BE Civil Engineering (IOE, TU) Fundamentals of Thermodynamics & Heat Transfer (IOE, ME 451) Question Paper 2080 Nepal

First Law statements

• For a cycle: $\oint \delta Q = \oint \delta W$. The cyclic integral of heat transfer equals the cyclic integral of work — net heat added to a system executing a cycle equals net work done by it.
• For a process: $Q - W = \Delta U$, i.e. $\delta Q = dU + \delta W$. Energy is conserved; heat supplied either raises internal energy or does work.

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

(a) Final temperature (polytropic relation):

$$\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}=\left(\frac{900}{300}\right)^{\frac{0.3}{1.3}}=3^{0.2308}$$

$3^{0.2308}=e^{0.2308\times\ln 3}=e^{0.2308\times1.0986}=e^{0.2536}=1.2887$

$$T_2 = 350\times1.2887 = \mathbf{451.0\ K}$$

(b) Boundary work for a polytropic process:

$$W=\frac{mR(T_1-T_2)}{n-1}=\frac{0.5\times0.287\times(350-451.0)}{1.3-1}=\frac{0.5\times0.287\times(-101.0)}{0.3}$$ $$W=\frac{-14.494}{0.3}=\mathbf{-48.31\ kJ}$$

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

(c) Heat transfer from the first law $Q=\Delta U + W$:

$$\Delta U = m c_v (T_2-T_1)=0.5\times0.718\times(451.0-350)=0.5\times0.718\times101.0=36.26\ \text{kJ}$$ $$Q = 36.26 + (-48.31) = \mathbf{-12.05\ kJ}$$

The negative sign means heat is rejected by the gas (transferred from the system to surroundings) — about $12.05\ \text{kJ}$ leaves the air during compression.

Kelvin-Planck statement: It is impossible to construct a device operating in a cycle that produces no effect other than the extraction of heat from a single reservoir and the production of an equivalent amount of work. (No engine can have 100% efficiency.)

Clausius statement: 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 up a temperature gradient.)

Equivalence (proof by contradiction): Suppose a device violates Clausius — it moves $Q$ from cold to hot reservoir with no work input. Couple it with an ordinary engine drawing $Q_H$ from the hot reservoir, doing work $W$, and rejecting $Q$ to the cold reservoir. The cold reservoir then has zero net exchange, and the combined device draws $(Q_H - Q)$ net from the hot reservoir and produces work $W$ — a Kelvin-Planck violator. Hence violating Clausius violates Kelvin-Planck. A symmetric argument proves the converse. The two statements are therefore equivalent.

Numerical part. Given $T_H=750\ \text{K}$, $T_L=300\ \text{K}$, $\dot W_{net}=40\ \text{kW}$.

(a) Efficiency:

$$\eta_{th}=1-\frac{T_L}{T_H}=1-\frac{300}{750}=1-0.4=\mathbf{0.60\ (60\%)}$$

(b) Heat supply:

$$\dot Q_H=\frac{\dot W_{net}}{\eta_{th}}=\frac{40}{0.60}=\mathbf{66.67\ kW}$$

(c) Heat rejection:

$$\dot Q_L=\dot Q_H-\dot W_{net}=66.67-40=\mathbf{26.67\ kW}$$

(d) Entropy rates:

$$\dot S_{source}=-\frac{\dot Q_H}{T_H}=-\frac{66.67}{750}=-0.0889\ \text{kW/K}$$ $$\dot S_{sink}=+\frac{\dot Q_L}{T_L}=+\frac{26.67}{300}=+0.0889\ \text{kW/K}$$

Net entropy change $=\dot S_{source}+\dot S_{sink}=-0.0889+0.0889=\mathbf{0\ kW/K}$, confirming the cycle is reversible (zero entropy generation).

Phase change of a pure substance (constant pressure heating of water):

 T |              superheated
   |            /  vapour
   |   sat.    /
   |  liquid  *----*  (sat. vapour, x=1)
   |  (x=0)   |wet |
   |    *-----*    region
   |   /  compressed/subcooled liquid
   |__/________________________ v

Starting as a compressed (subcooled) liquid, heating raises temperature until the saturated liquid state ($x=0$) is reached. Further heat causes boiling at constant temperature and pressure; the mixture in this wet region is a two-phase liquid-vapour mixture characterised by quality $x = m_g/(m_f+m_g)$, the mass fraction of vapour. When the last drop evaporates the saturated vapour state ($x=1$) is reached. Beyond this, adding heat produces superheated vapour at temperatures above saturation.

Given: $V=0.15\ \text{m}^3$, $p=200\ \text{kPa}$, $x=0.80$, $v_f=0.001061$, $v_g=0.8857\ \text{m}^3/\text{kg}$, $u_f=504.5$, $u_{fg}=2025.0\ \text{kJ/kg}$.

Specific volume of mixture:

$$v = v_f + x(v_g-v_f)=0.001061+0.80\times(0.8857-0.001061)$$ $$v=0.001061+0.80\times0.884639=0.001061+0.707711=0.708772\ \text{m}^3/\text{kg}$$

(a) Mass:

$$m=\frac{V}{v}=\frac{0.15}{0.708772}=\mathbf{0.2116\ kg}$$

(b) Specific internal energy:

$$u=u_f+x\,u_{fg}=504.5+0.80\times2025.0=504.5+1620.0=2124.5\ \text{kJ/kg}$$

Total internal energy:

$$U=m\,u=0.2116\times2124.5=\mathbf{449.6\ kJ}$$

Fourier's law and resistance. For steady 1-D conduction with no generation, Fourier's law states $\dot Q = -kA\dfrac{dT}{dx}$. With constant $k$ and area $A$, integrating across a wall of thickness $L$ from $T_1$ to $T_2$:

$$\dot Q\int_0^L dx = -kA\int_{T_1}^{T_2} dT \;\Rightarrow\; \dot Q\,L = kA(T_1-T_2)$$ $$\dot Q=\frac{kA(T_1-T_2)}{L}=\frac{T_1-T_2}{L/(kA)}=\frac{\Delta T}{R_{th}}$$

The thermal (conductive) resistance is $R_{th}=\dfrac{L}{kA}$ (analogous to electrical resistance, with $\dot Q$ as current and $\Delta T$ as voltage). Layers in series add resistances.

Given: $A=5\ \text{m}^2$; layer 1: $L_1=0.20\ \text{m}$, $k_1=1.0$; layer 2: $L_2=0.10\ \text{m}$, $k_2=0.20$; $T_{in}=1000\,^\circ\text{C}$, $T_{out}=100\,^\circ\text{C}$.

Resistances:

$$R_1=\frac{L_1}{k_1 A}=\frac{0.20}{1.0\times5}=0.04\ \text{K/W}$$ $$R_2=\frac{L_2}{k_2 A}=\frac{0.10}{0.20\times5}=\frac{0.10}{1.0}=0.10\ \text{K/W}$$ $$R_{total}=R_1+R_2=0.04+0.10=0.14\ \text{K/W}$$

(a) Heat loss:

$$\dot Q=\frac{\Delta T}{R_{total}}=\frac{1000-100}{0.14}=\frac{900}{0.14}=\mathbf{6428.6\ W}\approx 6.43\ \text{kW}$$

(b) Interface temperature $T_i$ (same $\dot Q$ through layer 1):

$$\dot Q=\frac{T_{in}-T_i}{R_1}\;\Rightarrow\;T_i=T_{in}-\dot Q R_1=1000-6428.6\times0.04$$ $$T_i=1000-257.14=\mathbf{742.9\,^\circ C}$$

Check via layer 2: $T_{out}=T_i-\dot Q R_2=742.9-6428.6\times0.10=742.9-642.9=100\,^\circ\text{C}$ ✓

Free vs forced convection. In free (natural) convection, fluid motion is driven solely by buoyancy arising from density differences caused by temperature gradients (no external mover). In forced convection, fluid is moved over the surface by an external agency (fan, pump, wind), giving generally higher $h$ and heat transfer rates.

Convective coefficient $h$: defined by Newton's law of cooling $\dot Q = hA(T_s-T_\infty)$. It is the heat transferred per unit surface area per unit temperature difference between surface and fluid (units $\text{W/m}^2\text{·K}$); it lumps the combined effects of fluid properties, flow regime and geometry.

Given: $D=0.10\ \text{m}$, $L=8\ \text{m}$, $T_s=180\,^\circ\text{C}=453.15\ \text{K}$, $T_\infty=25\,^\circ\text{C}=298.15\ \text{K}$, $h=12$, $\varepsilon=0.85$, $\sigma=5.67\times10^{-8}$.

Surface area:

$$A=\pi D L=\pi\times0.10\times8=2.5133\ \text{m}^2$$

Convective loss:

$$\dot Q_{conv}=hA(T_s-T_\infty)=12\times2.5133\times(180-25)=12\times2.5133\times155$$ $$\dot Q_{conv}=4674.7\ \text{W}\approx\mathbf{4.67\ kW}$$

Radiative loss (using absolute temperatures):

$$T_s^4=(453.15)^4=4.2166\times10^{10}\ \text{K}^4,\quad T_\infty^4=(298.15)^4=7.9013\times10^{9}\ \text{K}^4$$ $$T_s^4-T_\infty^4=4.2166\times10^{10}-0.79013\times10^{10}=3.4265\times10^{10}\ \text{K}^4$$ $$\dot Q_{rad}=\varepsilon\sigma A (T_s^4-T_\infty^4)=0.85\times5.67\times10^{-8}\times2.5133\times3.4265\times10^{10}$$ $$\dot Q_{rad}=0.85\times5.67\times10^{-8}\times8.612\times10^{10}=0.85\times4882.3=4150.0\ \text{W}\approx\mathbf{4.15\ kW}$$

(intermediate: $2.5133\times3.4265\times10^{10}=8.612\times10^{10}$; $5.67\times10^{-8}\times8.612\times10^{10}=4882.3$)

Total heat loss:

$$\dot Q_{total}=\dot Q_{conv}+\dot Q_{rad}=4674.7+4150.0=\mathbf{8824.7\ W}\approx 8.82\ \text{kW}$$

Given: $m=2\ \text{kg}$, $R=0.287\ \text{kJ/kg·K}$, $T=400\ \text{K}$, $p_1=500\ \text{kPa}$, $p_2=100\ \text{kPa}$.

For an isothermal process of an ideal gas, $\Delta U=0$, so $Q=W$, and the volume ratio equals the inverse pressure ratio: $V_2/V_1=p_1/p_2=5$.

(a) Work done by the gas:

$$W=mRT\ln\frac{p_1}{p_2}=2\times0.287\times400\times\ln\frac{500}{100}$$ $$=229.6\times\ln 5=229.6\times1.6094=\mathbf{369.5\ kJ}$$

(b) Heat transfer: since $\Delta U=0$,

$$Q=W=\mathbf{369.5\ kJ}\ \text{(supplied to the gas)}$$

(c) Entropy change (isothermal, ideal gas):

$$\Delta S=\frac{Q}{T}=\frac{369.5}{400}=\mathbf{0.924\ kJ/K}$$

Equivalently $\Delta S=mR\ln(p_1/p_2)=2\times0.287\times1.6094=0.924\ \text{kJ/K}$ ✓

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

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

where $h$ is specific enthalpy, $V$ velocity, $z$ elevation.

Assumptions for an adiabatic nozzle: no shaft work ($w=0$); adiabatic ($q=0$); negligible elevation change ($z_2\approx z_1$); inlet velocity negligible ($V_1\approx0$). The SFEE reduces to:

$$0=(h_2-h_1)+\frac{V_2^2}{2}\;\Rightarrow\;\frac{V_2^2}{2}=h_1-h_2=c_p(T_1-T_2)$$

Given: $T_1=500\ \text{K}$, $T_2=400\ \text{K}$, $c_p=1.005\ \text{kJ/kg·K}$, $V_1\approx0$.

$$\frac{V_2^2}{2}=c_p(T_1-T_2)=1.005\times(500-400)=100.5\ \text{kJ/kg}=100500\ \text{J/kg}$$ $$V_2=\sqrt{2\times100500}=\sqrt{201000}=\mathbf{448.3\ m/s}$$

(Unit note: $1\ \text{kJ/kg}=1000\ \text{J/kg}=1000\ \text{m}^2/\text{s}^2$.)

Definitions. For a refrigerator the desired effect is heat removed from the cold space: $COP_R=\dfrac{Q_L}{W_{net}}$. For a heat pump the desired effect is heat delivered to the warm space: $COP_{HP}=\dfrac{Q_H}{W_{net}}$. Since $Q_H=Q_L+W_{net}$, they are related by $COP_{HP}=COP_R+1$.

Given: $T_L=-10\,^\circ\text{C}=263.15\ \text{K}$, $T_H=30\,^\circ\text{C}=303.15\ \text{K}$, $\dot Q_L=2.5\ \text{kW}$.

(a) COP (reversed Carnot):

$$COP_R=\frac{T_L}{T_H-T_L}=\frac{263.15}{303.15-263.15}=\frac{263.15}{40}=\mathbf{6.579}$$

(b) Power input:

$$\dot W_{net}=\frac{\dot Q_L}{COP_R}=\frac{2.5}{6.579}=\mathbf{0.380\ kW}\ (380\ \text{W})$$

(c) Heat rejection:

$$\dot Q_H=\dot Q_L+\dot W_{net}=2.5+0.380=\mathbf{2.880\ kW}$$

Derivation. For steady radial conduction with no generation, the heat rate $\dot Q$ is constant through every cylindrical shell of radius $r$, length $L$, area $A=2\pi rL$. Fourier's law:

$$\dot Q=-kA\frac{dT}{dr}=-k(2\pi rL)\frac{dT}{dr}$$

Separating variables and integrating from $r_1$ to $r_2$ ($T_1$ to $T_2$):

$$\dot Q\int_{r_1}^{r_2}\frac{dr}{r}=-2\pi kL\int_{T_1}^{T_2}dT$$ $$\dot Q\ln\frac{r_2}{r_1}=2\pi kL(T_1-T_2)\;\Rightarrow\;\boxed{\dot Q=\frac{2\pi kL(T_1-T_2)}{\ln(r_2/r_1)}}$$

The thermal resistance of the cylindrical wall is $R_{th}=\dfrac{\ln(r_2/r_1)}{2\pi kL}$.

Given: $r_1=0.05$, $r_2=0.07\ \text{m}$, $L=10\ \text{m}$, $k=50\ \text{W/m·K}$, $T_1=120\,^\circ\text{C}$, $T_2=80\,^\circ\text{C}$.

$$\ln\frac{r_2}{r_1}=\ln\frac{0.07}{0.05}=\ln 1.4=0.33647$$ $$\dot Q=\frac{2\pi\times50\times10\times(120-80)}{0.33647}=\frac{2\pi\times50\times10\times40}{0.33647}$$

Numerator $=2\pi\times20000=125663.7\ \text{W·(dimensionless)}$ (since $50\times10\times40=20000$).

$$\dot Q=\frac{125663.7}{0.33647}=\mathbf{373{,}477\ W}\approx 373.5\ \text{kW}$$

Given: $m=1.5\ \text{kg}$, $T_1=300\ \text{K}$, $T_2=500\ \text{K}$, $\Delta T=200\ \text{K}$, $c_v=0.718$, $c_p=1.005\ \text{kJ/kg·K}$.

(a) Constant volume. Work $W=\int p\,dV=0$ (no volume change). By the first law $Q=\Delta U$:

$$Q_v=m c_v \Delta T=1.5\times0.718\times200=\mathbf{215.4\ kJ},\qquad W=\mathbf{0\ kJ}$$

(b) Constant pressure. Heat added:

$$Q_p=m c_p \Delta T=1.5\times1.005\times200=\mathbf{301.5\ kJ}$$

Work done by the gas $W=p\Delta V=mR\Delta T$:

$$W=1.5\times0.287\times200=\mathbf{86.1\ kJ}$$

(Check: $\Delta U=mc_v\Delta T=215.4\ \text{kJ}$, and $Q_p=\Delta U+W=215.4+86.1=301.5\ \text{kJ}$ ✓)

Explanation of difference. The internal energy rise is the same in both cases ($215.4\ \text{kJ}$) because $\Delta U$ depends only on temperature for an ideal gas. At constant pressure the gas additionally expands and does $86.1\ \text{kJ}$ of boundary work, so more heat ($301.5$ vs $215.4\ \text{kJ}$) must be supplied. The extra heat equals the work done, which is why $c_p>c_v$ and $c_p-c_v=R=0.287\ \text{kJ/kg·K}$.

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 the absolute temperature. For a real (grey) surface $E=\varepsilon\sigma T^4$.

Definitions. A black body is an ideal surface that absorbs all incident radiation and emits the maximum possible at every wavelength ($\varepsilon=1$). Emissivity $\varepsilon$ is the ratio of a surface's emissive power to that of a black body at the same temperature ($0\le\varepsilon\le1$). A grey body has emissivity independent of wavelength ($\varepsilon=\text{const}<1$).

Given: $A=1.5\times10^{-4}\ \text{m}^2$, $T=2500\ \text{K}$, $\varepsilon=0.40$, $\sigma=5.67\times10^{-8}$.

(a) Radiant power emitted:

$$T^4=(2500)^4=3.90625\times10^{13}\ \text{K}^4$$ $$\dot Q=\varepsilon\sigma A T^4=0.40\times5.67\times10^{-8}\times1.5\times10^{-4}\times3.90625\times10^{13}$$

Step: $5.67\times10^{-8}\times3.90625\times10^{13}=2.21484\times10^{6}\ \text{W/m}^2$ (black-body emissive power).

$$\dot Q=0.40\times2.21484\times10^{6}\times1.5\times10^{-4}=0.40\times332.23=\mathbf{132.9\ W}$$

(intermediate: $2.21484\times10^{6}\times1.5\times10^{-4}=332.23\ \text{W}$)

(b) Wien's displacement law:

$$\lambda_{max}=\frac{2.898\times10^{-3}}{T}=\frac{2.898\times10^{-3}}{2500}=1.1592\times10^{-6}\ \text{m}=\mathbf{1.159\ \mu m}$$

This lies in the near-infrared, consistent with an incandescent filament radiating mostly heat with some visible light.