BE Civil Engineering (IOE, TU) Fundamentals of Thermodynamics & Heat Transfer (IOE, ME 451) Question Paper 2078 Nepal

First Law statements

For a closed system undergoing a cycle: $\oint \delta Q = \oint \delta W$ — the net heat transfer equals the net work transfer over the cycle.

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

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

Energy is conserved; the difference between heat added and work done equals the change in internal energy.

Given: $m = 0.25\ \text{kg}$, $p_1 = 150\ \text{kPa}$, $T_1 = 300\ \text{K}$, $p_2 = 750\ \text{kPa}$, $n = 1.3$.

Initial volume

$$V_1 = \frac{mRT_1}{p_1} = \frac{0.25 \times 0.287 \times 300}{150} = \frac{21.525}{150} = 0.14350\ \text{m}^3$$

(a) Final temperature and volume

For a polytropic process:

$$\frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}$$ $$\frac{p_2}{p_1} = \frac{750}{150} = 5, \qquad \frac{n-1}{n} = \frac{0.3}{1.3} = 0.23077$$ $$T_2 = 300 \times 5^{0.23077} = 300 \times 1.4452 = \mathbf{433.6\ K}$$

Final volume:

$$V_2 = \frac{mRT_2}{p_2} = \frac{0.25 \times 0.287 \times 433.6}{750} = \frac{31.11}{750} = \mathbf{0.04148\ m^3}$$

(Check via $V_2 = V_1 (p_1/p_2)^{1/n} = 0.1435 \times 5^{-1/1.3} = 0.1435 \times 0.2890 = 0.04148\ \text{m}^3$ ✓)

(b) Polytropic work

$$W_{1-2} = \frac{p_2 V_2 - p_1 V_1}{1-n} = \frac{(750 \times 0.04148) - (150 \times 0.14350)}{1 - 1.3}$$ $$= \frac{31.11 - 21.525}{-0.3} = \frac{9.585}{-0.3} = -31.95\ \text{kJ}$$

The negative sign shows work is done on the air. $\mathbf{W_{1-2} = -31.95\ kJ}$ (i.e. $31.95\ \text{kJ}$ done on the gas).

(c) Heat transfer

$$\Delta U = m c_v (T_2 - T_1) = 0.25 \times 0.718 \times (433.6 - 300) = 0.25 \times 0.718 \times 133.6 = 23.98\ \text{kJ}$$ $$Q_{1-2} = \Delta U + W_{1-2} = 23.98 + (-31.95) = -7.97\ \text{kJ}$$

$\mathbf{Q_{1-2} = -7.97\ kJ}$. The negative sign means heat is rejected from the air to the surroundings.

(a) Second Law statements

Kelvin-Planck: It is impossible to construct a device operating in a cycle that produces no effect other than the production of work and the exchange of heat with a single reservoir. (No engine can be 100% efficient.)

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 body to a hotter body. (Heat cannot spontaneously flow from cold to hot.)

Equivalence: Suppose a device V violates Clausius, transferring $Q$ from the cold to the hot reservoir with no work input. Combine V with an ordinary engine E that draws $Q_H$ from the hot reservoir, rejects $Q$ to the cold reservoir, and produces work $W = Q_H - Q$. The cold reservoir then sees zero net heat exchange, so the composite device draws net heat $(Q_H - Q)$ from the hot reservoir alone and converts it entirely to work — a violation of Kelvin-Planck. The reverse argument holds similarly, proving the two statements are equivalent.

(b) Carnot engine

Given: $T_H = 700\ \text{K}$, $T_L = 310\ \text{K}$, $\dot W_{net} = 40\ \text{kW}$.

(i) Thermal efficiency:

$$\eta_{th} = 1 - \frac{T_L}{T_H} = 1 - \frac{310}{700} = 1 - 0.4429 = 0.5571 = \mathbf{55.71\%}$$

(ii) Heat supplied:

$$\dot Q_H = \frac{\dot W_{net}}{\eta_{th}} = \frac{40}{0.5571} = \mathbf{71.80\ kW}$$

(iii) Heat rejected:

$$\dot Q_L = \dot Q_H - \dot W_{net} = 71.80 - 40 = \mathbf{31.80\ kW}$$

(c) Net entropy change rate (source + sink)

Source loses heat at $T_H$:

$$\dot S_{source} = -\frac{\dot Q_H}{T_H} = -\frac{71.80}{700} = -0.10257\ \text{kW/K}$$

Sink gains heat at $T_L$:

$$\dot S_{sink} = +\frac{\dot Q_L}{T_L} = +\frac{31.80}{310} = +0.10258\ \text{kW/K}$$ $$\dot S_{net} = -0.10257 + 0.10258 \approx \mathbf{0\ kW/K}$$

Comment: The net entropy change of the reservoirs is zero (within rounding), confirming the Carnot cycle is reversible — there is no entropy generation. Any real (irreversible) engine between the same reservoirs would produce $\dot S_{net} > 0$.

Given: $V = 0.5\ \text{m}^3$ (rigid), $p_1 = 200\ \text{kPa}$, $x_1 = 0.6$, $p_2 = 800\ \text{kPa}$.

State 1 — specific volume

$$v_1 = v_f + x_1(v_g - v_f) = 0.001061 + 0.6(0.8857 - 0.001061)$$ $$= 0.001061 + 0.6 \times 0.884639 = 0.001061 + 0.530783 = 0.531844\ \text{m}^3/\text{kg}$$

(a) Mass of steam

$$m = \frac{V}{v_1} = \frac{0.5}{0.531844} = \mathbf{0.9401\ kg}$$

(b) Final dryness fraction

The tank is rigid and closed, so specific volume is constant: $v_2 = v_1 = 0.531844\ \text{m}^3/\text{kg}$. At $800\ \text{kPa}$:

$$x_2 = \frac{v_2 - v_f}{v_g - v_f} = \frac{0.531844 - 0.001115}{0.2404 - 0.001115} = \frac{0.530729}{0.239285} = \mathbf{2.218}$$

Since $x_2 > 1$, the final state is superheated (not a wet mixture). The mixture becomes fully dry and superheated before reaching $800\ \text{kPa}$. For this paper we treat the limiting saturated-vapour internal energy and note the state is superheated; using $u_2 \approx u_g$ at $800\ \text{kPa}$ as a conservative table-only estimate.

(c) Heat transfer

Internal energy at state 1:

$$u_1 = u_f + x_1(u_g - u_f) = 504.5 + 0.6(2529.5 - 504.5) = 504.5 + 0.6 \times 2025 = 504.5 + 1215 = 1719.5\ \text{kJ/kg}$$

For a rigid tank, boundary work $W = 0$, so by the First Law:

$$Q = m(u_2 - u_1)$$

Using the table-limited estimate $u_2 \approx u_g(800\ \text{kPa}) = 2576.8\ \text{kJ/kg}$:

$$Q = 0.9401 \times (2576.8 - 1719.5) = 0.9401 \times 857.3 = \mathbf{806.0\ kJ}$$

Heat is added to the system (positive).

Note: because the final state is actually superheated, $u_2$ would be slightly larger than $u_g$ and the exact $Q$ a little higher; with only saturation data the above is the best available estimate. The key learning point is recognising that constant-$v$ heating of wet steam can carry the fluid into the superheated region.

Resistance network (per $1\ \text{m}^2$)

 T_o ──R_conv,o── T1 ──R_brick── T2 ──R_ins── T3 ──R_plaster── T4 ──R_conv,i── T_i
     (outside)                                                          (inside)

(a) Individual resistances (units $\text{m}^2\text{K/W}$, area $=1\ \text{m}^2$):

$$R_{conv,o} = \frac{1}{h_o} = \frac{1}{20} = 0.05000$$ $$R_{brick} = \frac{L_1}{k_1} = \frac{0.20}{0.70} = 0.28571$$ $$R_{ins} = \frac{L_2}{k_2} = \frac{0.08}{0.025} = 3.20000$$ $$R_{plaster} = \frac{L_3}{k_3} = \frac{0.015}{0.50} = 0.03000$$ $$R_{conv,i} = \frac{1}{h_i} = \frac{1}{8} = 0.12500$$

Total resistance:

$$R_{tot} = 0.05000 + 0.28571 + 3.20000 + 0.03000 + 0.12500 = 3.69071\ \text{m}^2\text{K/W}$$

The insulation dominates (about 87% of total resistance), as expected.

(b) Heat gain per unit area

Driving temperature difference (heat flows from warm outside to cold inside):

$$\Delta T = T_o - T_i = 32 - (-5) = 37\ \text{K}$$ $$\dot q = \frac{\Delta T}{R_{tot}} = \frac{37}{3.69071} = \mathbf{10.03\ W/m^2}$$

(c) Brick-insulation interface temperature ($T_2$)

From outside air to the interface, the heat passes through $R_{conv,o}$ and $R_{brick}$:

$$T_2 = T_o - \dot q\,(R_{conv,o} + R_{brick}) = 32 - 10.03 \times (0.05000 + 0.28571)$$ $$= 32 - 10.03 \times 0.33571 = 32 - 3.367 = \mathbf{28.63\,^\circ C}$$

So most of the temperature drop occurs across the insulation, confirming its role in the cold-storage wall.

Given: $D = 0.080\ \text{m}$, $L = 12\ \text{m}$, $T_s = 180^\circ\text{C} = 453.15\ \text{K}$, $T_\infty = T_{surr} = 25^\circ\text{C} = 298.15\ \text{K}$.

Surface area

$$A = \pi D L = \pi \times 0.080 \times 12 = 3.01593\ \text{m}^2$$

(a) Convective heat loss

$$\dot Q_{conv} = h A (T_s - T_\infty) = 9.5 \times 3.01593 \times (180 - 25)$$ $$= 9.5 \times 3.01593 \times 155 = \mathbf{4441\ W} \approx 4.44\ \text{kW}$$

(b) Radiative heat loss

$$\dot Q_{rad} = \varepsilon \sigma A (T_s^4 - T_{surr}^4)$$ $$T_s^4 = (453.15)^4 = 4.2161\times10^{10}\ \text{K}^4$$ $$T_{surr}^4 = (298.15)^4 = 7.9006\times10^{9}\ \text{K}^4$$ $$T_s^4 - T_{surr}^4 = 4.2161\times10^{10} - 0.79006\times10^{10} = 3.4260\times10^{10}\ \text{K}^4$$ $$\dot Q_{rad} = 0.85 \times 5.67\times10^{-8} \times 3.01593 \times 3.4260\times10^{10}$$ $$= 0.85 \times 5.67\times10^{-8} \times 1.03326\times10^{11} = 0.85 \times 5858.6 = \mathbf{4980\ W} \approx 4.98\ \text{kW}$$

(c) Total heat loss and radiation coefficient

$$\dot Q_{total} = \dot Q_{conv} + \dot Q_{rad} = 4441 + 4980 = \mathbf{9421\ W} \approx 9.42\ \text{kW}$$

Equivalent radiation heat-transfer coefficient:

$$h_{rad} = \frac{\dot Q_{rad}}{A (T_s - T_\infty)} = \frac{4980}{3.01593 \times 155} = \frac{4980}{467.47} = \mathbf{10.65\ W/m^2K}$$

Note that $h_{rad} > h_{conv}$ here, so radiation is the larger loss mechanism at this surface temperature.

Given: $m_{N_2} = 0.4\ \text{kg}$, $m_{CO_2} = 0.6\ \text{kg}$, $T = 300\ \text{K}$, $p = 250\ \text{kPa}$.

Moles of each component

$$N_{N_2} = \frac{0.4}{28} = 0.014286\ \text{kmol}, \qquad N_{CO_2} = \frac{0.6}{44} = 0.013636\ \text{kmol}$$ $$N_{total} = 0.014286 + 0.013636 = 0.027922\ \text{kmol}$$

(a) Mole fractions and apparent molar mass

$$y_{N_2} = \frac{0.014286}{0.027922} = \mathbf{0.5117}, \qquad y_{CO_2} = \frac{0.013636}{0.027922} = \mathbf{0.4883}$$ $$M_m = y_{N_2}M_{N_2} + y_{CO_2}M_{CO_2} = 0.5117\times28 + 0.4883\times44$$ $$= 14.328 + 21.485 = \mathbf{35.81\ kg/kmol}$$

(Check: $M_m = m_{total}/N_{total} = 1.0/0.027922 = 35.81\ \text{kg/kmol}$ ✓)

(b) Partial pressures (Dalton's law: $p_i = y_i\,p$)

$$p_{N_2} = 0.5117 \times 250 = \mathbf{127.9\ kPa}$$ $$p_{CO_2} = 0.4883 \times 250 = \mathbf{122.1\ kPa}$$

(Sum $= 250\ \text{kPa}$ ✓)

(c) Volume of vessel

$$V = \frac{N_{total}\,\bar R\,T}{p} = \frac{0.027922 \times 8.314 \times 300}{250} = \frac{69.64}{250} = \mathbf{0.2786\ m^3}$$

(a) SFEE (per unit mass), single inlet/outlet:

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

For a nozzle: no shaft work ($w_s = 0$), adiabatic ($q = 0$), negligible elevation change ($\Delta z = 0$). It reduces to:

$$h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$$

(b) Exit velocity

For an ideal gas, $h_1 - h_2 = c_p(T_1 - T_2)$:

$$\frac{V_2^2 - V_1^2}{2} = c_p(T_1 - T_2)$$ $$\frac{V_2^2 - 30^2}{2} = 1005 \times (420 - 350) \quad [\text{J/kg}]$$ $$\frac{V_2^2 - 900}{2} = 1005 \times 70 = 70350$$ $$V_2^2 = 2\times70350 + 900 = 140700 + 900 = 141600$$ $$V_2 = \sqrt{141600} = \mathbf{376.3\ m/s}$$

(c) Mass flow rate

Inlet specific volume:

$$v_1 = \frac{R T_1}{p_1} = \frac{0.287 \times 420}{500} = \frac{120.54}{500} = 0.24108\ \text{m}^3/\text{kg}$$

Inlet area $A_1 = 90\ \text{cm}^2 = 90\times10^{-4}\ \text{m}^2 = 0.009\ \text{m}^2$.

$$\dot m = \frac{A_1 V_1}{v_1} = \frac{0.009 \times 30}{0.24108} = \frac{0.27}{0.24108} = \mathbf{1.120\ kg/s}$$

Given: $m_1 = m_2 = 3\ \text{kg}$, $T_1 = 80^\circ\text{C} = 353.15\ \text{K}$, $T_2 = 20^\circ\text{C} = 293.15\ \text{K}$, $c = 4.18\ \text{kJ/kg\cdot K}$.

(a) Final temperature

Energy balance (adiabatic, equal masses and $c$):

$$m c (T_f - T_1) + m c (T_f - T_2) = 0 \Rightarrow T_f = \frac{T_1 + T_2}{2}$$ $$T_f = \frac{353.15 + 293.15}{2} = \frac{646.30}{2} = 323.15\ \text{K} = \mathbf{50\,^\circ C}$$

(b) Entropy change of the universe

For a substance with constant $c$: $\Delta S = m c \ln(T_f/T_i)$.

Hot stream (cools $353.15 \to 323.15$):

$$\Delta S_{hot} = 3 \times 4.18 \times \ln\frac{323.15}{353.15} = 12.54 \times \ln(0.91505) = 12.54 \times (-0.08877) = -1.1132\ \text{kJ/K}$$

Cold stream (warms $293.15 \to 323.15$):

$$\Delta S_{cold} = 3 \times 4.18 \times \ln\frac{323.15}{293.15} = 12.54 \times \ln(1.10234) = 12.54 \times 0.097439 = 1.2219\ \text{kJ/K}$$

Net entropy change (container insulated, so surroundings unchanged):

$$\Delta S_{univ} = -1.1132 + 1.2219 = \mathbf{+0.1087\ kJ/K}$$

(c) Reversibility

Since $\Delta S_{univ} = +0.109\ \text{kJ/K} > 0$, entropy is generated. By the increase-of-entropy principle the process is irreversible — heat transfer across the finite temperature difference between the two water masses is an inherently irreversible (spontaneous) process.

(a) Derivation — radial conduction in a hollow cylinder

For steady, 1-D radial conduction with no heat generation, Fourier's law at radius $r$ (area $A = 2\pi r L$):

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

At steady state $\dot Q$ is constant. Separating variables:

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

Integrate from $r_1$ to $r_2$ and $T_1$ to $T_2$:

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

Equivalently, the conduction resistance is $R_{cyl} = \dfrac{\ln(r_2/r_1)}{2\pi k L}$.

(b) Numerical

Given: $r_1 = 0.010\ \text{m}$, $r_2 = 0.014\ \text{m}$, $L = 2.5\ \text{m}$, $T_1 - T_2 = 90 - 86 = 4\ \text{K}$, $k = 385\ \text{W/m\cdot K}$.

$$\ln\!\left(\frac{r_2}{r_1}\right) = \ln\!\left(\frac{0.014}{0.010}\right) = \ln(1.4) = 0.33647$$ $$\dot Q = \frac{2\pi \times 385 \times 2.5 \times 4}{0.33647} = \frac{24190.3}{0.33647} = \mathbf{71{,}894\ W} \approx 71.9\ \text{kW}$$

The very high conductivity of copper combined with the small log term gives a large heat-flow rate for only a $4^\circ\text{C}$ wall difference.

(a) Dimensionless numbers

Reynolds number $Re = \dfrac{\rho V L}{\mu} = \dfrac{V L}{\nu}$ — ratio of inertial to viscous forces; governs whether flow is laminar or turbulent.

Prandtl number $Pr = \dfrac{\mu c_p}{k} = \dfrac{\nu}{\alpha}$ — ratio of momentum diffusivity to thermal diffusivity; relates the velocity and thermal boundary-layer thicknesses (a fluid property).

Nusselt number $Nu = \dfrac{h L}{k}$ — ratio of convective to conductive heat transfer across the fluid layer; $Nu = 1$ means pure conduction, larger values indicate stronger convection.

(b) Convection coefficient and heat rate

From $\overline{Nu_L} = \dfrac{\bar h L}{k}$:

$$\bar h = \frac{\overline{Nu_L}\, k}{L} = \frac{235 \times 0.0285}{0.5} = \frac{6.6975}{0.5} = \mathbf{13.40\ W/m^2K}$$

Heat transfer per metre width ($A = L \times 1\ \text{m} = 0.5\ \text{m}^2$):

$$\dot Q = \bar h A (T_s - T_\infty) = 13.40 \times 0.5 \times (80 - 20) = 13.40 \times 0.5 \times 60 = \mathbf{401.9\ W per metre width}$$

(a) Definitions

Stefan-Boltzmann law: the total emissive power of a black body is $E_b = \sigma T^4$, where $\sigma = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$ and $T$ is absolute temperature.

Black body: an ideal surface that absorbs all incident radiation (absorptivity $=1$) and emits the maximum possible radiation at every wavelength and temperature.

Emissivity $\varepsilon$: the ratio of the radiation emitted by a real surface to that emitted by a black body at the same temperature ($0 \le \varepsilon \le 1$).

(b) Net exchange between large parallel plates

For two large parallel plates (view factor $=1$), the net heat exchange per unit area is:

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

Compute the temperature terms:

$$T_1^4 = (800)^4 = 4.0960\times10^{11}\ \text{K}^4, \quad T_2^4 = (500)^4 = 6.250\times10^{10}\ \text{K}^4$$ $$T_1^4 - T_2^4 = 4.0960\times10^{11} - 0.6250\times10^{11} = 3.4710\times10^{11}\ \text{K}^4$$ $$\sigma(T_1^4 - T_2^4) = 5.67\times10^{-8} \times 3.4710\times10^{11} = 19680.6\ \text{W/m}^2$$

Denominator:

$$\frac{1}{0.6} + \frac{1}{0.4} - 1 = 1.6667 + 2.5000 - 1 = 3.1667$$ $$\dot q_{12} = \frac{19680.6}{3.1667} = \mathbf{6215\ W/m^2} \approx 6.22\ \text{kW/m}^2$$

(c) Effect of a radiation shield

Inserting a low-emissivity shield ($\varepsilon_s = 0.05$) adds two more high-resistance radiation surfaces in the path. The total surface-resistance term grows greatly (each shield face contributes $\frac{2}{\varepsilon_s} - 2 \approx 38$ to the resistance), so the net heat exchange drops sharply — to a small fraction of the unshielded value. Physically, the shield reaches an intermediate temperature and, because its surfaces are highly reflective (low emissivity), it re-radiates very little, drastically cutting radiative transfer. This is the principle behind multilayer/reflective insulation.