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22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2017

Question 1 of 8: Adiabatic Tank Filling and an Adiabatic Air Blower

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

Notes on this paper

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, filling of evacuated vessels, air-standard dual and gas-turbine (turbojet) cycles, 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 composite-cylinder conduction with convection, external cross-flow over a cylinder, internal-flow temperature decay, radiation between concentric spheres with a shield, and the ε–NTU cross-flow heat-exchanger method. Ammonia and ideal-gas air properties are read from the tables appended to the examination paper; the turbojet uses cold-air-standard constant specific heats.

Paper format: National Examination 16-Mec-A1, May 2017, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three questions from one part and two from the other). All eight questions are solved in full below.

Question 1: Adiabatic Tank Filling and an Adiabatic Air Blower (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.

Given. Data for each part are tabulated below. Find. (a) the air mass and temperature in tanks A and B after equalisation; (b) the blower power and the discharge area.

Part (a) — two connected tanks, reversible & adiabatic

QuantityValue
Tank A volume / tank B volume$V_A=1.0$ m³, $V_B=0.3$ m³
Initial state (all air in A)$P_1=700$ kPa, $T_1=20$ °C $=293.15$ K; B evacuated
ProcessRigid overall, adiabatic ($Q=0$); air remaining in A expands reversibly (isentropically)
Air properties$R=0.287$ kJ/kg·K, $\gamma=1.4$, $c_v=0.718$ kJ/kg·K
Tank A$V_A$ = 1.0 m³700 kPa, 20 °CTank B$V_B$ = 0.3 m³evacuatedvalve
Figure 1a — Air is released from A into the evacuated B until both reach the common pressure $P_2$. The gas that stays in A cools (isentropic expansion); the gas swept into B is compressed and heats up.

Approach. The whole two-tank system is rigid and adiabatic, so its total internal energy is unchanged; for an ideal gas this fixes the equilibrium pressure directly, and the reversible-adiabatic relation applied to the gas that remains in A gives its temperature — the rest follows from mass and the ideal-gas law.

  1. Total mass of air. $m=\dfrac{P_1V_A}{RT_1}=\dfrac{700(1.0)}{0.287(293.15)}=8.32$ kg.
  2. Equilibrium pressure from energy conservation. With $Q=0$ and no work crossing the (fixed) outer boundary, $U_2=U_1$. For an ideal gas $U=\frac{c_v}{R}\sum P_iV_i$, so $\sum P_iV_i$ is conserved: $$P_2(V_A+V_B)=P_1V_A\ \Rightarrow\ P_2=\frac{P_1V_A}{V_A+V_B}=\frac{700(1.0)}{1.3}$$ $P_2 = 538.5$ kPa
  3. Temperature of the air left in A (reversible adiabatic expansion). $T_{A2}=T_1\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}=293.15\left(\dfrac{538.5}{700}\right)^{0.2857}=272.0$ K.
  4. Mass and quantities in A. $m_A=\dfrac{P_2V_A}{RT_{A2}}=\dfrac{538.5(1.0)}{0.287(272.0)}=6.90$ kg, so $m_B=m-m_A=8.32-6.90=1.42$ kg.
  5. Temperature in B (ideal-gas law). $T_{B2}=\dfrac{P_2V_B}{R\,m_B}=\dfrac{538.5(0.3)}{0.287(1.42)}=395.9$ K. Tank A: $m_A=6.90$ kg at $T_{A2}=272.0$ K ($-1.2$ °C); Tank B: $m_B=1.42$ kg at $T_{B2}=395.9$ K ($122.7$ °C)

The energy check closes: $m_AT_{A2}+m_BT_{B2}=6.90(272.0)+1.42(395.9)=2439\approx mT_1$, confirming the internal energy is conserved. Physically the gas that ends up in B behaves like gas filling an evacuated bottle — flow work delivered by the gas behind it raises its temperature — while the gas remaining in A does work pushing that flow and cools.

Part (b) — adiabatic blower

StateData
Inlet (1)$P_1=90$ kPa, $T_1=20$ °C, $u_1=209.3$ kJ/kg, $V_1\approx0$, $\dot V_1=12.3$ m³/min
Outlet (2)$P_2=106$ kPa, $T_2=30$ °C, printed "$u=303.4$ kJ/kg" (read as $h_2$ — see the callout), $V_2=24$ m/s
ProcessSteady flow, adiabatic ($\dot Q=0$); $R=0.287$ kJ/kg·K

Approach. Get the specific volumes from the ideal-gas law to convert the volumetric intake to a mass flow and to form the inlet enthalpy $h_1=u_1+P_1v_1$, then apply the steady-flow energy equation (with the exit kinetic energy) for the power, and continuity for the discharge area. The discharge property printed on the paper needs one interpretation step, set out in the callout below.

  1. Specific volumes. $v_1=\dfrac{RT_1}{P_1}=\dfrac{0.287(293.15)}{90}=0.9348$ m³/kg, $\;v_2=\dfrac{RT_2}{P_2}=\dfrac{0.287(303.15)}{106}=0.8208$ m³/kg.
  2. Mass flow rate. $\dot m=\dfrac{\dot V_1}{v_1}=\dfrac{12.3/60}{0.9348}=0.2193$ kg/s.
  3. Enthalpies. Inlet: $h_1=u_1+P_1v_1=209.3+90(0.9348)=293.4$ kJ/kg. Discharge: the printed 303.4 kJ/kg is the enthalpy at 30 °C — it is the $h$ entry of the appended air table at 303 K ($h=303.4$, against $u=216.3$), so $h_2=303.4$ kJ/kg. The enthalpy rise is therefore $h_2-h_1=9.97$ kJ/kg, which matches $c_p\Delta T=1.005(10)=10.05$ kJ/kg as it must for an ideal gas.
  4. Steady-flow energy balance (power in). With $V_1\approx0$, $$\dot W_{in}=\dot m\!\left[(h_2-h_1)+\tfrac12 V_2^{2}\right]=0.2193\!\left[9.97+\tfrac{24^{2}}{2000}\right]=0.2193(10.25)$$ $\dot W_{in}\approx2.25$ kW
  5. Discharge area (continuity). $\dot m=\dfrac{A_2V_2}{v_2}\Rightarrow A_2=\dfrac{\dot m\,v_2}{V_2}=\dfrac{0.2193(0.8208)}{24}$. $A_2=7.50\times10^{-3}$ m² $=75.0$ cm²
Check — the discharge property printed as "$u=303.4$ kJ/kg".
The paper labels both bracketed properties $u$, but the two cannot both be internal energies: 303.4 − 209.3 = 94.1 kJ/kg would need $\Delta T=\Delta u/c_v=94.1/0.718=131$ °C, not the 10 °C the question itself states. The appended ideal-gas air table settles it — at 303 K it lists $h=303.4$ and $u=216.3$ kJ/kg, so 303.4 is the enthalpy column (and 209.3 is correctly the $u$ column at 293 K, whose enthalpy 293.4 kJ/kg the $u_1+P_1v_1$ step reproduces). Reading $h_2=303.4$ gives $h_2-h_1=9.97$ kJ/kg $\approx c_p\Delta T$, hence the 2.25 kW above. Taking the label literally instead ($h_2=u_2+P_2v_2=390.4$) would give 21.3 kW, i.e. 97 kJ/kg of work into an adiabatic stream that warms by only 10 °C — impossible for an ideal gas, and inconsistent with the 30 °C used for $v_2$ in the same calculation. The discharge area is unaffected either way.
QuantityResult
(a) Equilibrium pressure538.5 kPa
(a) Tank A$m_A=6.90$ kg, $T_{A2}=272.0$ K ($-1.2$ °C)
(a) Tank B$m_B=1.42$ kg, $T_{B2}=395.9$ K ($122.7$ °C)
(b) Blower power≈ 2.25 kW (21.3 kW on the literal reading of the printed "$u_2$" — see callout)
(b) Discharge area75.0 cm² ($7.50\times10^{-3}$ m²)
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