Polytropic Gas Expansion (Part a) and Ammonia Throttling (Part b): 17-Phys-B6 Applied Thermodynamics and Heat Transfer, Undated paper, Q1

Question 1: Polytropic Gas Expansion (Part a) and Ammonia Throttling (Part b)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Given. A fixed mass of gas, closed system, undergoes an expansion between two states whose end-point pressures are not both given directly — instead the FIRST-LAW terms (work and heat, both "by the gas") are given, which together with the gas's own $u(T)$ relation is enough to pin down $T_2$ without ever needing the polytropic exponent.

Given data
Quantity	Symbol	Value
Gas constant (from $pv=RT_{abs}$)	$R$	0.317 kJ/kg·K
Specific heat, constant volume (from $u=u_o+c_vT$)	$c_v$	0.846 kJ/kg·K
Cylinder volume	$V_1$	0.025 m³
Initial pressure	$P_1$	350 kN/m²
Initial temperature	$T_1$	$80\,{}^{\circ}\text{C}$ (353 K)
Work done BY the gas during expansion	$W_{12}$	2.85 kJ (see the check note)
Heat transferred BY the gas during expansion	$Q_{12}$	1.90 kJ (see the check note)

Check: the source prints "2.85 J" and "1.90 J" for the work and heat of a process moving 0.025 m³ of gas at 350 kN/m² (a $Pv$ scale of $350\times0.025=8.75$ kJ) — a change three orders of magnitude smaller than the system's own energy scale is not physically sensible, and a dropped k prefix is a common printing slip. Both figures are read as kilojoules here, which is what makes the resulting $\Delta T$ and mass land in an ordinary engineering range.

Part (a) — Find. The temperature $T_2$ after the expansion, and the work $W_{12,adiabatic}$ that a DIFFERENT process between the same two end states would do if no heat were transferred.

Part (a) — Approach. Get the (constant) mass from the initial ideal-gas state, apply the closed-system first law $Q_{12}-W_{12}=\Delta U=mc_v(T_2-T_1)$ to the actual process to find $T_2$, then note that $\Delta U$ between $T_1$ and $T_2$ is a STATE function — independent of path — so the same $\Delta U$ applies to the adiabatic process, giving its work directly from $Q=0$.

Mass of gas from the initial state. $$m=\frac{P_1V_1}{RT_1}=\frac{350\times0.025}{0.317\times353}$$ $$\boxed{m=0.07819\text{ kg}}$$
First law on the actual process (sign convention: $Q,W$ positive when leaving the gas, since both are stated "by the gas"). $$\Delta U = -Q_{12}-W_{12} = -1.90-2.85=-4.75\text{ kJ}$$ $$\Delta T=\frac{\Delta U}{mc_v}=\frac{-4.75}{0.07819\times0.846}$$ $$\boxed{\Delta T=-71.80\,{}^{\circ}\text{C}}$$
Temperature after expansion. $$T_2=T_1+\Delta T=80.0-71.80$$ $$\boxed{T_2=8.20\,{}^{\circ}\text{C}}$$
Work of an adiabatic process between the same two states. $\Delta U$ is fixed by $T_1$ and $T_2$ alone (an ideal-gas-like state function via $u=u_o+c_vT$), so it is the SAME $-4.75$ kJ regardless of path. With $Q=0$, the first law gives $$W_{12,adiabatic}=-\Delta U=-(-4.75)$$ $$\boxed{W_{12,adiabatic}=4.75\text{ kJ}}$$ — the gas does MORE work adiabatically than in the actual process, because none of its internal-energy drop is diverted to heat loss.

Question 1(a) — results
Quantity	Value
Gas mass $m$	0.0782 kg
Temperature after expansion $T_2$	$8.20\,{}^{\circ}\text{C}$
Work if the process were adiabatic instead	4.75 kJ

Topic. Closed-system first law for a non-ideal (but ideal-gas-LIKE) working fluid defined by its own printed $p$-$v$-$T$ and $u(T)$ relations.

Key relations. Equation of state $pv=R T_{abs}$; caloric equation $u=u_o+c_vT$; first law for a closed system $Q_{12}-W_{12}=\Delta U=m(u_2-u_1)$.

Why this works. Because $u$ depends on $T$ ALONE for this gas (no $v$ or $p$ dependence is printed), $\Delta U$ is a pure function of the end-point temperatures and is therefore identical for every process connecting $T_1$ to $T_2$ — this is exactly what licenses reusing the same $\Delta U=-4.75$ kJ for the hypothetical adiabatic process in the second part without knowing its polytropic index at all.

Common pitfall. Mixing up the sign convention on "work/heat BY the gas" (both positive quantities LEAVING the system) with the more common "heat added, work done ON the system" convention; and reaching for the printed $pv^{1.30}=C$ exponent, which is not needed once $W_{12}$ and $Q_{12}$ are given directly — it would only be needed to find the FINAL pressure, which is not asked.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4 (closed-system energy balance, moving-boundary work).

Part (b) — Given. Ammonia is throttled from a known saturated-liquid state to a lower, known pressure — a valve is adiabatic, does no shaft work and has negligible kinetic/potential-energy change, so enthalpy is conserved across it.

Given data
Quantity	Symbol	Value
Inlet state	—	saturated liquid ammonia, $50\,{}^{\circ}\text{C}$
Exit pressure	$P_2$	0.45 MPa

Part (b) — Find. The temperature $T_2$ and quality $x_2$ after the expansion.

T–s sketch: state 1 is saturated liquid on the LEFT (low-entropy) branch of the dome at $50\,{}^{\circ}\text{C}$; the isenthalpic throttling line drops nearly vertically in $T$ down to state 2, a low-quality two-phase point at $0.45$ MPa's saturation temperature, well inside the dome.

Part (b) — Approach. Read $h_f$ at $50\,{}^{\circ}\text{C}$ (state 1), set $h_2=h_1$ (throttling), then locate state 2 on the $0.45$ MPa isobar: its saturation temperature is $T_2$ directly, and its quality follows from $h_2=h_f+x_2h_{fg}$ at that pressure.

Inlet enthalpy (saturated liquid, $50\,{}^{\circ}\text{C}$). $$h_1=h_f@50\,{}^{\circ}\text{C}$$ $$\boxed{h_1=586.14\text{ kJ/kg}}$$
Throttling: $h_2=h_1$. Locating $0.45$ MPa on the ammonia saturation table gives the exit temperature directly, since a two-phase mixture's temperature is fixed by its pressure alone: $$\boxed{T_2=1.27\,{}^{\circ}\text{C}}$$
Quality at state 2. At $0.45$ MPa, $h_f=351.54$ kJ/kg and $h_g=1608.80$ kJ/kg ($h_{fg}=1257.26$ kJ/kg): $$x_2=\frac{h_2-h_f}{h_{fg}}=\frac{586.14-351.54}{1257.26}$$ $$\boxed{x_2=0.187\ (18.7\%)}$$

Question 1(b) — results
Quantity	Value
Temperature after the valve $T_2$	$1.27\,{}^{\circ}\text{C}$
Quality after the valve $x_2$	0.187 (18.7%)
Entropy rise (irreversibility check)	$s_1=2.282\to s_2=2.360$ kJ/kg·K

Topic. Adiabatic throttling of a saturated liquid into the two-phase dome (the "flash" that every refrigeration expansion device performs).

Key relations. Steady-flow energy balance with $\dot W=0$, $\dot Q=0$, negligible $\Delta ke,\Delta pe$ reduces to $h_2=h_1$; two-phase quality $x=(h-h_f)/h_{fg}$.

Why this works. A partially-open valve is the textbook constant-enthalpy device: no useful work crosses its boundary and heat transfer over the short residence time is negligible, so the single equation $h_2=h_1$ (an isenthalpic process, NOT isentropic — entropy strictly increases, confirmed above) fully determines the downstream two-phase state once the exit pressure is fixed.

Common pitfall. Assuming the exit is still saturated liquid or superheated vapour rather than checking $h_2$ against $h_f$ and $h_g$ at $P_2$ first; and confusing throttling (isenthalpic, entropy-generating) with isentropic expansion (a turbine or nozzle).

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4–5 (throttling devices, quality).

Question 1 of 8: Polytropic Gas Expansion (Part a) and Ammonia Throttling (Part b)

Question 1: Polytropic Gas Expansion (Part a) and Ammonia Throttling (Part b)