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22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2018

Question 2 of 8: Throttling calorimeter, turbine power and an isentropic-with-heat-loss process

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas mixtures, the air-standard Otto cycle, wet-region steam properties, the throttling calorimeter, the steady-flow energy equation, the regenerative gas-turbine (Brayton) cycle and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radial conduction through composite cylinders, conduction with internal heat generation, internal-flow convection with a constant surrounding-fluid temperature, and the effectiveness–NTU method for s​hell-and-tube exchangers. Steam properties are IAPWS-consistent (equivalent to the steam tables); ammonia properties are read from the saturated- and superheated-ammonia tables appended to the examination; air and combustion gases are treated as ideal gases with constant specific heats ($\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$).

Question 2 — Throttling calorimeter, turbine power and an isentropic-with-heat-loss process (Part A, equal value)

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) — Quality from the throttling calorimeter

Given. Wet steam at $P_1=1$ MPa is throttled to atmospheric pressure ($P_2=101.325$ kPa) where it is superheated at $T_2=150\ ^\circ\text{C}$. A throttle is adiabatic with negligible work and negligible kinetic-energy change, so it is a constant-enthalpy (isenthalpic) process, $h_1=h_2$.

Find. the quality $x_1$ of the wet steam at 1 MPa.

Approach. Read $h_2$ at the superheated exit, set $h_1=h_2$, and invert the wet-mixture relation $h_1=h_f+x_1h_{fg}$ at 1 MPa.

  1. Exit enthalpy (superheated, 101.325 kPa, 150 °C). From the steam tables, $h_2=2776.5\ \text{kJ/kg}$.
  2. Saturation properties at 1 MPa. $h_f=762.5$, $h_g=2777.1\ \text{kJ/kg}$, hence $h_{fg}=2014.6\ \text{kJ/kg}$ ($T_\text{sat}=179.9\ ^\circ\text{C}$).
  3. Invert for quality. With $h_1=h_2=2776.5\ \text{kJ/kg}$: $$x_1=\frac{h_1-h_f}{h_{fg}}=\frac{2776.5-762.5}{2014.6}=0.9997$$ $x_1\approx 0.9997$ (about 99.97 % dry)

The sample is very nearly saturated vapour — exactly the situation a throttling calorimeter is designed to measure: because the throttle superheats even high-quality steam, the small distance of state 2 into the superheat region maps back to a quality extremely close to unity.

Part (b) — Power developed by the turbine

Given. The same steam (state 1 = 1 MPa, $x_1=0.9997$, $h_1=2776.5\ \text{kJ/kg}$, $V_1=1.5$ m/s) expands through a turbine to $P_2=7.5$ kPa, $x_2=0.783$, $V_2=90$ m/s. Heat loss $\dot Q_\text{loss}=527$ W; mass flow $\dot m=0.45$ kg/s.

State$P$$x$$h$ (kJ/kg)$V$ (m/s)
1 (inlet)1 MPa0.99972776.51.5
2 (exit)7.5 kPa0.7832052.190

Find. the power developed $\dot W$.

Turbine 1: 1 MPa, x≈1$V_1=1.5$ m/s 2: 7.5 kPa, x=0.783$V_2=90$ m/s $\dot Q_\text{loss}=527$ W $\dot W\approx 324$ kW
Figure 2 — Turbine control volume. The steady-flow energy balance keeps the kinetic-energy change (large at exit, $V_2=90$ m/s) and the small 527 W heat loss.

Approach. Apply the steady-flow energy equation with kinetic energy retained and potential energy neglected.

  1. Exit enthalpy. At 7.5 kPa, $h_f=168.8$, $h_{fg}=2405.3\ \text{kJ/kg}$, so $$h_2=h_f+x_2h_{fg}=168.8+0.783(2405.3)=2052.1\ \text{kJ/kg}$$
  2. Steady-flow energy equation. Heat leaves the control volume, so $\dot Q=-527$ W: $$\dot W=\dot m\!\left[(h_1-h_2)+\tfrac{1}{2}(V_1^2-V_2^2)\right]+\dot Q$$ The kinetic term is $\tfrac12(1.5^2-90^2)=-4049\ \text{J/kg}$, small but not negligible against the enthalpy drop of $724{,}400\ \text{J/kg}$.
  3. Evaluate. $$\dot W=0.45\big[(2776.5-2052.1)\times10^3-4049\big]-527=325{,}980-1822-527$$ $\dot W\approx 323{,}600\ \text{W}=323.6\ \text{kW}$
ContributionValue
Enthalpy-drop power $\dot m(h_1-h_2)$+326.0 kW
Kinetic-energy change $\tfrac12\dot m(V_1^2-V_2^2)$−1.82 kW
Heat loss $\dot Q_\text{loss}$−0.53 kW
Power developed $\dot W$≈ 323.6 kW

Part (c) — Isentropic despite a heat loss

Comparing the tabulated entropies at the two end states shows $s_1=s_f+x_1s_{fg}\approx 6.584$ kJ/kg·K at 1 MPa and $s_2=s_f+x_2s_{fg}\approx 6.585$ kJ/kg·K at 7.5 kPa — essentially equal, so the steam undergoes an isentropic change of state even though heat was rejected. This is possible because "isentropic" (constant fluid entropy) is not the same as "reversible and adiabatic." The entropy of the steam changes for two competing reasons captured by the entropy balance for the control volume,

$s_2-s_1=\underbrace{-\dfrac{\dot Q_\text{loss}}{\dot m\,T_b}}_{\text{entropy carried out with heat}}+\underbrace{s_\text{gen}}_{\text{internal irreversibility}}$

Real turbine expansion is irreversible — friction, throttling across blade rows and turbulence generate entropy inside the machine ($s_\text{gen}>0$, which alone would raise the exit entropy). At the same time, heat rejected to the surroundings removes entropy from the fluid ($-\dot Q_\text{loss}/T_b<0$). In this particular case the two effects happen to be equal and opposite, so their sum is zero and the fluid's entropy is unchanged. The equality is therefore a numerical coincidence of the operating point, not evidence that the expansion was reversible. (A genuinely reversible adiabatic turbine would have $s_\text{gen}=0$ and $\dot Q=0$; here both are non-zero but cancel.)