BE Civil Engineering (IOE, TU) Fundamentals of Thermodynamics & Heat Transfer (IOE, ME 451) Question Paper 2076 Nepal

First Law statements

• For a system undergoing a cycle: $\oint \delta Q = \oint \delta W$ — the cyclic integral of heat transfer equals the cyclic integral of work.
• For a process (change of state): $Q - W = \Delta U$, i.e. energy is conserved; heat added minus work done by the system equals the change in internal energy.

Given: $m = 0.5\ \text{kg}$, $p_1 = 300\ \text{kPa}$, $T_1 = 350\ \text{K}$, $p_2 = 900\ \text{kPa}$, $n = 1.3$, $R = 0.287\ \text{kJ/kg·K}$, $c_v = 0.718\ \text{kJ/kg·K}$.

(a) Final temperature

For a polytropic process:

$$\frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}$$ $$\frac{T_2}{T_1} = \left(\frac{900}{300}\right)^{\frac{0.3}{1.3}} = 3^{0.23077}$$ $$3^{0.23077} = e^{0.23077 \times \ln 3} = e^{0.23077 \times 1.09861} = e^{0.25353} = 1.2886$$ $$T_2 = 350 \times 1.2886 = \mathbf{451.0\ K}$$

(b) Boundary work

For a polytropic process the work is

$$W = \frac{mR(T_1 - T_2)}{n - 1} = \frac{0.5 \times 0.287 \times (350 - 451.0)}{1.3 - 1}$$ $$W = \frac{0.5 \times 0.287 \times (-101.0)}{0.3} = \frac{-14.4935}{0.3} = \mathbf{-48.31\ kJ}$$

The negative sign confirms work is done on the gas (compression).

(c) Heat transfer

$$\Delta U = m c_v (T_2 - T_1) = 0.5 \times 0.718 \times (451.0 - 350) = 0.5 \times 0.718 \times 101.0 = 36.26\ \text{kJ}$$ $$Q = \Delta U + W = 36.26 + (-48.31) = \mathbf{-12.05\ kJ}$$

The negative sign means heat is rejected from the gas to the surroundings during compression.

Second Law statements

• Kelvin–Planck: It is impossible to construct a device that, operating in a cycle, produces no effect other than the extraction of heat from a single reservoir and the production of an equivalent amount of work. (No heat engine can have 100% efficiency.)
• Clausius: It is impossible to construct a device that, operating in a cycle, transfers heat from a cooler body to a hotter body without any external work input.

Equivalence: A violation of one leads to a violation of the other. If a Kelvin–Planck-violating engine existed, its work could drive a refrigerator, the combination transferring heat from cold to hot with no net work — violating Clausius, and vice versa.

Given: $T_H = 800\ \text{K}$, $T_L = 320\ \text{K}$, $\dot{W}_{net} = 40\ \text{kW}$.

(a) Thermal efficiency

$$\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{320}{800} = 1 - 0.40 = \mathbf{0.60\ (60\%)}$$

(b) Heat supplied

$$\dot{Q}_H = \frac{\dot{W}_{net}}{\eta} = \frac{40}{0.60} = \mathbf{66.67\ kW}$$

(c) Heat rejected

$$\dot{Q}_L = \dot{Q}_H - \dot{W}_{net} = 66.67 - 40 = \mathbf{26.67\ kW}$$

(d) Entropy rates

$$\dot{S}_{source} = -\frac{\dot{Q}_H}{T_H} = -\frac{66.67}{800} = -0.08333\ \text{kW/K}$$ $$\dot{S}_{sink} = +\frac{\dot{Q}_L}{T_L} = +\frac{26.67}{320} = +0.08333\ \text{kW/K}$$ $$\dot{S}_{universe} = -0.08333 + 0.08333 = \mathbf{0\ kW/K}$$

Comment: For the reversible (Carnot) cycle the net entropy change of the universe is zero, as expected for a fully reversible process. Any real engine would show $\dot{S}_{universe} > 0$.

Definitions

• Saturation temperature: the temperature at which a pure substance changes phase (boils/condenses) at a given pressure.
• Dryness fraction (quality), $x$: the ratio of the mass of vapour to the total mass of a liquid–vapour mixture, $x = m_g/(m_f + m_g)$.
• Degree of superheat: the amount by which the actual temperature of a superheated vapour exceeds the saturation temperature at the same pressure, $T - T_{sat}$.

Given: $V = 0.15\ \text{m}^3$, $p = 200\ \text{kPa}$, $x = 0.85$.

(a) Mass of steam

Specific volume of the mixture:

$$v = v_f + x\,(v_g - v_f) = 0.001061 + 0.85\,(0.8857 - 0.001061)$$ $$v = 0.001061 + 0.85 \times 0.884639 = 0.001061 + 0.751943 = 0.753004\ \text{m}^3/\text{kg}$$ $$m = \frac{V}{v} = \frac{0.15}{0.753004} = \mathbf{0.1992\ kg}$$

(b) Specific internal energy

$$u = u_f + x\,u_{fg} = 504.5 + 0.85 \times 2025.0 = 504.5 + 1721.25 = \mathbf{2225.75\ kJ/kg}$$

(c) Total internal energy

$$U = m\,u = 0.1992 \times 2225.75 = \mathbf{443.4\ kJ}$$

Derivation (electrical analogy)

For steady 1-D conduction through a single slab, Fourier's law gives

$$\dot{Q} = \frac{kA\,\Delta T}{L} = \frac{\Delta T}{R}, \qquad R = \frac{L}{kA}$$

where $R$ is the conductive thermal resistance (analogous to electrical resistance, with $\dot{Q}\leftrightarrow I$ and $\Delta T \leftrightarrow V$). For layers in series carrying the same $\dot{Q}$, resistances add:

$$\dot{Q} = \frac{T_{in} - T_{out}}{R_1 + R_2 + R_3}, \qquad R_i = \frac{L_i}{k_i A}$$

Given (per unit area, $A = 1\ \text{m}^2$): $T_{in} = 1000\ ^\circ\text{C}$, $T_{out} = 50\ ^\circ\text{C}$.

Individual resistances:

$$R_1 = \frac{0.20}{1.04 \times 1} = 0.19231\ \text{K/W}$$ $$R_2 = \frac{0.10}{0.15 \times 1} = 0.66667\ \text{K/W}$$ $$R_3 = \frac{0.01}{45 \times 1} = 0.000222\ \text{K/W}$$ $$R_{total} = 0.19231 + 0.66667 + 0.000222 = 0.85920\ \text{K/W}$$

(a) Heat flux

$$q = \frac{T_{in} - T_{out}}{R_{total}} = \frac{1000 - 50}{0.85920} = \frac{950}{0.85920} = \mathbf{1105.7\ W/m^2}$$

(b) Interface temperature (firebrick–insulating brick)

Applying the same $q$ across the first layer:

$$T_{12} = T_{in} - q\,R_1 = 1000 - 1105.7 \times 0.19231 = 1000 - 212.6 = \mathbf{787.4\ ^\circ C}$$

Convection modes: In forced convection fluid motion is driven by an external agent (pump, fan, wind); in free (natural) convection fluid motion arises from buoyancy due to density differences caused by temperature gradients. Both obey Newton's law of cooling, $\dot{Q} = hA(T_s - T_\infty)$.

Radiation law (small grey body in large surroundings):

$$\dot{Q}_{rad} = \varepsilon\,\sigma\,A\,(T_s^4 - T_{sur}^4)$$

with absolute temperatures.

Given: $D = 0.10\ \text{m}$, $L = 8\ \text{m}$, $T_s = 180\ ^\circ\text{C} = 453.15\ \text{K}$, $T_\infty = T_{sur} = 25\ ^\circ\text{C} = 298.15\ \text{K}$, $h = 12\ \text{W/m}^2\text{K}$, $\varepsilon = 0.85$.

Surface area

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

Convective loss

$$\dot{Q}_{conv} = hA(T_s - T_\infty) = 12 \times 2.5133 \times (180 - 25) = 12 \times 2.5133 \times 155 = \mathbf{4674.7\ W}$$

Radiative loss

$$T_s^4 = 453.15^4 = 4.2161\times10^{10}\ \text{K}^4$$ $$T_{sur}^4 = 298.15^4 = 0.79003\times10^{10}\ \text{K}^4$$ $$\dot{Q}_{rad} = 0.85 \times 5.67\times10^{-8} \times 2.5133 \times (4.2161 - 0.79003)\times10^{10}$$ $$= 0.85 \times 5.67\times10^{-8} \times 2.5133 \times 3.4261\times10^{10}$$ $$= 0.85 \times 5.67\times10^{-8} \times 8.6113\times10^{10} = 0.85 \times 4882.6 = \mathbf{4150.2\ W}$$

Total heat loss

$$\dot{Q}_{total} = \dot{Q}_{conv} + \dot{Q}_{rad} = 4674.7 + 4150.2 = \mathbf{8824.9\ W \approx 8.82\ kW}$$

Given: $m = 2\ \text{kg}$, $T = 400\ \text{K}$, $p_1 = 500\ \text{kPa}$, $p_2 = 150\ \text{kPa}$, $R = 0.287\ \text{kJ/kg·K}$.

Why $\Delta U = 0$: For an ideal gas, internal energy depends only on temperature. The process is isothermal ($T$ constant), so $\Delta U = m c_v \Delta T = 0$.

(a) Work done (isothermal, reversible)

$$W = mRT\,\ln\!\frac{p_1}{p_2} = 2 \times 0.287 \times 400 \times \ln\!\frac{500}{150}$$ $$\ln\!\frac{500}{150} = \ln 3.3333 = 1.20397$$ $$W = 229.6 \times 1.20397 = \mathbf{276.4\ kJ}$$

(b) Heat transfer

From the First Law, $Q = \Delta U + W = 0 + 276.4 = \mathbf{276.4\ kJ}$ (heat added to the gas).

SFEE (per unit mass), one inlet–one outlet:

$$h_1 + \frac{V_1^2}{2} + gz_1 + q = h_2 + \frac{V_2^2}{2} + gz_2 + w$$

where $h$ = specific enthalpy, $V^2/2$ = specific kinetic energy, $gz$ = specific potential energy, $q$ = heat added per unit mass, $w$ = shaft work per unit mass.

Given (nozzle): adiabatic ($q = 0$), no shaft work ($w = 0$), $V_1 \approx 0$, $\Delta z = 0$, $T_1 = 500\ \text{K}$, $T_2 = 380\ \text{K}$, $c_p = 1.005\ \text{kJ/kg·K}$.

The SFEE reduces to

$$h_1 = h_2 + \frac{V_2^2}{2} \implies \frac{V_2^2}{2} = h_1 - h_2 = c_p(T_1 - T_2)$$ $$\frac{V_2^2}{2} = 1.005 \times (500 - 380) = 1.005 \times 120 = 120.6\ \text{kJ/kg} = 120600\ \text{J/kg}$$ $$V_2 = \sqrt{2 \times 120600} = \sqrt{241200} = \mathbf{491.1\ m/s}$$

Distinction: A refrigerator removes heat from a low-temperature space (useful effect = $\dot{Q}_L$, cooling). A heat pump delivers heat to a high-temperature space (useful effect = $\dot{Q}_H$, heating). They are the same cycle; only the desired output differs.

Given: $T_L = -8\ ^\circ\text{C} = 265.15\ \text{K}$, $T_H = 27\ ^\circ\text{C} = 300.15\ \text{K}$, $\dot{Q}_L = 3\ \text{kW}$.

(a) COP of refrigerator (Carnot)

$$\text{COP}_R = \frac{T_L}{T_H - T_L} = \frac{265.15}{300.15 - 265.15} = \frac{265.15}{35} = \mathbf{7.576}$$

(b) Power input

$$\dot{W} = \frac{\dot{Q}_L}{\text{COP}_R} = \frac{3}{7.576} = \mathbf{0.396\ kW}$$

(c) COP as heat pump

$$\text{COP}_{HP} = \text{COP}_R + 1 = 7.576 + 1 = \mathbf{8.576}$$

(Check: $\text{COP}_{HP} = T_H/(T_H - T_L) = 300.15/35 = 8.576$.)

Derivation (radial conduction, hollow cylinder)

For steady 1-D radial conduction with no heat generation, Fourier's law on a cylindrical shell of length $L$ at radius $r$:

$$\dot{Q} = -k\,(2\pi r L)\frac{dT}{dr}$$

Since $\dot{Q}$ is constant, separate and integrate from $r_1$ to $r_2$:

$$\dot{Q}\int_{r_1}^{r_2}\frac{dr}{r} = -2\pi k L \int_{T_1}^{T_2} dT$$ $$\dot{Q}\,\ln\!\frac{r_2}{r_1} = 2\pi k L (T_1 - T_2)$$ $$\boxed{\dot{Q} = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)}}$$

Given: $r_1 = 0.025\ \text{m}$, $r_2 = 0.035\ \text{m}$, $L = 5\ \text{m}$, $k = 50\ \text{W/m·K}$, $T_1 = 120\ ^\circ\text{C}$, $T_2 = 80\ ^\circ\text{C}$.

$$\ln\!\frac{r_2}{r_1} = \ln\!\frac{0.035}{0.025} = \ln 1.4 = 0.33647$$ $$\dot{Q} = \frac{2\pi \times 50 \times 5 \times (120 - 80)}{0.33647} = \frac{2\pi \times 50 \times 5 \times 40}{0.33647}$$ $$= \frac{62831.85}{0.33647} = \mathbf{186{,}740\ W \approx 186.7\ kW}$$

Given: $m = 1.5\ \text{kg}$, $p = 250\ \text{kPa}$ (constant), $T_1 = 300\ \text{K}$, $T_2 = 560\ \text{K}$, $R = 0.297$, $c_p = 1.040$, $c_v = 0.743\ \text{kJ/kg·K}$.

(a) Heat added (constant pressure)

$$Q = m c_p (T_2 - T_1) = 1.5 \times 1.040 \times (560 - 300) = 1.5 \times 1.040 \times 260 = \mathbf{405.6\ kJ}$$

(b) Work done (constant pressure)

$$W = p\,\Delta V = mR(T_2 - T_1) = 1.5 \times 0.297 \times 260 = \mathbf{115.83\ kJ}$$

(c) Change in internal energy

$$\Delta U = m c_v (T_2 - T_1) = 1.5 \times 0.743 \times 260 = \mathbf{289.77\ kJ}$$

Check (First Law): $Q = \Delta U + W = 289.77 + 115.83 = 405.6\ \text{kJ}$ ✓

Stefan–Boltzmann law: The total emissive power of a black body is proportional to the fourth power of its absolute temperature, $E_b = \sigma T^4$.

• Emissivity ($\varepsilon$): ratio of the radiation emitted by a surface to that emitted by a black body at the same temperature ($0 \le \varepsilon \le 1$).
• Black body: an ideal surface that absorbs all incident radiation and emits the maximum possible at every wavelength ($\varepsilon = 1$).
• Grey body: a surface whose emissivity is constant (less than 1) and independent of wavelength.

Given: $T_s = 727\ ^\circ\text{C} = 1000.15\ \text{K}$, $T_{sur} = 27\ ^\circ\text{C} = 300.15\ \text{K}$. (Using rounded absolute values $T_s = 1000\ \text{K}$, $T_{sur} = 300\ \text{K}$.)

$$T_s^4 = 1000^4 = 1.000\times10^{12}\ \text{K}^4$$ $$T_{sur}^4 = 300^4 = 8.1\times10^{9}\ \text{K}^4$$

(a) Black body net flux ($\varepsilon = 1$)

$$q = \sigma (T_s^4 - T_{sur}^4) = 5.67\times10^{-8}\,(1.000\times10^{12} - 0.0081\times10^{12})$$ $$= 5.67\times10^{-8} \times 0.9919\times10^{12} = 5.67\times10^{-8} \times 9.919\times10^{11} = \mathbf{56{,}240\ W/m^2 \approx 56.24\ kW/m^2}$$

(b) Grey body net flux ($\varepsilon = 0.6$)

$$q = \varepsilon\,\sigma (T_s^4 - T_{sur}^4) = 0.6 \times 56{,}240 = \mathbf{33{,}744\ W/m^2 \approx 33.74\ kW/m^2}$$