22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2015
Question 1 of 8: Bursting Capsule and Compressor Power
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2015. Open-book, 3-hour paper. 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, all of equal value. Full worked solutions to all eight questions are given below.
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, vapour and gas power 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) — composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU / LMTD-correction heat-exchanger methods. Freon-12 property data are taken from the appendix supplied with the exam; steam, air and water data from standard tables.
Question 1: Bursting Capsule and Compressor Power (equal value)
Given. A rigid, closed container of total volume $V=0.025\ \text{m}^3$ holds a small capsule of compressed liquid water; after the capsule bursts the water fills the whole container.
Quantity
Value
Capsule (liquid) volume
$5\ \text{cm}^3=5\times10^{-6}\ \text{m}^3$
Initial state
compressed liquid, 7.5 MPa, 60 °C
Container volume (evacuated)
0.025 m³
Final temperature (heat exchange to surroundings)
60 °C
Find. The final density of the water/vapour mixture and the heat $Q$ exchanged with the surroundings.
Approach. Mass is conserved, so the final density is simply the fixed mass divided by the fixed container volume; the closed-system first law (rigid walls, $W=0$) then gives the heat as the change in internal energy between the compressed-liquid start and the wet-mixture end at 60 °C.
Mass of water. For compressed liquid at 7.5 MPa, 60 °C the specific volume is close to the saturated-liquid value, $v_1\approx0.001004\ \text{m}^3/\text{kg}$, so$$m=\frac{V_\text{cap}}{v_1}=\frac{5\times10^{-6}}{0.001004}=\boxed{4.98\times10^{-3}\ \text{kg}}.$$
Final specific volume and density. The same mass now occupies the full container:$$v_2=\frac{V}{m}=\frac{0.025}{4.98\times10^{-3}}=5.02\ \tfrac{\text{m}^3}{\text{kg}}\;\Rightarrow\;\rho_2=\frac{1}{v_2}=\boxed{0.199\ \text{kg/m}^3}.$$
Final state (two-phase at 60 °C). At 60 °C, $v_f=0.001017$ and $v_g=7.667\ \text{m}^3/\text{kg}$; since $v_f<v_2<v_g$ the water is a wet mixture with quality$$x_2=\frac{v_2-v_f}{v_g-v_f}=\frac{5.02-0.001017}{7.666}=\boxed{0.655}.$$
First law for the rigid container. With $W=0$, $Q=m(u_2-u_1)$. Using $u_1\approx250.1$ kJ/kg (compressed liquid) and $u_2=u_f+x_2u_{fg}=251.16+0.655(2205.4)=1695$ kJ/kg,$$Q=4.98\times10^{-3}(1695-250.1)=\boxed{+7.2\ \text{kJ}}.$$ The positive sign means about 7.2 kJ must be absorbed from the surroundings — evaporation is endothermic.
Check
The density result is independent of temperature: it follows purely from the conserved mass and the fixed container volume. The heat, however, depends on the final temperature; with heat exchange to the surroundings the mixture settles at the initial 60 °C (the only temperature specified). The compressed-liquid approximation $v_1\approx v_f(60\,{}^\circ\text{C})$ shifts $m$ by under 1.5 %.
Quantity
Result
Water mass
$4.98\times10^{-3}$ kg
Final density
≈ 0.199 kg/m³
Final quality at 60 °C
0.655
Heat absorbed from surroundings
≈ +7.2 kJ
Part (b) — power to drive the compressor
Given. Air (ideal gas, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$, $k=1.4$) enters at $P_1=1\ \text{atm}=101.3\ \text{kPa}$, $T_1=20\ ^\circ\text{C}$ with negligible velocity; it leaves through a 1 cm bore at $V_2=7\ \text{m/s}$ and $P_2=3.5\ \text{atm}=354.6\ \text{kPa}$. Compression is adiabatic (taken reversible).
Find. The compressor shaft power $\dot W$.
Approach. Fix the discharge temperature from the isentropic relation, get the mass flow from the discharge density–area–velocity, then apply the steady-flow energy equation (retaining the small exit kinetic-energy term, inlet velocity negligible).
Discharge temperature (isentropic). $$T_2=T_1\!\left(\frac{P_2}{P_1}\right)^{\!(k-1)/k}=293.15(3.5)^{0.2857}=\boxed{419.3\ \text{K}\ (146\,{}^\circ\text{C})}.$$
Mass flow from the discharge state. $\rho_2=P_2/RT_2=354.6/(0.287\cdot419.3)=2.947\ \text{kg/m}^3$; with $A_2=\pi(0.01)^2/4=7.85\times10^{-5}\ \text{m}^2$,$$\dot m=\rho_2A_2V_2=2.947(7.85\times10^{-5})(7)=\boxed{1.62\times10^{-3}\ \text{kg/s}}.$$
Energy balance (adiabatic, $V_1\approx0$). $$\dot W=\dot m\!\left[c_p(T_2-T_1)+\tfrac12V_2^2\right]=1.62\times10^{-3}\!\left[126.8+\frac{7^2}{2000}\right]=\boxed{0.205\ \text{kW}\approx205\ \text{W}}.$$
Check
The exit kinetic-energy term (0.0245 kJ/kg) is negligible against the 126.8 kJ/kg enthalpy rise, so the power is essentially set by the temperature lift. Modelling the compression as reversible-adiabatic (isentropic) is the natural reading of "adiabatically" with no efficiency stated.
