ME-302 Thermodynamics - Dec 2023 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
   aDefine Thermodynamic system and write various thermodynamic system used in thermodynamics, and also explain briefly Zeroth law of thermodynamics.
   How to answer

   • Define thermodynamic system, boundary, and surroundings.
   • List and describe types of systems: closed, open, and isolated systems with examples.
   • State the Zeroth law of thermodynamics.
   • Explain the significance of Zeroth law as the basis for temperature measurement.
2. 7 Marks
   bTo a closed system $150\text{ kJ}$ of work is supplied. If the initial volume is $0.6\text{ m}^3$ and pressure of the system change as $p = 8 - 4V$, where $p$ is pressure in bar and $V$ is volume in $\text{m}^3$, determine the final volume and pressure of the system.
   How to answer

   • Note work supplied to the system is negative: $W = -150\text{ kJ}$.
   • Convert pressure from bar to $\text{kPa}$: $p = 100(8 - 4V)\text{ kPa}$.
   • Set up integral $W = \int_{0.6}^{V_2} 100(8 - 4V) dV = -150$.
   • Integrate to obtain a quadratic equation in terms of $V_2$.
   • Solve for valid physical root $V_2$ and substitute to find final pressure $p_2$.
3. 7 Marks
   aWrite the limitations of first law of thermodynamics and apply the steady flow energy equation to water turbine.
   How to answer

   • State limitations of First Law (no directionality, no statement on feasibility or completeness of conversion of heat to work).
   • Write general Steady Flow Energy Equation (SFEE).
   • Apply assumptions for water turbine: adiabatic process ($q = 0$), incompressible fluid ($u_1 \approx u_2$).
   • Deduce simplified work output equation.
4. 7 Marks
   bAir at $1.02\text{ bar}$, $22^\circ\text{C}$, initially occupying a cylinder volume of $0.015\text{ m}^3$, is compressed reversibly and adiabatically by a piston to a pressure of $6.8\text{ bar}$. Calculate: i) The final temperature and final volume ii) The work done
   How to answer

   • Convert $T_1 = 22 + 273.15 = 295.15\text{ K}$, take $\gamma = 1.4$.
   • Calculate final volume using $V_2 = V_1 (p_1 / p_2)^{1/\gamma}$.
   • Calculate final temperature using $T_2 = T_1 (p_2 / p_1)^{(\gamma - 1)/\gamma}$.
   • Calculate work done: $W = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1}$ or $W = \frac{m R (T_1 - T_2)}{\gamma - 1}$.
5. 7 Marks
   aDefine Heat engine, refrigerator and heat pump and what do you mean by Clausius inequality explain.
   How to answer

   • Define Heat Engine with schematic and thermal efficiency $\eta = W/Q_H$.
   • Define Refrigerator and Heat Pump with schematic and their respective COP expressions.
   • State and write Clausius Inequality: $\oint \frac{\delta Q}{T} \le 0$.
   • Explain physical meaning for reversible ($=0$), irreversible ($<0$), and impossible ($>0$) cycles.
6. 7 Marks
   bA Carnot engine work in series between the source and the sink temperature of $550\text{ K}$ and $350\text{ K}$. If both engines develop equal power determine the intermediate temperature.
   How to answer

   • State relation for two reversible engines in series with equal work: $W_1 = W_2$.
   • Express work in terms of temperatures and heat absorbed: $Q_1(1 - T_2/T_1) = Q_2(1 - T_3/T_2)$ where $Q_2/Q_1 = T_2/T_1$.
   • Show this simplifies to $T_2 = \frac{T_1 + T_3}{2}$.
   • Substitute $T_1 = 550\text{ K}$ and $T_3 = 350\text{ K}$ to find $T_2 = 450\text{ K}$.
7. 14 Marks
   One mole of air (ideal gas with $\gamma = 1.4$) at pressure $P$ and temperature $T$, is compressed at constant volume till its pressure is doubled. Then it is allowed to expand reversibly and isothermally to the original pressure and finally restored to the original temperature by cooling at constant pressure. Sketch the path followed by the gas, on a P-V diagram and calculate the network done by the gas.
   How to answer

   • Identify the 3 processes: 1-2 constant volume to $2P$ (temp rises to $2T$), 2-3 isothermal expansion at $2T$ to $P$, 3-1 isobaric cooling at $P$ to $T$.
   • Sketch cycle 1-2-3 on P-V diagram.
   • Calculate work for each process: $W_{1-2} = 0$, $W_{2-3} = R(2T)\ln(2P/P) = 2RT\ln 2$, $W_{3-1} = P(V_1 - V_3) = R(T - 2T) = -RT$.
   • Sum the work to get net work: $W_{net} = RT(2\ln 2 - 1) \approx 0.386 RT$.
8. 7 Marks
   aCalculate the internal energy per kg of superheated steam at a pressure of $10\text{ bar}$ and temperature of $300^\circ\text{C}$. Also find the change of internal energy if this steam is expanded to $1.4\text{ bar}$ and dryness fraction $0.8$.
   How to answer

   • Use steam tables to find $h_1$ and $v_1$ (or $u_1$) at $10\text{ bar}$ and $300^\circ\text{C}$.
   • Calculate $u_1 = h_1 - p_1 v_1$.
   • From steam tables at $1.4\text{ bar}$, find saturation properties ($h_f, h_{fg}, v_f, v_g$ or $u_f, u_{fg}$).
   • Calculate $h_2 = h_f + x_2 h_{fg}$ and $v_2 \approx x_2 v_g$, then $u_2 = h_2 - p_2 v_2$.
   • Compute change in internal energy $\Delta u = u_2 - u_1$.
9. 7 Marks
   bDefine dryness fraction and entropy of evaporation.
   How to answer

   • Define dryness fraction $x = \frac{m_v}{m_v + m_l}$ as the ratio of mass of dry steam to total mass of wet mixture.
   • Define entropy of evaporation $s_{fg} = \frac{h_{fg}}{T_{sat}}$, the change in entropy during isothermal phase transition from saturated liquid to saturated vapor.
10. 7 Marks
    aWhat is meant by Air standard efficiency of cycle?
    How to answer

    • Define air standard efficiency as the thermal efficiency of an ideal cycle operating with air as the working fluid under air-standard assumptions.
    • List key air-standard assumptions (working fluid is air behaving as ideal gas, constant specific heats, closed cycle, reversible processes, combustion replaced by external heat addition).
11. 7 Marks
    bDerive the expression of Air standard efficiency of Otto cycle.
    How to answer

    • Draw P-V and T-s diagrams for Otto cycle showing the four processes.
    • Write expressions for heat added $Q_{in} = m c_v (T_3 - T_2)$ and heat rejected $Q_{out} = m c_v (T_4 - T_1)$.
    • Express thermal efficiency: $\eta = 1 - \frac{T_4 - T_1}{T_3 - T_2}$.
    • Use isentropic temperature-volume relations with compression ratio $r = V_1/V_2$ to deduce $\eta = 1 - \frac{1}{r^{\gamma-1}}$.
12. 7 Marks
    aWhat is adiabatic flame temperature? Discuss
    How to answer

    • Define adiabatic flame temperature as the maximum temperature attained by combustion products when the reaction is complete and no heat is lost to surroundings ($Q=0, W=0$).
    • State the governing enthalpy equation: $H_{reactants}(T_i) = H_{products}(T_{ad})$.
    • Discuss factors affecting it (excess air, dissociation, fuel composition, initial temperature).
13. 7 Marks
    bThe percentage composition of sample of liquid fuel by weight is $\text{C} = 84.8$ and $\text{H}_2 = 15.2$. Calculate: i) The weight of air needed for the combustion of $1\text{kg}$ of fuel ii) The volumetric composition of the product of combustion if $15\%$ excess air is supplied.
    How to answer

    • Find theoretical $O_2$ required: $O_2 = 0.848 \times (32/12) + 0.152 \times 8 = 2.2613 + 1.216 = 3.4773\text{ kg}$.
    • Calculate stoichiometric air by weight: $\frac{3.4773}{0.23} \approx 15.12\text{ kg}$ of air per kg of fuel.
    • Calculate actual air with $15\%$ excess: $1.15 \times 15.12 = 17.388\text{ kg}$.
    • Determine the mass and mole of combustion products: $\text{CO}_2, \text{H}_2\text{O}, \text{O}_2\text{ (excess)}, \text{N}_2$.
    • Calculate the mole fraction or percentage by volume for each constituent.
14. 14 Marks
    Two identical steel blocks each of mass $1000\text{ kg}$ are available at two different temperatures. The first block is at $1200\text{ K}$ while the second block is at $600\text{ K}$. The ambient temperature is $300\text{ K}$. The specific heat of steel is $0.5\text{ kJ/kg K}$. The blocks can be used as a source of energy to obtain work, till they are cooled to $300\text{ K}$. Calculate the available energy of each block.
    How to answer

    • State available energy formula for finite heat capacity body cooled to ambient: $AE = m c [(T - T_0) - T_0 \ln(T / T_0)]$.
    • Calculate heat capacity $m c = 1000 \times 0.5 = 500\text{ kJ/K}$.
    • For block 1 ($T_1 = 1200\text{ K}$): $AE_1 = 500 [(1200 - 300) - 300 \ln(1200/300)] = 500 [900 - 300 \ln 4]$.
    • For block 2 ($T_2 = 600\text{ K}$): $AE_2 = 500 [(600 - 300) - 300 \ln(600/300)] = 500 [300 - 300 \ln 2]$.