BE Civil Engineering (IOE, TU) Fundamentals of Thermodynamics & Heat Transfer (IOE, ME 451) Question Paper 2079 Nepal

(a) First Law for a closed system

For a closed system undergoing a process between states 1 and 2:

$$Q_{1-2} - W_{1-2} = \Delta U = U_2 - U_1$$

• $Q_{1-2}$ = net heat transferred to the system (positive when added to the system),
• $W_{1-2}$ = net work done by the system (positive when the system does work on surroundings),
• $\Delta U$ = change in internal energy, a property depending only on end states.

The law expresses conservation of energy: heat added minus work done equals the rise in stored internal energy.

(b) Final volume and work

For a polytropic process $pV^n = \text{const}$ with $n = 1.3$:

$$V_2 = V_1\left(\frac{p_1}{p_2}\right)^{1/n} = 0.04\left(\frac{800}{120}\right)^{1/1.3}$$ $$\frac{800}{120} = 6.6667,\qquad 6.6667^{1/1.3} = 6.6667^{0.7692} = 4.231$$ $$V_2 = 0.04 \times 4.231 = \mathbf{0.1692\ m^3}$$

Boundary work for a polytropic process:

$$W_{1-2} = \frac{p_1 V_1 - p_2 V_2}{n-1}$$ $$p_1 V_1 = 800 \times 0.04 = 32\ \text{kJ}$$ $$p_2 V_2 = 120 \times 0.1692 = 20.30\ \text{kJ}$$ $$W_{1-2} = \frac{32 - 20.30}{1.3 - 1} = \frac{11.70}{0.3} = \mathbf{39.0\ kJ}$$

(c) Internal energy change and heat transfer

Temperatures from the ideal gas law $pV = mRT$:

$$T_1 = \frac{p_1 V_1}{mR} = \frac{800 \times 0.04}{0.5 \times 0.287} = \frac{32}{0.1435} = 222.99\ \text{K}$$ $$T_2 = \frac{p_2 V_2}{mR} = \frac{120 \times 0.1692}{0.5 \times 0.287} = \frac{20.30}{0.1435} = 141.49\ \text{K}$$

Change in internal energy:

$$\Delta U = m c_v (T_2 - T_1) = 0.5 \times 0.718 \times (141.49 - 222.99)$$ $$\Delta U = 0.5 \times 0.718 \times (-81.50) = \mathbf{-29.26\ kJ}$$

Heat transfer from the First Law:

$$Q_{1-2} = \Delta U + W_{1-2} = -29.26 + 39.0 = \mathbf{+9.74\ kJ}$$

Heat is added to the system (≈ 9.74 kJ) even though it expands and cools, because the work output exceeds the drop in internal energy.

(a) Second Law statements

Kelvin–Planck: It is impossible to construct a device operating in a cycle that produces no effect other than the transfer of heat from a single reservoir and the production of an equal amount of work (no 100 %-efficient heat engine).

Clausius: It is impossible to construct a device operating in a cycle that produces no effect other than the transfer of heat from a cooler to a hotter body (heat does not flow spontaneously from cold to hot without work input).

Equivalence (Clausius violation → Kelvin–Planck violation): Suppose a device transfers heat $Q$ from cold to hot with no work (violates Clausius). Couple it with an ordinary engine that takes $Q_H$ from the hot reservoir, does work $W$, and rejects $Q$ to the cold reservoir. The cold reservoir then has zero net exchange, and the combined device draws $(Q_H - Q)$ from the hot reservoir alone while producing work $W$ — a Kelvin–Planck violation. The reverse implication follows symmetrically, so the two statements are equivalent.

(b) Carnot engine

Thermal efficiency:

$$\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{750} = 1 - 0.4 = 0.60 = \mathbf{60\%}$$

Net work output:

$$W_{net} = \eta\, Q_H = 0.60 \times 1200 = \mathbf{720\ kJ}$$

Heat rejected:

$$Q_L = Q_H - W_{net} = 1200 - 720 = \mathbf{480\ kJ}$$

(c) Carnot refrigerator driven by the engine

Reservoir temperatures: $T_{cold} = -10 + 273 = 263\ \text{K}$, $T_{hot} = 30 + 273 = 303\ \text{K}$.

COP of a Carnot refrigerator:

$$\text{COP}_R = \frac{T_{cold}}{T_{hot} - T_{cold}} = \frac{263}{303 - 263} = \frac{263}{40} = \mathbf{6.575}$$

The work input to the refrigerator is the engine's net work, $W = 720\ \text{kJ}$ per cycle.

Heat extracted from the cold space:

$$Q_{cold} = \text{COP}_R \times W = 6.575 \times 720 = \mathbf{4734\ kJ\ per\ cycle}$$

Thus each engine cycle of 1200 kJ input enables about 4734 kJ to be pumped out of the cold space.

(a) Process on the p–v diagram

The vessel is rigid, so specific volume $v$ is constant (the mass and volume do not change). On the $p$–$v$ diagram the process is a vertical line moving upward from state 1 to state 2, both lying inside the two-phase dome (since the steam remains wet). As pressure rises along constant $v$, the state moves toward the saturated-vapour line but here stays wet.

 p |        .-''-.  (sat. dome)
   |  2 *  /      \
   |    | /        \
   |  1 *           \
   |____\____________\___ v
        v = const

(b) Mass and final dryness fraction

State 1 specific volume:

$$v_1 = v_f + x_1(v_g - v_f) = 0.001061 + 0.7(0.8857 - 0.001061)$$ $$v_1 = 0.001061 + 0.7 \times 0.884639 = 0.001061 + 0.619247 = 0.62031\ \text{m}^3/\text{kg}$$

Mass of steam:

$$m = \frac{V}{v_1} = \frac{0.15}{0.62031} = \mathbf{0.2418\ kg}$$

Since the vessel is rigid, $v_2 = v_1 = 0.62031\ \text{m}^3/\text{kg}$.

Final dryness fraction at 500 kPa:

$$x_2 = \frac{v_2 - v_{f2}}{v_{g2} - v_{f2}} = \frac{0.62031 - 0.001093}{0.3749 - 0.001093} = \frac{0.619217}{0.373807} = \mathbf{1.657}$$

A quality greater than 1 is impossible, which tells us the steam has become superheated before reaching 500 kPa. For the level of steam-table data provided we report that the final state is superheated vapour (the mixture dries out and continues heating). Using the available data, the limiting saturated state where the vessel just becomes dry-saturated is where $v_g = 0.62031\ \text{m}^3/\text{kg}$, which occurs at roughly $300\ \text{kPa}$; beyond that the steam is superheated. We therefore continue using the closest tabulated values and note the result for completeness.

(c) Heat added

For a rigid vessel the boundary work is zero ($W = \int p\,dV = 0$ since $V$ is constant). The First Law reduces to:

$$Q = \Delta U = m(u_2 - u_1)$$

State 1 specific internal energy:

$$u_1 = u_f + x_1(u_g - u_f) = 504.5 + 0.7(2529.5 - 504.5)$$ $$u_1 = 504.5 + 0.7 \times 2025.0 = 504.5 + 1417.5 = 1922.0\ \text{kJ/kg}$$

State 2: Because the steam dries out and becomes superheated, take the dry-saturated value at 500 kPa as a conservative estimate of $u_2 \approx u_{g2} = 2561.2\ \text{kJ/kg}$ (a small superheat correction would raise it slightly):

$$u_2 \approx 2561.2\ \text{kJ/kg}$$

Heat added:

$$Q = m(u_2 - u_1) = 0.2418 \times (2561.2 - 1922.0)$$ $$Q = 0.2418 \times 639.2 = \mathbf{\approx 154.6\ kJ}$$

So approximately 155 kJ of heat is required (the value would increase slightly once full superheat data are applied).

(a) Fourier's law and plane-wall conduction

Fourier's law of conduction in one dimension:

$$\dot{Q} = -kA\frac{dT}{dx}$$

For steady state with no heat generation and constant $k$, $\dot{Q}$ is constant. Separating variables across a wall of thickness $L$ between surfaces $T_1$ and $T_2$:

$$\dot{Q}\int_0^L dx = -kA\int_{T_1}^{T_2} dT \;\Rightarrow\; \dot{Q}L = kA(T_1 - T_2)$$ $$\boxed{\dot{Q} = \frac{kA(T_1 - T_2)}{L} = \frac{T_1 - T_2}{L/(kA)}}$$

where $R = L/(kA)$ is the conductive thermal resistance.

(b) Heat loss using thermal resistances

Individual resistances ($R = L/(kA)$, with $A = 5\,\text{m}^2$):

$$R_1 = \frac{0.20}{1.2 \times 5} = \frac{0.20}{6.0} = 0.033333\ \text{K/W}$$ $$R_2 = \frac{0.10}{0.15 \times 5} = \frac{0.10}{0.75} = 0.133333\ \text{K/W}$$ $$R_3 = \frac{0.01}{45 \times 5} = \frac{0.01}{225} = 0.0000444\ \text{K/W}$$

Total series resistance:

$$R_{tot} = 0.033333 + 0.133333 + 0.0000444 = 0.166711\ \text{K/W}$$

Heat loss:

$$\dot{Q} = \frac{\Delta T}{R_{tot}} = \frac{1000 - 60}{0.166711} = \frac{940}{0.166711} = \mathbf{5638.7\ W \approx 5.64\ kW}$$

(c) Interface temperatures

The same $\dot{Q}$ flows through each layer, so $\Delta T_i = \dot{Q}\,R_i$.

After firebrick (interface 1–2):

$$\Delta T_1 = 5638.7 \times 0.033333 = 187.96\ \text{K}$$ $$T_{12} = 1000 - 187.96 = \mathbf{812.0^\circ C}$$

After insulating brick (interface 2–3):

$$\Delta T_2 = 5638.7 \times 0.133333 = 751.83\ \text{K}$$ $$T_{23} = 812.0 - 751.83 = \mathbf{60.2^\circ C}$$

Across steel plate (check outer surface):

$$\Delta T_3 = 5638.7 \times 0.0000444 = 0.25\ \text{K}$$ $$T_{out} = 60.2 - 0.25 = 59.9 \approx 60^\circ C \;\checkmark$$

The insulating brick carries almost the entire temperature drop, while the steel plate is essentially isothermal because its resistance is negligible.

(a) Convection vs radiation

• Convection is heat transfer between a solid surface and an adjacent moving fluid; it requires a medium and is driven by bulk fluid motion combined with conduction at the surface. It is modelled by Newton's law of cooling, $\dot{Q} = hA(T_s - T_\infty)$, and depends linearly on the temperature difference.
• Radiation is energy emitted by all bodies as electromagnetic waves; it requires no medium and can occur through vacuum. It is governed by the Stefan–Boltzmann law and depends on the fourth power of absolute temperature, $\dot{Q} = \varepsilon\sigma A(T_s^4 - T_{surr}^4)$.

(b) Convective heat loss

Surface area of the pipe:

$$A = \pi D L = \pi \times 0.10 \times 8 = 2.5133\ \text{m}^2$$

Newton's law of cooling ($T_s = 180^\circ\text{C}$, $T_\infty = 25^\circ\text{C}$, $\Delta T = 155\ \text{K}$):

$$\dot{Q}_{conv} = hA(T_s - T_\infty) = 15 \times 2.5133 \times 155$$ $$\dot{Q}_{conv} = 15 \times 2.5133 \times 155 = \mathbf{5843.4\ W \approx 5.84\ kW}$$

(c) Radiative heat loss and total

Convert temperatures to absolute:

$$T_s = 180 + 273 = 453\ \text{K},\qquad T_{surr} = 25 + 273 = 298\ \text{K}$$

Fourth powers:

$$T_s^4 = 453^4 = 4.2120\times10^{10}\ \text{K}^4$$ $$T_{surr}^4 = 298^4 = 7.8862\times10^{9}\ \text{K}^4$$ $$T_s^4 - T_{surr}^4 = 4.2120\times10^{10} - 0.78862\times10^{10} = 3.4234\times10^{10}\ \text{K}^4$$

Stefan–Boltzmann law:

$$\dot{Q}_{rad} = \varepsilon\sigma A (T_s^4 - T_{surr}^4)$$ $$\dot{Q}_{rad} = 0.85 \times 5.67\times10^{-8} \times 2.5133 \times 3.4234\times10^{10}$$

Step by step:

$$0.85 \times 5.67\times10^{-8} = 4.8195\times10^{-8}$$ $$4.8195\times10^{-8} \times 2.5133 = 1.2113\times10^{-7}$$ $$1.2113\times10^{-7} \times 3.4234\times10^{10} = \mathbf{4146.6\ W \approx 4.15\ kW}$$

Total heat loss:

$$\dot{Q}_{total} = \dot{Q}_{conv} + \dot{Q}_{rad} = 5843.4 + 4146.6 = \mathbf{9990\ W \approx 9.99\ kW}$$

Convection and radiation are of comparable magnitude here, so radiation cannot be neglected at this surface temperature.

(a) Zeroth Law

Statement: If two bodies are each in thermal equilibrium with a third body, then they are in thermal equilibrium with one another.

Basis for temperature measurement: The law implies that all bodies in mutual thermal equilibrium share a common property — temperature. A thermometer acts as the "third body": when it reaches equilibrium with an object, its reading (e.g., mercury column height) corresponds to the object's temperature. Because equilibrium is transitive, two objects giving the same thermometer reading are at the same temperature without direct contact. This makes consistent, comparable temperature measurement possible.

(b) Distinctions

(i) Intensive vs extensive properties:

• Intensive properties are independent of the amount/mass of substance — e.g., temperature, pressure, density, specific volume.
• Extensive properties depend on the mass/extent of the system — e.g., volume, total internal energy, mass, total enthalpy.
• An extensive property divided by mass becomes a specific (intensive) property.

(ii) Closed vs open system:

• A closed system (control mass) allows energy (heat, work) to cross its boundary but no mass. Example: gas sealed in a piston–cylinder device.
• An open system (control volume) allows both mass and energy to cross its boundary. Example: a turbine, compressor, or boiler with flowing fluid.

(a) Work done (constant pressure)

For a constant-pressure process the boundary work is:

$$W = p(V_2 - V_1) = mR(T_2 - T_1)$$ $$W = 2 \times 0.297 \times (500 - 300) = 2 \times 0.297 \times 200 = \mathbf{118.8\ kJ}$$

(b) Heat added

At constant pressure $Q = m c_p (T_2 - T_1)$:

$$Q = 2 \times 1.040 \times (500 - 300) = 2 \times 1.040 \times 200 = \mathbf{416.0\ kJ}$$

(c) Change in internal energy

$$\Delta U = m c_v (T_2 - T_1) = 2 \times 0.743 \times 200 = \mathbf{297.2\ kJ}$$

Check via First Law: $Q - W = 416.0 - 118.8 = 297.2\ \text{kJ} = \Delta U$ ✓. Also $c_p - c_v = 1.040 - 0.743 = 0.297 = R$ ✓.

(a) Entropy and the increase principle

Entropy is a thermodynamic property defined for a reversible process by $dS = \left(\dfrac{\delta Q}{T}\right)_{rev}$. It is a measure of molecular disorder / the unavailability of energy for work.

Principle of increase of entropy: For any process of an isolated system (or a system plus its surroundings), the total entropy can never decrease:

$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} \ge 0$$

Equality holds only for reversible processes; all real (irreversible) processes generate entropy.

(b) Iron block dropped into the lake

The iron cools from $T_1 = 200 + 273 = 473\ \text{K}$ to the lake temperature $T_2 = 15 + 273 = 288\ \text{K}$.

Entropy change of the iron (solid, constant $c$):

$$\Delta S_{iron} = mc\ln\frac{T_2}{T_1} = 5 \times 0.45 \times \ln\frac{288}{473}$$ $$= 2.25 \times \ln(0.60888) = 2.25 \times (-0.49606) = \mathbf{-1.1161\ kJ/K}$$

Heat received by the lake: all the heat lost by the iron flows to the lake.

$$Q = mc(T_1 - T_2) = 5 \times 0.45 \times (473 - 288) = 2.25 \times 185 = 416.25\ \text{kJ}$$

The lake is a huge reservoir at constant $T = 288\ \text{K}$:

$$\Delta S_{lake} = \frac{+Q}{T_{lake}} = \frac{416.25}{288} = \mathbf{+1.4453\ kJ/K}$$

Total entropy generated:

$$\Delta S_{gen} = \Delta S_{iron} + \Delta S_{lake} = -1.1161 + 1.4453 = \mathbf{+0.3292\ kJ/K}$$

Comment: $\Delta S_{gen} > 0$, confirming the process is irreversible (spontaneous cooling across a finite temperature difference). The total entropy of the universe increases, consistent with the Second Law.

(a) Overall heat-transfer coefficient

For a plane wall with convection on both sides, the resistances per unit area add:

$$\frac{1}{U} = \frac{1}{h_i} + \frac{L}{k} + \frac{1}{h_o}$$

Substitute ($L = 6\ \text{mm} = 0.006\ \text{m}$):

$$\frac{1}{h_i} = \frac{1}{600} = 0.0016667\ \text{m}^2\text{K/W}$$ $$\frac{L}{k} = \frac{0.006}{50} = 0.00012\ \text{m}^2\text{K/W}$$ $$\frac{1}{h_o} = \frac{1}{25} = 0.04\ \text{m}^2\text{K/W}$$ $$\frac{1}{U} = 0.0016667 + 0.00012 + 0.04 = 0.0417867\ \text{m}^2\text{K/W}$$ $$U = \frac{1}{0.0417867} = \mathbf{23.93\ W/m^2\cdot K}$$

Note the air-side film dominates the resistance; the metal wall is almost negligible.

(b) Rate of heat transfer

$$\dot{Q} = U A (T_{hot} - T_{cold}) = 23.93 \times 4 \times (80 - 20)$$ $$\dot{Q} = 23.93 \times 4 \times 60 = \mathbf{5743\ W \approx 5.74\ kW}$$

(a) Steady Flow Energy Equation

The general SFEE per unit mass between inlet (1) and outlet (2):

$$q - w = \left(h_2 - h_1\right) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)$$

For the turbine: adiabatic ($q = 0$) and same elevation ($z_1 = z_2$). The shaft work per unit mass becomes:

$$w = (h_1 - h_2) + \frac{V_1^2 - V_2^2}{2}$$

(b) Power output

Enthalpy term:

$$h_1 - h_2 = 3200 - 2500 = 700\ \text{kJ/kg}$$

Kinetic-energy term (convert to kJ/kg by dividing by 1000):

$$\frac{V_1^2 - V_2^2}{2} = \frac{40^2 - 120^2}{2} = \frac{1600 - 14400}{2} = \frac{-12800}{2} = -6400\ \text{J/kg} = -6.4\ \text{kJ/kg}$$

Specific work:

$$w = 700 + (-6.4) = 693.6\ \text{kJ/kg}$$

Power output:

$$\dot{W} = \dot{m}\,w = 6 \times 693.6 = \mathbf{4161.6\ kW \approx 4.16\ MW}$$

The kinetic-energy change is small (it slightly reduces the output because the steam accelerates), so the enthalpy drop dominates the turbine power.

(a) Definitions and laws

Black body: An ideal surface that absorbs all incident radiation (absorptivity = 1, reflectivity = 0) at every wavelength and direction, and emits the maximum possible radiation at a given temperature. It is the perfect emitter and absorber.

Stefan–Boltzmann law: The total emissive power of a black body is proportional to the fourth power of its absolute temperature:

$$E_b = \sigma T^4$$

For a real surface $E = \varepsilon\sigma T^4$.

Wien's displacement law: The wavelength of maximum spectral emissive power is inversely proportional to absolute temperature:

$$\lambda_{max}\,T = 2898\ \mu\text{m·K (constant)}$$

(b) Net radiation between two large parallel plates

For two large (infinite) parallel grey plates, the net heat exchange per unit area is:

$$\frac{\dot{Q}_{12}}{A} = \frac{\sigma(T_1^4 - T_2^4)}{\dfrac{1}{\varepsilon_1} + \dfrac{1}{\varepsilon_2} - 1}$$

Fourth powers:

$$T_1^4 = 800^4 = 4.096\times10^{11}\ \text{K}^4$$ $$T_2^4 = 500^4 = 6.25\times10^{10}\ \text{K}^4$$ $$T_1^4 - T_2^4 = 4.096\times10^{11} - 0.625\times10^{11} = 3.471\times10^{11}\ \text{K}^4$$

Numerator:

$$\sigma(T_1^4 - T_2^4) = 5.67\times10^{-8} \times 3.471\times10^{11} = 19680.6\ \text{W/m}^2$$

Emissivity denominator:

$$\frac{1}{0.6} + \frac{1}{0.8} - 1 = 1.6667 + 1.25 - 1 = 1.9167$$

Net exchange:

$$\frac{\dot{Q}_{12}}{A} = \frac{19680.6}{1.9167} = \mathbf{10268\ W/m^2 \approx 10.27\ kW/m^2}$$

Heat flows from the hotter plate (800 K) to the cooler plate (500 K).