (Bourgoyne et al., Applied Drilling Engineering, chapter 4) assumes the parasitic pressure loss rises with flow rate to a power of about 1.75 in turbulent flow. On that basis the bit should take about 65% of the available pump pressure to maximise bit hydraulic horsepower, or about 48% to maximise jet impact force. In the example, with 3,200 psi available:
| Criterion | Target ΔPbit | Required TFA | Nearest nozzles |
|---|---|---|---|
| Maximum bit hydraulic horsepower (≈ 65%) | 2,080 psi | 0.443 in² | 3 × 14/32 (0.451 in²) |
| Maximum jet impact force (≈ 48%) | 1,536 psi | 0.516 in² | 3 × 15/32 (0.518 in²) |
These ratios are starting points, not rules. In practice the flow rate is often fixed first for hole cleaning, and the nozzles are then sized so the standpipe pressure stays inside the pump liner rating and surface equipment limits, with a margin. PDC bits are commonly run for cooling and cleaning the cutters as well as the hole, and bit suppliers publish their own recommended HSI or flow ranges for each design.
5. Checks before you change nozzles
- Pump limits. Standpipe pressure at the planned flow rate must stay below the liner rating and the lowest-rated item in the circulating system. Smaller nozzles raise ΔPbit with the square of the flow rate.
- Hole cleaning. Cutting the flow rate to gain bit pressure can leave cuttings in the annulus, especially in deviated hole. Check annular velocity against the section's minimum.
- ECD. More flow raises annular friction and ECD; compare with the fracture gradient and leak-off test results.
- Mud weight changes. ΔPbit is proportional to mud weight, so a weight-up after a kick also raises bit pressure drop and standpipe pressure at the same rate.
- Plugged or washed nozzles. A sudden standpipe rise at constant rate can mean a plugged nozzle; a gradual fall can mean a washed nozzle or a washout in the string. Record pressures at slow pump rates as well; they are used on the kill sheet.
6. In DrillX
DrillX brings daily drilling reports with IADC-style time logs, a rig fleet view, NPT insights and engineering calculators into one tool. The calculators cover hydrostatic pressure, ECD, annular pressure loss, triplex pump output, annular capacity and velocity, buoyancy and the kill sheet. This page explains the bit hydraulics arithmetic; it does not describe a DrillX nozzle calculator. DrillX is in final testing; email info@drillx.app for a demo or early access.
Questions
How do you calculate nozzle TFA?
Total flow area (TFA) is the sum of the areas of all the bit nozzles. Nozzle sizes are given in 32nds of an inch, so the area of one nozzle is π ÷ 4 × (size ÷ 32)² in². One 14/32 nozzle is 0.1503 in², so three 14s give a TFA of 0.451 in². Mixed sizes are added one by one.
What is the formula for pressure drop across the bit?
The field formula is ΔPbit (psi) = MW × Q² ÷ (10,858 × TFA²), with mud weight in ppg, flow rate in gpm and TFA in square inches. It assumes a nozzle discharge coefficient of 0.95. With 10.5 ppg mud, 650 gpm and a TFA of 0.451 in² the bit pressure drop is about 2,010 psi.
How do you calculate jet velocity?
Nozzle (jet) velocity in ft/s is Vn = 0.3208 × Q ÷ TFA, with Q in gpm and TFA in in². At 650 gpm through a TFA of 0.451 in² the jet velocity is about 462 ft/s.
What is HSI in drilling?
HSI is hydraulic horsepower per square inch of hole area. First work out bit hydraulic horsepower, HHPb = ΔPbit × Q ÷ 1714, then divide by the bit face area: HSI = HHPb × 1.27 ÷ bit diameter². In the worked example 762 hhp on a 12¼ in bit gives about 6.4 hp/in². The target value depends on the bit, formation and the operator's practice.
What share of standpipe pressure should be used at the bit?
The classical optimisation in Applied Drilling Engineering gives about 65% of the available pump pressure across the bit for maximum bit hydraulic horsepower and about 48% for maximum jet impact force, assuming parasitic losses rise with flow rate to the power 1.75. These are starting points; hole cleaning, pump limits and bit type also decide the nozzle selection.
How do you choose nozzle sizes for a target bit pressure drop?
Rearrange the bit pressure drop formula: TFA = √(MW × Q² ÷ (10,858 × ΔPbit)). Divide the TFA by the number of nozzles to get the area of each, then convert to 32nds with size = 32 × √(4 × area ÷ π) and round to available sizes. For 1,536 psi at 650 gpm and 10.5 ppg the TFA is 0.516 in², which is about three 15/32 nozzles.
Sources and further reading
- A. T. Bourgoyne Jr., K. K. Millheim, M. E. Chenevert and F. S. Young Jr., Applied Drilling Engineering, SPE Textbook Series Vol. 2 (1986), chapter 4: bit nozzle pressure drop (8.311 × 10⁻⁵ ρ q² ÷ (Cd² At²)), hydraulic horsepower, jet impact force and the maximum-horsepower and maximum-impact-force criteria.
- N. J. Lapeyrouse, Formulas and Calculations for Drilling, Production and Workover, 2nd ed. (Gulf Professional Publishing, 2002): field-unit bit hydraulics (10,858 constant, 0.32086 jet velocity factor, 1714, 1.27, impact force ÷ 1930).
- Rocky Mountain Bit, TFA reference table: nozzle total flow area by size and count, used as a cross-check for the table above.
- API RP 13D, Rheology and Hydraulics of Oil-well Drilling Fluids: bit pressure loss and hydraulics optimisation with rheological models.
Reference summary only. Worked examples are illustrative arithmetic. Nozzle selection and flow rates are set by your operator's drilling programme, the bit supplier and rig equipment ratings.