Throttling Calorimeter, Steam Turbine Power, and an Apparently Isentropic Process: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2018, Q2

Question 2: Throttling Calorimeter, Steam Turbine Power, and an Apparently Isentropic Process

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.

Given. A calorimeter throttles a sample of wet steam from 1 MPa down to atmospheric pressure (101.325 kPa), where it is measured superheated at $150\,{}^{\circ}\text{C}$. The SAME 1 MPa wet steam (not the throttled sample — the calorimeter only taps a measuring side-stream) then enters a turbine and expands to 7.5 kPa, 78.3% quality, with the velocities, heat loss and flow rate below.

Given data
Quantity	Symbol	Value
Calorimeter inlet pressure	$P_1$	1 MPa
Calorimeter exit (atm., superheated)	$P_2,T_2$	101.325 kPa, $150\,{}^{\circ}\text{C}$
Turbine inlet velocity	$V_1$	1.5 m/s
Turbine exit pressure, quality	$P_4,x_4$	7.5 kPa, 78.3%
Turbine exit velocity	$V_4$	90 m/s
Turbine heat loss	$\dot Q_{loss}$	527 W
Mass flow rate	$\dot m$	0.45 kg/s

Find. (a) Quality of the wet steam at 1 MPa; (b) power developed by the turbine; (c) an explanation for why $s_{in}\approx s_{out}$ across the turbine despite the heat loss.

$T$–$s$ states: 1 (turbine inlet, 1 MPa, wet — same enthalpy as the calorimeter's throttled sample), 2 (calorimeter exit, atm., $150\,{}^{\circ}\text{C}$), 3 (turbine exit, 7.5 kPa, $x=0.783$). Throttling is isenthalpic, so $h_1=h_2$; the turbine expansion 1→3 turns out to be nearly a vertical (constant-$s$) line despite the heat loss — see part (c).

Approach. (a) Throttling is isenthalpic ($h_1=h_2$, no work, negligible $\Delta$KE/PE across a throttle), so the measured superheated enthalpy at atmospheric pressure equals the wet-steam enthalpy at 1 MPa; back out the quality from $h_1=h_f+x_1h_{fg}$ at 1 MPa. (b) This 1 MPa wet-steam state is ALSO the turbine's inlet state (same $h$), so apply the steady-flow energy equation across the turbine with both kinetic-energy terms and the stated heat loss. (c) Compute $s_1$ (from $x_1$ at 1 MPa) and $s_3$ (from $x_4=0.783$ at 7.5 kPa) and compare.

Part (a) — calorimeter exit enthalpy. $$h_2=h(101.325\text{ kPa},150\,{}^{\circ}\text{C})=2776.5\text{ kJ/kg}$$ At 1 MPa: $h_f=762.5\text{ kJ/kg}$, $h_{fg}=2014.6\text{ kJ/kg}$. Since throttling is isenthalpic, $h_1=h_2=2776.5\text{ kJ/kg}$: $$x_1=\frac{h_1-h_f}{h_{fg}}=\frac{2776.5-762.5}{2014.6}$$ $$\boxed{x_1=0.9997\approx99.97\%}$$ The steam entering the turbine is thus very nearly (but not quite) saturated vapour at 1 MPa.
Part (b) — turbine exit enthalpy at 7.5 kPa, $x=0.783$. From the steam tables at 7.5 kPa: $h_f=168.7\text{ kJ/kg}$, $h_{fg}=2405.3\text{ kJ/kg}$: $$h_3=h_f+x_4h_{fg}=168.7+0.783\times2405.3$$ $$\boxed{h_3=2052.1\text{ kJ/kg}}$$
Part (b) — steady-flow energy equation across the turbine. $$h_1+\frac{V_1^2}{2}=h_3+\frac{V_4^2}{2}+w_s+q_{loss}$$ $$w_s=(h_1-h_3)+\frac{V_1^2-V_4^2}{2}-\frac{\dot Q_{loss}}{\dot m}$$ $$w_s=(2776.5-2052.1)+\frac{1.5^2-90^2}{2\times1000}-\frac{0.527}{0.45}$$ $$w_s=724.4-4.05-1.171$$ $$\boxed{w_s=719.2\text{ kJ/kg}}$$ $$\dot W_t=\dot m\,w_s=0.45\times719.2$$ $$\boxed{\dot W_t=323.6\text{ kW}}$$
Part (c) — entropy check. $$s_1=s_f+x_1s_{fg}\Big|_{1\text{ MPa}}=2.1381+0.9997\times4.4470=6.584\text{ kJ/kg}\cdot\text{K}$$ $$s_3=s_f+x_4s_{fg}\Big|_{7.5\text{ kPa}}=0.5763+0.783\times7.6738=6.585\text{ kJ/kg}\cdot\text{K}$$ $$\boxed{s_1=6.584\text{ kJ/kg}\cdot\text{K}\approx s_3=6.585\text{ kJ/kg}\cdot\text{K}}$$ The two entropies match to within the tables' own rounding — the process reads as isentropic even though it is neither adiabatic (there IS a heat loss) nor reversible (a real turbine has friction/eddy losses). The explanation: an entropy balance on the turbine is $s_3-s_1=-q_{loss}/T_b+\sigma_{gen}$, where $T_b$ is the boundary temperature the heat leaves through and $\sigma_{gen}\geq0$ is entropy generated by irreversibility. Heat LEAVING the control volume removes entropy ($-q_{loss}/T_b\lt0$), while internal irreversibility (friction in the blading, mixing losses) ADDS entropy ($\sigma_{gen}\gt0$). For THIS particular turbine the two effects happen to cancel almost exactly, so $\Delta s\approx0$ — a numerical coincidence of the given data, not a sign that the process is reversible and adiabatic. "Isentropic" describes the net entropy change; it does not by itself imply either zero heat transfer or zero irreversibility individually.

Question 2 — results
Quantity	Value
(a) Quality of wet steam at 1 MPa, $x_1$	0.9997
(b) Turbine specific work $w_s$	719.2 kJ/kg
(b) Turbine power $\dot W_t$	323.6 kW
(c) $s_1$ vs. $s_3$	6.584 vs. 6.585 kJ/kg·K (equal within rounding)

Topic. Throttling calorimeters, the steady-flow energy equation with kinetic energy and heat loss, and the distinction between an isentropic PROCESS and an isentropic RESULT.

Key relations. Throttling: $h_{in}=h_{out}$ (SFEE with $w=0$, $\Delta$KE/PE negligible, adiabatic); general SFEE $h_1+V_1^2/2=h_2+V_2^2/2+w_s+q$; entropy balance $\Delta s=\int\delta q/T+\sigma_{gen}$.

Why this works. A throttling calorimeter exists precisely BECAUSE quality is hard to measure directly on a wet mixture but easy on superheated steam (a single $T$ at known $P$ fixes the state) — throttling to a low enough pressure that the result is superheated, while conserving $h$, transfers that same enthalpy value to a state where it is trivially readable. The turbine problem then reuses that same enthalpy as its OWN inlet condition, which is what lets part (c) be checked at all: both $s_1$ and $s_3$ come from the same two data points already computed in (a) and (b).

Common pitfall. Treating the calorimeter's exit state (atmospheric, $150\,{}^{\circ}\text{C}$) as the turbine's OWN inlet state and re-solving the turbine at atmospheric pressure — the turbine inlet is the ORIGINAL 1 MPa wet steam, which merely happens to share the same enthalpy as the throttled sample; also concluding from part (c) that the turbine is reversible, when the entropy balance shows heat loss and irreversibility are both present and simply offsetting.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4 & Ch. 7 (throttling devices; entropy balance for control volumes).

Question 2 of 8: Throttling Calorimeter, Steam Turbine Power, and an Apparently Isentropic Process

Question 2: Throttling Calorimeter, Steam Turbine Power, and an Apparently Isentropic Process