Calculating Throttling Temperature Change in Air Ducts

Across a well-insulated 1 m² duct carrying air at 10 m/s and 20 °C, a 1000 Pa obstruction changes the air temperature by about 0.002 °C for real air and by zero for an ideal gas. The +0.83 °C figure from ΔP·Q/(ṁ·cp) is wrong for a gas. The same method predicts +414 °C for a 5 atm drop, where the Joule-Thomson coefficient gives cooling of roughly 1–2 °C. The deciding quantity is enthalpy: an adiabatic obstruction with no shaft work conserves stagnation enthalpy, and the enthalpy of an ideal gas depends on temperature alone.

Reading the three conflicting results

Three methods give +414 °C, −2 °C and 3.33 °C for the same 6 atm to 1 atm restriction because each carries a different hidden assumption. Only the enthalpy-based methods are valid for a flowing gas through an insulated restriction.

Input check before comparing methods: the source writes the flow as 10 m3/2, which reads as 10 m³/s (1 m² × 10 m/s). The stated 1.2 kg/m³ matches air at about 1 atm absolute and 20 °C (ρ = P/(R·T) with R = 287 J/kg·K gives 1.204 kg/m³, R being an assumed air constant). The upstream pressure is never stated, so fix absolute pressure at both stations before any calculation. Mass flow is ṁ = 1.2 × 1 × 10 = 12 kg/s.

Enthalpy balance across the obstruction

An insulated obstruction with no moving parts is a steady-flow control volume with zero heat transfer and zero shaft work, so h1 + V1²/2 = h2 + V2²/2. The obstruction is not a turbine and does not transmit shaft work across the boundary. Insulation removes heat transfer. No energy leaves the stream, which is why a temperature drop from "lost energy" does not occur.

The 10 kW in Q·ΔP = 10 m³/s × 1000 Pa is mechanical energy dissipated by turbulence. It does not appear as extra sensible heat, because enthalpy is h = u + p/ρ and the flow-work term p/ρ falls as the internal energy rises. The general relation is:

dh = cp·dT + [ v − T·(∂v/∂T)p ]·dp

• Ideal gas: v = R·T/p, so the bracket equals v − v = 0. Constant h means constant T.
• Incompressible fluid: (∂v/∂T)p ≈ 0, the bracket equals v, and ΔT = ΔP/(ρ·c). This is the form of the power equation. With the source numbers it is 1000/(1.2 × 1005) = 0.829 K, valid for a liquid and wrong for air.

The power equation does apply where shaft work crosses the boundary, such as a fan or blower in the control volume. There the shaft power delivered to the air enters the balance as ΔT = P_shaft/(ṁ·cp). A passive restriction between two stations delivers no shaft power.

Joule-Thomson correction for real air

The Joule-Thomson coefficient μJT = (∂T/∂P)h = −[v − T·(∂v/∂T)p]/cp is exactly zero for an ideal gas, so any nonzero value measures the departure of the real gas from ideality. Air at 20 °C has μJT ≈ 0.2 °C/atm, a positive value, so air cools on throttling at room temperature.

For a small ΔP, ΔT = μJT·ΔP is sufficient. For a multi-atmosphere drop, hold enthalpy constant between the two states and solve for the downstream temperature from real-gas enthalpy data, because μJT is not constant across the range.

Choked flow in the 6 atm to 1 atm case

A 6:1 pressure ratio exceeds the critical ratio for air, so the flow through the obstruction chokes. With γ = 1.4 (assumed for air), p*/p0 = (2/(γ+1))^(γ/(γ−1)) = 0.528, which places the choke at about 3.2 atm from a 6 atm upstream. A shock or pressure discontinuity forms downstream of the throat, and static temperature swings well below the stagnation value in the expansion jet before mixing restores it. The enthalpy balance conserves stagnation enthalpy, so the outlet temperature is read after the jet has mixed and the velocity profile has recovered.

The 5 atm case is also not the 1000 Pa case scaled up. The drop is about 500 times larger, and the stated 10 m/s and 1.2 kg/m³ cannot hold at both stations. At 6 atm and 20 °C the density is roughly 7.2 kg/m³ (proportional to absolute pressure), so 12 kg/s in 1 m² runs at about 1.7 m/s upstream. The kinetic-energy term ΔV²/2 ≈ 49 J/kg is still only about 0.05 K.

A reversible, work-extracting expansion gives a different bound: T2 = T1·(P2/P1)^((γ−1)/γ) = 293.15 K × (1/6)^0.2857 ≈ 176 K (about −97 °C). Throttling extracts no work, so it does not follow this line. The 6/20 = 1/x ratio fails even as a constant-volume estimate: in kelvin it gives 293.15/6 = 48.9 K, which shows a closed constant-volume relation cannot describe a flowing, expanding gas.

Procedure for the temperature change across a duct obstruction

1. Record absolute upstream and downstream pressure, upstream temperature, and gas composition. Convert gauge readings to absolute.
2. Compute density at each station from ρ = P/(R·T) and the velocity at each station from V = ṁ/(ρ·A).
3. Draw the control volume from the upstream station to a downstream station past the mixing length. List every boundary crossing: shaft work (none for a passive obstruction, nonzero if a fan is inside), heat transfer (zero if insulated, otherwise estimate it), and mass flow.
4. Write h1 + V1²/2 = h2 + V2²/2 and evaluate the kinetic-energy difference. For 1000 Pa near 1 atm it is about 1 J/kg (about 0.001 K); this assumes upstream pressure near 1 atm because the source omits it.
5. For an ideal gas set T2 = T1 after correcting for the kinetic term. For real air use ΔT = μJT·ΔP at small ΔP, or the constant-enthalpy solution from real-gas property data at large ΔP.
6. If P2/P1 is below the critical ratio (about 0.528 for γ = 1.4), treat the obstruction as choked and reference stagnation properties, then recheck the density and velocity at each station.
7. Add heat gain or loss through the duct wall and any fan or motor heat separately as boundary terms; do not fold them into the throttling term.

Expected magnitudes and sensor limits

An expected 0.002 °C change across a 1000 Pa obstruction sits far below the accuracy of a duct temperature sensor. Read the sensor accuracy class, resolution, and self-heating in its datasheet and compare against the calculated ΔT before trusting any measured difference.

1. Compare the calculated ΔT against the stated accuracy of both sensors. When ΔT is smaller, no measurement across the obstruction can confirm the calculation.
2. Take a baseline with the obstruction removed or fully open at the same flow; any reading difference at zero restriction is sensor offset, stratification, or wall heat exchange.
3. Swap the two sensor positions to separate sensor offset from real air temperature difference.
4. Use total (stagnation) temperature comparisons, or correct the static reading with the dynamic term (0.05 K at 10 m/s), when velocities differ between stations.
5. Check duct insulation and ambient gradient. Heat exchange through the wall of a few tens of watts is large compared with the JT effect at 1000 Pa.
6. For the multi-atmosphere case, compare the measured downstream temperature with the constant-enthalpy real-gas prediction, allowing for the choked-flow mixing length.

Recurring calculation and measurement pitfalls

• Applying ΔP·Q/(ṁ·cp) to gas throttling. It reproduces the incompressible-fluid answer and overpredicts by orders of magnitude for the 5 atm case.
• Using °C in gas-law ratios. Convert to kelvin, and remember P ∝ T holds only at constant volume.
• Treating the ideal gas law as a state equation that fixes T from P alone. It ties P, V and T together; the process path (throttling, isentropic, isochoric) fixes the outcome.
• Omitting the upstream absolute pressure. Density stated without pressure cannot define both stations.
• Holding μJT constant over a multi-atmosphere drop. Use enthalpy data or integrate.
• Ignoring choked flow above the critical pressure ratio.
• Mixing up throttling with fan work. Fan shaft power heats the stream as P_shaft/(ṁ·cp); a passive restriction does not.

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