22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2018
Question 1 of 8: Gas mixing in connected vessels & an air-standard Otto cycle
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 shell-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 1 — Gas mixing in connected vessels & an air-standard Otto cycle (Part A, equal value)
Find. the volume $V_B$ of the argon vessel and the temperature $T_B$ of the argon before mixing (1 atm = 101.325 kPa throughout).
Approach. Get the air mass from $PV=mRT$ in vessel A; write the mixture as the sum of two ideal-gas partial contributions (Dalton) to fix the total volume $V_A+V_B$; then close vessel B's own equation of state for $T_B$.
Mass of air in vessel A. With $P_A=40(101.325)=4053\ \text{kPa}$:
$$m_\text{air}=\frac{P_A V_A}{R_\text{air}T_A}=\frac{4053(0.06)}{0.287(313.15)}=2.706\ \text{kg}$$
Total volume from the mixture state. After mixing, each species still obeys the ideal-gas law at $T_m$ and shares the total volume $V=V_A+V_B$; summing partial pressures (Dalton) gives $P_m V=(m_\text{air}R_\text{air}+m_\text{Ar}R_\text{Ar})T_m$:
$$V=\frac{(m_\text{air}R_\text{air}+m_\text{Ar}R_\text{Ar})T_m}{P_m}=\frac{\big[2.706(0.287)+1.35(0.2081)\big](303.15)}{1823.85}=0.1758\ \text{m}^3$$
$V_B=V-V_A=0.1758-0.06=0.116\ \text{m}^3$
Temperature of the argon before mixing. Apply $PV=mRT$ to vessel B alone, with $P_B=7(101.325)=709.3\ \text{kPa}$:
$$T_B=\frac{P_B V_B}{m_\text{Ar}R_\text{Ar}}=\frac{709.3(0.1158)}{1.35(0.2081)}=292.3\ \text{K}$$
$T_B=292\ \text{K}=19.1\ ^\circ\text{C}$
Quantity
Result
Mass of air (vessel A)
2.71 kg
Volume of argon vessel $V_B$
0.116 m³
Temperature of argon before mixing $T_B$
292 K (19.1 °C)
Part (b) — Air-standard Otto cycle
Given. A closed ideal cycle: isentropic compression 1→2, constant-volume heat addition 2→3, isentropic expansion 3→4, constant-volume heat rejection 4→1 — the air-standard Otto cycle. Air with $\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$ (so $c_v=R/(\gamma-1)=0.7175$, $c_p=1.0045\ \text{kJ/kg}\cdot\text{K}$); $P_1=138$ kPa, $T_1=37\ ^\circ\text{C}=310.15$ K, compression ratio $r=V_1/V_2=8$, and $T_3=T_\text{max}=1772$ K.
Find. the remaining temperatures and pressures, the thermal efficiency, and the power output for a heat-addition rate of 370 W.
Figure 1 — The Otto cycle on P–v and T–s planes. 1→2 isentropic compression, 2→3 constant-volume heat addition (to $T_\text{max}=1772$ K), 3→4 isentropic expansion, 4→1 constant-volume heat rejection. Pressures are in kPa.
Approach. Use the isentropic relations $T v^{\gamma-1}=\text{const}$ and $P v^{\gamma}=\text{const}$ across the compression and expansion, the constant-volume relation $P/T=\text{const}$ across the heat-transfer legs, and the closed-form Otto efficiency $\eta=1-r^{1-\gamma}$.
End of compression (1→2, isentropic).
$$T_2=T_1\,r^{\gamma-1}=310.15(8)^{0.4}=712.5\ \text{K},\qquad P_2=P_1\,r^{\gamma}=138(8)^{1.4}=2536\ \text{kPa}$$
End of heat addition (2→3, constant volume). With $T_3=1772$ K and $P/T$ constant:
$$P_3=P_2\,\frac{T_3}{T_2}=2536\left(\frac{1772}{712.5}\right)=6308\ \text{kPa}$$
End of expansion (3→4, isentropic). The expansion ratio equals $r$:
$$T_4=T_3\,r^{1-\gamma}=\frac{1772}{8^{0.4}}=771.3\ \text{K},\qquad P_4=P_3\,r^{-\gamma}=\frac{6308}{8^{1.4}}=343\ \text{kPa}$$
As a check, the constant-volume leg 4→1 gives $P_4=P_1(T_4/T_1)=138(771.3/310.15)=343$ kPa. ✓
Thermal efficiency. For the air-standard Otto cycle the efficiency depends only on the compression ratio:
$$\eta_\text{th}=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{8^{0.4}}=1-0.4353=0.565$$
$\eta_\text{th}=56.5\%$
Power output for a 370 W heat-addition rate. The net power is the efficiency times the heat-input rate:
$$\dot W_\text{net}=\eta_\text{th}\,\dot Q_\text{in}=0.565(370)=209\ \text{W}$$
$\dot W_\text{net}\approx 209\ \text{W}$
Check — magnitude of the heat-addition rate.
The printed heat-addition rate "370 W" is taken as given, yielding $\dot W_\text{net}=209$ W. For a real engine this is implausibly small (a fraction of a watt-hour), and the figure was very likely intended as 370 kW, in which case $\dot W_\text{net}=209$ kW. The efficiency and all state values are unaffected by this scaling; only the absolute power scales linearly with $\dot Q_\text{in}$.
