A compound pneumatic locomotive carries compressed air in receivers and expands it through high- and low-pressure cylinders. Explore a schematic tank locomotive and calculate the air lost through a pinhole in its receiver. The calculator estimates a steady leak, not locomotive power, range or charge duration.
Compound Pneumatic Locomotive Interactive Calculator
Adjust pinhole diameter and receiver gauge pressure to estimate a constant-pressure air leak. The locomotive and its coupled running gear give context; the enlarged leak section responds to the selected hole size. Engine speed and time between charges are not calculated.
Equation Used
- Pressure and temperature remain constant while the leak rate is evaluated.
- Discharge coefficient 0.50 is assumed, not a measured property of a particular hole.
- One-hour loss extrapolates this steady rate; a sealed receiver would lose pressure instead.
- No tank volume, valve timing, engine efficiency, load or operating speed is provided.
- Locomotive silhouette and running gear are schematic; dimensions and playback speed are not engineering results.
The locomotive motion explains the connected running gear. The calculated results describe air leaking through a pinhole at constant pressure.
Receivers, cylinders and coupled wheels
The historical arrangement in Hiscox Figure 663 places long air receivers above the running gear. An intermediate-pressure receiver supplies the high-pressure cylinders. Their exhaust passes to the low-pressure cylinders before discharge near atmospheric pressure.
The illustration shows a cab, receiver supports, frame, guided piston rod, connecting rod and coupled driving wheels. The near-side cylinder is shown in cutaway; the opposite compound stage is hidden. These parts remain connected as the wheels turn. Playback speed is illustrative and does not change with the leak inputs.
The separate enlarged section shows air escaping through the receiver wall. Changing hole diameter changes the opening; changing pressure changes the calculated flow. The tank does not visibly empty because the calculation holds pressure constant.
Using the leak estimate
The calculator compares leak rates for selected hole diameters and receiver pressures. It can help explain why a small opening can waste a substantial amount of stored air, but it does not diagnose a particular locomotive or specify a pressure rating.
A measured discharge coefficient, actual temperature and real-gas behavior can affect the result. Determining remaining operating time also requires tank capacity, the changing pressure and engine consumption, which are not inputs here.
What the calculator computes
Hole area is A = πd²/4. Gauge pressure becomes absolute pressure by adding 14.7psi. For the allowed 50–800psig range, discharge to atmosphere is choked. The mass-flow expression is ṁ = CdA Pabs√(γ/RT)[2/(γ+1)](γ+1)/(2(γ−1)), using Cd =0.50, γ=1.4, R=287.05J/(kg·K), and T=293.15K.
Standard volumetric flow is mass flow divided by air density at 20°C and 101325Pa. Multiplying SCFM by 60 gives a one-hour volume at an unchanged leak rate. It is not the amount a finite sealed receiver will necessarily lose in an hour.
The power-equivalent output uses ṁRT ln(Pabs/Patm). This is the ideal isothermal expansion-power equivalent of the lost air. It is not a prediction of shaft power or electrical energy consumed by a compressor.
Compare a small and large leak
At a fixed pressure and temperature, this model makes mass flow proportional to hole area. Increasing diameter from 1mm to 2mm therefore multiplies all four outputs by four.
Pressure enters as absolute pressure. Raising the gauge setting from 350 to 700psig does not exactly double flow: the multiplier is 714.7/364.7, about 1.96. The isothermal power equivalent also changes with the logarithm of the pressure ratio.
Use the two sliders to make these comparisons. Neither comparison establishes locomotive endurance without a tank-volume and consumption model.
What the drawing and calculation leave out
The visual simplifies a locomotive to expose its main connected parts. It does not simulate valve timing, leakage through piston rings, changing receiver pressure, temperature changes during expansion or traction at the wheels.
The pinhole model is likewise a simplified steady-flow estimate. Real openings may not behave like a circular orifice with the assumed discharge coefficient. The article therefore gives no universal efficiency, runtime, recharge pressure or capacity claim.
Questions about the leak calculator
Is the pressure gauge or absolute?
The slider is gauge pressure. The calculation adds 14.7psi to obtain absolute pressure.
Why does doubling hole size quadruple flow?
Circular area grows with diameter squared, and choked mass flow is proportional to that area at fixed pressure and temperature.
Does the animation predict locomotive speed?
No. The coupled-wheel motion illustrates the mechanism. Neither wheel speed nor traction is calculated.
What does the horsepower output mean?
It is an ideal isothermal expansion-power equivalent for the lost air, not measured engine output or compressor electrical demand.
Can the hourly loss predict time between charges?
No. Pressure is held constant, and neither tank capacity nor engine consumption is entered.
References
Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, printed page 178, Figure 663 (PDF page 188): receiver locomotive layout and compound-stage flow order.
NASA Glenn: Mass Flow Choking: compressible mass-flow relation. The leak calculation adds an assumed discharge coefficient; the drawing is an original schematic interpretation.
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