BE Civil Engineering (IOE, TU) Fundamentals of Thermodynamics & Heat Transfer (IOE, ME 451) Question Paper 2077 Nepal

(a) First Law of Thermodynamics

For a cycle (Joule's statement): when a closed system undergoes a complete cycle, the cyclic integral of heat equals the cyclic integral of work:

$$\oint \delta Q = \oint \delta W$$

For a process (between states 1 and 2): the net heat supplied to the system equals the increase in stored (internal) energy plus the net work done by the system:

$$Q_{1\text{-}2} = \Delta U + W_{1\text{-}2}, \qquad \delta Q = dU + \delta W$$

Why $U$ is a property but $Q$ and $W$ are path functions: For any cycle $\oint(\delta Q-\delta W)=0$, so $\oint dU = 0$. A quantity whose cyclic integral is always zero depends only on the end states, hence $U$ is a point function (property) with an exact differential $dU$. In contrast, the amount of heat and work exchanged between the same two states differs with the path followed (e.g. isothermal vs. adiabatic), so $\delta Q$ and $\delta W$ are inexact differentials — path functions.

(b) Polytropic compression of air

Given: $m=2\ \text{kg}$, $p_1=100\ \text{kPa}$, $T_1=300\ \text{K}$, $p_2=600\ \text{kPa}$, $n=1.3$.

1. Volumes

$$V_1=\frac{mRT_1}{p_1}=\frac{2\times287\times300}{100\,000}=\mathbf{1.722\ m^3}$$

2. Final temperature (polytropic relation)

$$T_2=T_1\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}=300\,(6)^{0.3/1.3}=300\times1.5121=\mathbf{453.6\ K}$$ $$V_2=\frac{mRT_2}{p_2}=\frac{2\times287\times453.6}{600\,000}=\mathbf{0.4340\ m^3}$$

3. Boundary work (polytropic)

$$W=\frac{p_1V_1-p_2V_2}{n-1}=\frac{(100\,000)(1.722)-(600\,000)(0.4340)}{1.3-1}$$ $$W=\frac{172\,200-260\,400}{0.3}=\frac{-88\,200}{0.3}=-294\,000\ \text{J}=\mathbf{-294.0\ kJ}$$

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

4. Change in internal energy

$$\Delta U = m c_v (T_2-T_1)=2\times718\times(453.6-300)=2\times718\times153.6=\mathbf{+220.6\ kJ}$$

5. Heat transfer (First Law, $Q=\Delta U+W$)

$$Q = 220.6 + (-294.0)=\mathbf{-73.4\ kJ}$$

The negative sign means heat is rejected by the gas to the surroundings (about $73.4\ \text{kJ}$).

(Cross-check using $Q=mc_n\Delta T$ with $c_n=c_v\frac{\gamma-n}{1-n}=718\frac{1.4-1.3}{1-1.3}=-239.3\ \text{J/kg·K}$ gives $Q=2(-239.3)(153.6)=-73.5\ \text{kJ}$, in agreement.)

(a) Second Law statements and their equivalence

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

Clausius statement: It is impossible to construct a device operating in a cycle whose sole effect is the transfer of heat from a colder body to a hotter body (heat cannot flow uphill unaided).

Equivalence (violation of one ⇒ violation of the other):

   HOT reservoir T1
   |  Q1            ^ Q (Clausius violator pumps Q up with no work)
   v                |
 [ENGINE]---W--->   [PMM-II / heat from single source]
   |  Q2            
   v                
   COLD reservoir T2

Suppose a Clausius violator moves heat $Q$ from cold to hot with no work. Couple it with an ordinary engine that takes $Q_1$ from the hot reservoir, produces work $W$, and rejects $Q_2=Q$ to the cold reservoir. The cold reservoir then has zero net exchange, and the combined device draws net heat $(Q_1-Q)$ only from the hot reservoir while delivering work $W$ — a Kelvin–Planck violator. The reverse coupling likewise shows a K–P violator produces a Clausius violator. Hence the two statements are equivalent.

(b) Reversible machine between 900 K and 300 K

1. As a heat engine ($Q_1=1500\ \text{kJ}$, $T_1=900\ \text{K}$, $T_2=300\ \text{K}$)

$$\eta_{Carnot}=1-\frac{T_2}{T_1}=1-\frac{300}{900}=0.6667=\mathbf{66.67\%}$$ $$W=\eta\,Q_1=0.6667\times1500=\mathbf{1000\ kJ}$$ $$Q_2=Q_1-W=1500-1000=\mathbf{500\ kJ\ (rejected)}$$

2. As a heat pump (delivering to the $900\ \text{K}$ space, $W=1000\ \text{kJ}$ input) For a reversed Carnot machine the COP of a heat pump is

$$\text{COP}_{HP}=\frac{T_H}{T_H-T_C}=\frac{900}{900-300}=\frac{900}{600}=\mathbf{1.5}$$ $$Q_{delivered}=\text{COP}_{HP}\times W=1.5\times1000=\mathbf{1500\ kJ}$$

It extracts $Q_C=Q_{delivered}-W=1500-1000=500\ \text{kJ}$ from the $300\ \text{K}$ surroundings — exactly the mirror of the engine, as expected for a reversible machine.

(a) Formation of steam at constant pressure

 T |                         superheated
   |                       ./ region
   |        f      g    ../   (single-phase vapour)
   |  .----o======o.../
   | /     |wet   |
   |/  sub-|steam |
   |  cooled (two-phase)
   +---------------------------- v

Heating sub-cooled (compressed) liquid at constant pressure first raises its temperature to the saturation temperature $T_{sat}$ (point $f$, saturated liquid). Continued heating evaporates the liquid at constant $T$ and $p$; any state between $f$ and $g$ is wet steam (a liquid–vapour mixture). At point $g$ all liquid has just evaporated — dry saturated steam. Beyond $g$, heating raises the temperature above $T_{sat}$ giving superheated steam.

• Dryness fraction: $x=\dfrac{m_g}{m_g+m_f}$ = mass of vapour ÷ total mass of mixture ($0\le x\le1$; $x=0$ at $f$, $x=1$ at $g$).

(b) Exit velocity from the nozzle (SFEE)

Steady-flow energy equation, adiabatic ($q=0$), no work ($w=0$), inlet velocity $\approx 0$, negligible elevation change:

$$h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\;\Rightarrow\; V_2=\sqrt{2\,(h_1-h_2)}$$ $$V_2=\sqrt{2\times(3200-2900)\times10^3}=\sqrt{2\times300\,000}=\sqrt{600\,000}$$ $$\boxed{V_2 \approx \mathbf{774.6\ m/s}}$$

(c) Properties of wet steam at 1 MPa, $x=0.9$

Specific enthalpy:

$$h=h_f+x\,h_{fg}=762+0.9\times2015=762+1813.5=\mathbf{2575.5\ kJ/kg}$$

Specific volume:

$$v=v_f+x\,(v_g-v_f)=0.001+0.9\,(0.194-0.001)=0.001+0.1737=\mathbf{0.1747\ m^3/kg}$$

Specific internal energy ($u=h-pv$, with $p=1000\ \text{kPa}$):

$$u=2575.5-1000\times0.1747=2575.5-174.7=\mathbf{2400.8\ kJ/kg}$$

(a) 1-D steady conduction through a plane wall

Fourier's law (rate of heat conduction):

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

where the minus sign reflects heat flow down the temperature gradient. Thermal conductivity $k$ (W/m·K) is the property measuring a material's ability to conduct heat — the heat rate per unit area per unit temperature gradient.

For steady state with no internal generation, energy balance on a slab gives $\dfrac{d\dot Q}{dx}=0$, so $\dot Q$ is constant and (constant $k$, $A$):

$$\frac{d^2T}{dx^2}=0 \;\Rightarrow\; T \text{ varies linearly across the wall.}$$

Integrating Fourier's law across thickness $L$ with faces at $T_1,T_2$:

$$\dot Q=\frac{kA(T_1-T_2)}{L}=\frac{T_1-T_2}{R_{cond}},\qquad R_{cond}=\frac{L}{kA}\ (\text{K/W})$$

Electrical analogy: heat rate $\dot Q \leftrightarrow$ current $I$, temperature difference $\Delta T \leftrightarrow$ voltage $V$, thermal resistance $R \leftrightarrow$ electrical resistance. Layers in series add their resistances, exactly like resistors in series.

(b) Composite cold-storage wall

Resistances ($A=10\ \text{m}^2$):

$$R_1=\frac{L_1}{k_1 A}=\frac{0.20}{0.70\times10}=0.02857\ \text{K/W}$$ $$R_2=\frac{L_2}{k_2 A}=\frac{0.05}{0.04\times10}=0.1250\ \text{K/W}$$ $$R_{total}=R_1+R_2=0.02857+0.1250=0.15357\ \text{K/W}$$

1. Heat-transfer rate (overall $\Delta T = 30-(-5)=35\ \text{K}$):

$$\dot Q=\frac{\Delta T}{R_{total}}=\frac{35}{0.15357}=\mathbf{227.9\ W}$$

(heat flows inward, from the $30\,^\circ\text{C}$ brick face toward the cold room.)

2. Brick–insulation interface temperature (drop across the brick layer):

$$T_{int}=30-\dot Q\,R_1=30-227.9\times0.02857=30-6.51=\mathbf{23.49\ ^\circ C}$$

3. Comment: $R_2$ (insulation) is about 4.4 times $R_1$ (brick), carrying nearly the whole $35\ \text{K}$ drop; the thin polyurethane layer controls (governs) the heat flow, which is exactly why insulation is so effective despite its small thickness.

(a) Free vs forced convection; $h$ and $Nu$

Convective coefficient $h$ (W/m²·K) is the proportionality constant in Newton's law of cooling, $\dot Q = hA(T_s-T_\infty)$; it measures how effectively a moving fluid carries heat from a surface and depends on the fluid, flow regime and geometry.

Nusselt number $Nu=\dfrac{hL_c}{k_{fluid}}$ is the dimensionless ratio of convective to (pure) conductive heat transfer across the fluid layer. $Nu=1$ means heat moves by conduction alone (stagnant fluid); $Nu>1$ quantifies the enhancement due to bulk fluid motion.

(b) Heat loss from the steam pipe

Surface area (cylindrical side):

$$A=\pi D L=\pi\times0.10\times10=3.1416\ \text{m}^2$$

1. Convection ($\Delta T = 200-25=175\ \text{K}$):

$$\dot Q_{conv}=hA(T_s-T_\infty)=15\times3.1416\times175=\mathbf{8246.7\ W}\approx 8.25\ \text{kW}$$

2. Radiation (use absolute temperatures $T_s=473\ \text{K}$, $T_{sur}=298\ \text{K}$):

$$\dot Q_{rad}=\varepsilon\sigma A\,(T_s^4-T_{sur}^4)$$ $$T_s^4=(473)^4=5.006\times10^{10},\quad T_{sur}^4=(298)^4=7.886\times10^{9}$$ $$\dot Q_{rad}=0.8\times5.67\times10^{-8}\times3.1416\times(5.006\times10^{10}-0.789\times10^{10})$$ $$=0.8\times5.67\times10^{-8}\times3.1416\times4.217\times10^{10}=\mathbf{6010\ W}\approx 6.01\ \text{kW}$$

3. Total heat loss:

$$\dot Q_{total}=\dot Q_{conv}+\dot Q_{rad}=8246.7+6010=\mathbf{14\,257\ W}\approx 14.26\ \text{kW}$$

Both modes are significant here; radiation contributes roughly 42 % of the loss because of the high surface temperature, which is why bare hot pipes are usually lagged.

Thermodynamic system: a quantity of matter or a region in space chosen for study; everything outside it is the surroundings, and the real or imaginary surface separating the two is the boundary.

Zeroth Law: 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 common property shared by bodies in mutual thermal equilibrium is temperature. The third body can therefore act as a thermometer: when it is brought to equilibrium with a system, its measurable property (e.g. mercury column length, thermocouple emf) gives the system's temperature, and any two systems showing the same thermometer reading must be at the same temperature. Without the Zeroth Law, a consistent temperature scale could not be defined.

Given: $m=0.5\ \text{kg}$, constant pressure, $T_1=300\ \text{K}$, $T_2=600\ \text{K}$, $\Delta T=300\ \text{K}$.

Heat supplied (constant-pressure process):

$$Q=m c_p \Delta T=0.5\times1.005\times300=\mathbf{150.75\ kJ}$$

Work done (constant pressure, $W=p\,\Delta V=mR\,\Delta T$):

$$W=mR\,\Delta T=0.5\times0.287\times300=\mathbf{43.05\ kJ}$$

Change in internal energy:

$$\Delta U=m c_v \Delta T=0.5\times0.718\times300=\mathbf{107.70\ kJ}$$

First-law check: $Q=\Delta U+W$

$$150.75 \stackrel{?}{=} 107.70+43.05=150.75\ \text{kJ}\;\checkmark$$

The First Law is satisfied.

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

Principle of increase of entropy: For any process of an isolated system (or system + surroundings, i.e. the universe), the total entropy can never decrease:

$$\Delta S_{universe}\ge 0$$

the equality holding only for a reversible process and the inequality for an irreversible (real) process.

Constant-volume entropy change of air: For an ideal gas at constant volume, $dS=mc_v\dfrac{dT}{T}$, so

$$\Delta S=m c_v \ln\frac{T_2}{T_1}=1\times0.718\times\ln\frac{500}{300}$$ $$\Delta S=0.718\times\ln(1.6667)=0.718\times0.5108=\mathbf{+0.3668\ kJ/K}$$

The positive value reflects the heat addition raising the air's entropy.

Radial steady conduction through a cylindrical shell (per metre, $L=1\ \text{m}$):

The conduction resistance of a hollow cylinder is

$$R_{cyl}=\frac{\ln(r_2/r_1)}{2\pi k L}=\frac{\ln(0.075/0.05)}{2\pi\times0.05\times1}=\frac{\ln 1.5}{0.31416}=\frac{0.4055}{0.31416}=1.2906\ \text{K/W}$$

Heat loss per metre ($\Delta T=120-25=95\ \text{K}$):

$$\dot Q=\frac{\Delta T}{R_{cyl}}=\frac{95}{1.2906}=\mathbf{73.6\ W/m}$$

Equivalently, using $\dot Q=\dfrac{2\pi k L (T_1-T_2)}{\ln(r_2/r_1)}=\dfrac{2\pi(0.05)(1)(95)}{0.4055}=73.6\ \text{W/m}$, the same result.

Stefan–Boltzmann law: the total energy radiated per unit area by a black body is proportional to the fourth power of its absolute temperature, $E_b=\sigma T^4$. For a grey body the emissive power is $\varepsilon\sigma T^4$.

Net radiant exchange with large surroundings:

$$\dot Q=\varepsilon\sigma A\,(T_s^4-T_{sur}^4)$$ $$T_s^4=(400)^4=2.560\times10^{10},\quad T_{sur}^4=(300)^4=0.810\times10^{10}$$ $$\dot Q=0.8\times5.67\times10^{-8}\times2\times(2.560-0.810)\times10^{10}$$ $$\dot Q=0.8\times5.67\times10^{-8}\times2\times1.750\times10^{10}=\mathbf{1587.6\ W}$$

Ideal-gas equation of state: $pV=mRT$ (or $pv=RT$ per unit mass), where $p$ is absolute pressure, $V$ volume, $m$ mass, $R$ the specific gas constant and $T$ absolute temperature.

Assumptions: the gas molecules are point masses occupying negligible volume; there are no intermolecular attractive/repulsive forces except during perfectly elastic collisions; the model is accurate at low pressure and high temperature (far from the saturation/critical region).

Mass of nitrogen:

$$m=\frac{pV}{RT}=\frac{300\times0.5}{0.297\times290}=\frac{150}{86.13}=\mathbf{1.741\ kg}$$