ME-302 Thermodynamics - Nov 2022 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aShow that heat is a path function and not a property.
   How to answer

   • Define point function vs path function with mathematical criteria (exact vs inexact differential).
   • Draw a P-V diagram showing two states connected by different reversible paths (e.g., path A and path B) forming a cycle.
   • Apply the First Law: $\oint \delta Q = \oint \delta W$.
   • Show that since work done $\delta W = P dV$ is path-dependent (area under the curve), heat interaction $\delta Q$ must also depend on the path.
   • Conclude that $\int \delta Q$ depends on path, hence heat is a path function and not a thermodynamic property.
2. 7 Marks
   bA new scale N of temperature is divided in such a way that the freezing point of ice is $100^{\circ}\text{N}$ and the boiling point is $400^{\circ}\text{N}$. What is the temperature reading on this new scale when the temperature is $150^{\circ}\text{C}$? At what temperature both the Celsius and the new scale reading would be the same?
   How to answer

   • Establish linear relation between Celsius scale ($T_C$) and scale N ($T_N$): $\frac{T_N - 100}{400 - 100} = \frac{T_C - 0}{100 - 0}$.
   • Simplify equation: $T_N = 3T_C + 100$.
   • Substitute $T_C = 150^{\circ}\text{C}$: $T_N = 3(150) + 100 = 550^{\circ}\text{N}$.
   • Set $T_N = T_C = T$: $T = 3T + 100 \implies -2T = 100 \implies T = -50^{\circ}$.
   • State final answers: $550^{\circ}\text{N}$ and $-50^{\circ}$ (or $-50^{\circ}\text{C} = -50^{\circ}\text{N}$).
3. 7 Marks
   aDiscuss why does the enthalpy of air-vapour mixture remains constant during an adiabatic saturation process.
   How to answer

   • Describe the schematic of an adiabatic saturator chamber where unsaturated air enters and saturated air leaves over a pool of water maintained at saturation temperature.
   • Apply the Steady Flow Energy Equation (SFEE): $m_{a1} h_{1} + m_{w} h_{f} = m_{a2} h_{2}$ with no heat transfer ($Q = 0$) and no work done ($W = 0$).
   • Express per unit mass of dry air: $h_1 + (\omega_2 - \omega_1) h_w = h_2$.
   • Explain that the sensible heat required to evaporate additional water comes entirely from the sensible cooling of the air-vapour mixture.
   • Since the liquid enthalpy term $(\omega_2 - \omega_1)h_w$ is negligible compared to the total mixture enthalpy, $h_1 \approx h_2$, meaning total enthalpy remains approximately constant.
4. 7 Marks
   bA rigid close tank of volume $3\text{ m}^3$ contains $5\text{ kg}$ of wet steam at a pressure of $200\text{ kPa}$. The tank is heated until the steam becomes dry saturated. Determine final pressure and heat transfer to the tank.
   How to answer

   • Compute specific volume: $v_1 = v_2 = \frac{V}{m} = \frac{3}{5} = 0.6\text{ m}^3/\text{kg}$.
   • From steam tables at $P_1 = 200\text{ kPa}$, lookup $v_{f1}, v_{g1}, u_{f1}, u_{fg1}$ and calculate dryness fraction $x_1 = \frac{v_1 - v_{f1}}{v_{fg1}}$ and initial internal energy $u_1 = u_{f1} + x_1 u_{fg1}$.
   • For final state: steam is dry saturated with $v_{g2} = 0.6\text{ m}^3/\text{kg}$. Lookup saturation tables to find final pressure $P_2$ corresponding to $v_g = 0.6\text{ m}^3/\text{kg}$ (approximately $300\text{ kPa}$) and note $u_2 = u_{g2}$.
   • Apply First Law for rigid tank ($W = 0$): $Q = \Delta U = m(u_2 - u_1)$.
   • Calculate and state numerical values for $P_2$ and $Q$.
5. 14 Marks
   A piston-cylinder device operates $1\text{ kg}$ of fluid at $20\text{ atm}$. pressure The initial volume is $0.04\text{ m}^3$. The fluid is allowed to expand reversibly following a process $pV^{1.4} = \text{constant}$ so that the volume becomes double. The fluid is then cooled at a constant pressure until the piston comes back to the original position. Keeping the piston unaltered, heat is added reversibly to restore it to the initial pressure. Calculate the work done in the cycle.
   How to answer

   • Sketch the cycle on a P-V diagram: 1-2 reversible polytropic expansion ($PV^{1.4} = C$), 2-3 constant pressure cooling ($P = C$), 3-1 constant volume heating ($V = C$).
   • Convert pressure: $P_1 = 20\text{ atm} = 20 \times 1.01325 \times 10^5 = 2.0265 \times 10^6\text{ Pa}$.
   • Calculate $P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{1.4} = P_1 (0.5)^{1.4} \approx 0.3789 P_1 = 7.679 \times 10^5\text{ Pa}$.
   • Compute process works: $W_{1-2} = \frac{P_1 V_1 - P_2 V_2}{1.4 - 1}$, $W_{2-3} = P_2(V_1 - V_2) = -P_2(V_1)$, and $W_{3-1} = 0$.
   • Calculate net work: $W_{\text{net}} = W_{1-2} + W_{2-3}$ and state final answer in kJ.
6. 7 Marks
   aWrite short notes on "Mollier diagram". Why do isobars on the Mollier Diagram diverge from one another?
   How to answer

   • Define Mollier diagram (h-s chart) and describe its key features: saturation line, constant pressure, constant temperature, and constant dryness fraction lines.
   • From $dh = Tds + v dP$, derive that the slope of an isobar on an h-s diagram is $\left(\frac{\partial h}{\partial s}\right)_P = T$.
   • Explain that in the superheated region, as entropy increases along a constant pressure line, temperature $T$ also increases, causing the slope to continuously increase.
   • For two different constant pressure lines $P_1 > P_2$, at any given entropy, higher pressure corresponds to higher temperature, so the higher pressure isobar is steeper.
   • Hence, conclude that isobars diverge from one another from left to right as entropy increases.
7. 7 Marks
   bA pressure cooker holding $2\text{ kg}$ of steam at $5\text{ bar}$ and $90\%$ dry is being cooled slowly. What quantity of heat must be extracted so as to reduce the steam quality down to $60\%$? Also calculate the pressure and temperature of the steam that remains in the pressure cooker after the heat rejection.
   How to answer

   • Recognize constant volume process inside rigid pressure cooker: $v_1 = v_2$.
   • Look up steam table at $P_1 = 5\text{ bar}$: obtain $v_{g1}, u_{f1}, u_{fg1}$ and calculate $v_1 \approx x_1 v_{g1} = 0.90 \times 0.3749 = 0.3374\text{ m}^3/\text{kg}$ and $u_1 = u_{f1} + x_1 u_{fg1}$.
   • Find state 2 properties: $v_2 = v_1 \implies x_2 v_{g2} \approx v_1 \implies v_{g2} = \frac{0.3374}{0.60} = 0.5623\text{ m}^3/\text{kg}$.
   • From steam tables, find saturation pressure $P_2$ and saturation temperature $T_2$ corresponding to $v_g \approx 0.5623\text{ m}^3/\text{kg}$ (around $3.2\text{ bar}, 135^{\circ}\text{C}$).
   • Find $u_2 = u_{f2} + x_2 u_{fg2}$ at $P_2$, then calculate heat rejected: $Q = m(u_1 - u_2)$.
8. 7 Marks
   aExplain the free expansion process.
   How to answer

   • Define free expansion as an irreversible, unresisted expansion of a gas into an evacuated space (vacuum).
   • Draw a well-insulated chamber divided into two compartments by a partition or valve, one with gas and the other evacuated.
   • Apply First Law for the isolated system: $Q = 0$, $W = 0$ (no boundary displacement against resistance).
   • Show that $\Delta U = 0 \implies U_1 = U_2$.
   • For an ideal gas, state that since $U = f(T)$, internal energy constant implies $T_1 = T_2$ (constant temperature).
   • Highlight that the process is highly irreversible and entropy generation is positive.
9. 7 Marks
   bAn engine works on a diesel cycle with an Inlet pressure and temperature of $1\text{ bar}$ and $17^{\circ}\text{C}$. The pressure at the end of the adiabatic compression is $35\text{ bar}$. The ratio of expansion, i.e. after constant pressure heat addition is $5$. Calculate the heat addition, heat rejection and efficiency of the cycle. Assume $\gamma = 1.4$; $C_p = 1.005\text{ kJ/kg}\cdot\text{K}$; $C_v = 0.717\text{ kJ/kg}\cdot\text{K}$
   How to answer

   • Calculate compression ratio: $r = (P_2/P_1)^{1/\gamma} = (35/1)^{1/1.4} \approx 12.67$.
   • Compute $T_2 = T_1 \cdot (P_2/P_1)^{(\gamma-1)/\gamma} = 290 \times (35)^{0.2857} \approx 799.3\text{ K}$.
   • Find cut-off ratio: $r_c = \frac{r}{r_e} = \frac{12.67}{5} = 2.534$.
   • Calculate $T_3 = T_2 \cdot r_c \approx 2025.4\text{ K}$, and $T_4 = T_3 / (r_e)^{\gamma - 1} = 2025.4 / (5)^{0.4} \approx 1063.8\text{ K}$.
   • Compute heat added: $q_{\text{in}} = C_p(T_3 - T_2) = 1.005 \times (2025.4 - 799.3) = 1232.2\text{ kJ/kg}$.
   • Compute heat rejected: $q_{\text{out}} = C_v(T_4 - T_1) = 0.717 \times (1063.8 - 290) = 554.8\text{ kJ/kg}$.
   • Compute thermal efficiency: $\eta = 1 - \frac{q_{\text{out}}}{q_{\text{in}}} \approx 54.98\%$.
10. 7 Marks
    aProve that the COP of the reversible refrigerator operating between two given temperatures is the maximum.
    How to answer

    • State Carnot's theorem corollary for refrigeration cycles.
    • Assume an irreversible refrigerator $R_I$ has a higher COP than a reversible refrigerator $R_R$ between reservoirs $T_H$ and $T_L$: $\text{COP}_I > \text{COP}_R$.
    • Couple $R_I$ with a reversed $R_R$ (operating as a reversible heat engine between the same reservoirs).
    • Show that the composite system transfers net heat from the low-temperature reservoir to the high-temperature reservoir without any external work.
    • Explain that this violates the Clausius statement of the Second Law of Thermodynamics.
    • Conclude by contradiction that no refrigerator can have a higher COP than a reversible refrigerator operating between the same temperature limits.
11. 7 Marks
    bA steam pressure of holding capacity $4\text{ m}^3$ contains a mixture of saturated water and saturated steam at $250^{\circ}\text{C}$. The mass of the liquid present is $1\text{ ton}$. Determine: i) Quality ii) Specific Volume iii) Specific Enthalpy iv) Specific Entropy v) Specific Internal Energy of steam
    How to answer

    • From steam tables at $250^{\circ}\text{C}$, read $v_f, v_g, h_f, h_{fg}, s_f, s_{fg}, u_f, u_{fg}$ (e.g., $v_f \approx 0.001251\text{ m}^3/\text{kg}, v_g \approx 0.05013\text{ m}^3/\text{kg}$).
    • Compute liquid volume: $V_f = m_f v_f = 1000 \times 0.001251 = 1.251\text{ m}^3$.
    • Compute vapour volume: $V_g = V - V_f = 4 - 1.251 = 2.749\text{ m}^3$.
    • Compute vapour mass: $m_g = V_g / v_g = 2.749 / 0.05013 \approx 54.84\text{ kg}$; total mass $m = 1054.84\text{ kg}$.
    • i) Quality $x = m_g / m = 54.84 / 1054.84 \approx 0.052$.
    • ii) Specific volume $v = V / m = 4 / 1054.84 \approx 0.00379\text{ m}^3/\text{kg}$.
    • iii) Specific enthalpy $h = h_f + x h_{fg}$.
    • iv) Specific entropy $s = s_f + x s_{fg}$.
    • v) Specific internal energy $u = u_f + x u_{fg}$.
12. 7 Marks
    aWhat is a constant volume gas thermometer? Why is it preferred to a constant pressure gas thermometer?
    How to answer

    • Define constant volume gas thermometer: a device that measures temperature based on the variation of pressure of a fixed mass of gas at constant volume ($P \propto T$).
    • Describe its working equation: $T = 273.16 \lim_{P_{tp} \to 0} \left(\frac{P}{P_{tp}}\right)$.
    • Explain reasons for preference over constant pressure gas thermometer:
    • 1. Simpler to maintain constant volume than keeping strictly constant pressure during temperature changes.
    • 2. Avoids large dead space volume and stem errors present in constant pressure gas thermometers.
    • 3. Pressure measurement via manometers is more accurate, direct, and sensitive than measuring volume changes of expanding gas.
13. 7 Marks
    bA balloon which is initially collapsed and flat is slowly filled with hydrogen at $100\text{ kPa}$ so as to form it into a sphere of radius $1\text{ m}$. Determine the work done by the gas in the balloon during the filling process.
    How to answer

    • State that initial volume $V_1 = 0$ since the balloon is initially collapsed and flat.
    • Calculate final volume of the spherical balloon: $V_2 = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1)^3 \approx 4.1888\text{ m}^3$.
    • Express boundary work at constant pressure $P = 100\text{ kPa} = 100 \times 10^3\text{ N/m}^2$:
    • $W = P(V_2 - V_1) = 100 \times 10^3 \times (4.1888 - 0) = 418.88\text{ kJ}$.
    • State final work done by the gas: $418.88\text{ kJ}$.
14. 7 Marks
    aA reversible heat engine operates between two reservoirs at temperatures of $600^{\circ}\text{C}$ and $40^{\circ}\text{C}$. The engine drives a reversible refrigerator which operates between reservoirs at temperatures of $40^{\circ}\text{C}$ and $-20^{\circ}\text{C}$. The heat transfer to the engine is $2000\text{ kJ}$ and the network output of the combined engine-refrigerator plant is $360\text{ kJ}$. Evaluate the heat transfer to the refrigerant and the net heat transfer to the reservoir at $40^{\circ}\text{C}$.
    How to answer

    • Convert temperatures to Kelvin: $T_1 = 873.15\text{ K}$, $T_2 = 313.15\text{ K}$, $T_3 = 253.15\text{ K}$.
    • Calculate Carnot efficiency of engine: $\eta_E = 1 - \frac{313.15}{873.15} = 0.6414$.
    • Engine work: $W_E = \eta_E Q_1 = 0.6414 \times 2000 = 1282.74\text{ kJ}$; heat rejected $Q_2 = Q_1 - W_E = 717.26\text{ kJ}$.
    • Work input to refrigerator: $W_R = W_E - W_{\text{net}} = 1282.74 - 360 = 922.74\text{ kJ}$.
    • Calculate COP of refrigerator: $\text{COP}_R = \frac{T_3}{T_2 - T_3} = \frac{253.15}{60} = 4.219$.
    • Heat absorbed by refrigerant: $Q_4 = \text{COP}_R \times W_R = 4.219 \times 922.74 = 3893.2\text{ kJ}$.
    • Heat rejected by refrigerator to $40^{\circ}\text{C}$: $Q_3 = Q_4 + W_R = 3893.2 + 922.74 = 4815.94\text{ kJ}$.
    • Net heat transferred to the $40^{\circ}\text{C}$ reservoir: $Q_{\text{net}, 40^{\circ}\text{C}} = Q_2 + Q_3 = 717.26 + 4815.94 = 5533.2\text{ kJ}$.
15. 7 Marks
    bReconsider (i) given that the efficiency of the heat engine and the COP of the refrigerator are each $40\%$ of their maximum possible value.
    How to answer

    • Calculate actual engine efficiency: $\eta_{\text{actual}} = 0.40 \times 0.6414 = 0.25656$.
    • Calculate actual engine work output: $W_E = 0.25656 \times 2000 = 513.12\text{ kJ}$.
    • Calculate heat rejected by engine: $Q_2 = 2000 - 513.12 = 1486.88\text{ kJ}$.
    • Work available for refrigerator: $W_R = W_E - W_{\text{net}} = 513.12 - 360 = 153.12\text{ kJ}$.
    • Calculate actual COP of refrigerator: $\text{COP}_{\text{actual}} = 0.40 \times 4.219 = 1.6876$.
    • Compute heat transfer to refrigerant: $Q_4 = \text{COP}_{\text{actual}} \times W_R = 1.6876 \times 153.12 = 258.4\text{ kJ}$.
    • Compute heat rejected by refrigerator to $40^{\circ}\text{C}$: $Q_3 = Q_4 + W_R = 258.4 + 153.12 = 411.52\text{ kJ}$.
    • Compute net heat rejected to $40^{\circ}\text{C}$ reservoir: $Q_{\text{net}, 40^{\circ}\text{C}} = 1486.88 + 411.52 = 1898.4\text{ kJ}$.