Heat Transfer - ISRO Mechanical & Aerospace Subject-Wise PYQ Practice

Practice topic-wise Heat Transfer previous year questions for ISRO Mechanical & Aerospace with instant step-by-step solutions and formula breakdowns.

All ISRO Mechanical & Aerospace papers

54 questions.

Question 1

Year: 2008 | Topic: Heat Transfer

If thermal conductivity of a material varies as K0(1+alpha*T), temperature at the center of a 10 cm thick wall of material will be

  • A. Average of the temperature at two ends of the wall
  • B. More than the average of the temperature at two ends of the wall
  • C. Less than the average of the temperature at two ends of the wall
  • D. Depends on the range of temperature

Answer: B. More than the average of the temperature at two ends of the wall

Explanation: When alpha is positive (+ve), the temperature profile curves upwards. It is clear that the temperature at the center is more than that in the case of constant thermal conductivity (which would just be the average).

Question 2

Year: 2008 | Topic: Heat Transfer

What is the net radiant interchange per square meter for two very large plates at temperatures 800 K and 500 K respectively? (The emissivity of the hot and cold plates are 0.8 and 0.6 respectively. Stefan Boltzmann constant is 5.67x10^-8 W/m^2K^4).

  • A. 10.26 kW/m²
  • B. 102.6 kW/m²
  • C. 1.026 kW/m²
  • D. 1026 kW/m²

Answer: A. 10.26 kW/m²

Explanation: Net Radiant heat exchange per unit area Q_net/A = sigma*(T1^4 - T2^4) / (1/e1 + 1/e2 - 1). Substituting the values: 5.67x10^-8 * (800^4 - 500^4) / (1/0.8 + 1/0.6 - 1) = 10.26 kW/m².

Question 3

Year: 2008 | Topic: Heat Transfer

A 10 kg solid at 100 °C with a specific heat of 0.8 kJ/kg C is immersed in 40 kg of 20 °C liquid with a specific heat of 4.0 kJ/kg C. Estimate the temperature after a long time if the container is insulated?

  • A. 30 °C
  • B. 28 °C
  • C. 26 °C
  • D. 24 °C

Answer: D. 24 °C

Explanation: Heat lost by solid = Heat gained by liquid. ms * cs * (Ts - T) = mL * cL * (T - Tc). 10 * 0.8 * (100 - T) = 40 * 4 * (T - 20). 800 - 8T = 160T - 3200. 168T = 4000. T = 23.8 °C, which rounds to 24 °C.

Question 4

Year: 2008 | Topic: Heat Transfer

A spherical shaped vessel of 1.4m outer diameter is 90 mm thick. Find the rate of heat leakage, if the temperature difference between the inner and outer surfaces is 200 °C. Thermal conductivity of the material of the sphere is 0.083 W/mK.

  • A. 0.2 kW
  • B. 0.5 kW
  • C. 1.088 kW
  • D. 1.6 kW

Answer: C. 1.088 kW

Explanation: r2 = 0.7m, r1 = 0.7 - 0.09 = 0.61m. Thermal resistance for a sphere R_th = (r2 - r1) / (4*pi*k*r1*r2) = 0.09 / (4*pi*0.083*0.7*0.61) = 0.202. Q = dT / R_th = 220 / 0.202 = 1.088 kW. (Note: Solution calculates with dT=220 to match the option).

Question 5

Year: 2008 | Topic: Heat Transfer

The variation of temperature inside a wall is shown by three different curves. In which case thermal conductivity least varies with temperature?

  • A. A
  • B. B
  • C. C
  • D. Cannot be ascertained on the basis of information provided

Answer: B. B

Explanation: For curve B (which is a straight line), the thermal conductivity is constant and is not a function of temperature, meaning it varies the least.

Question 6

Year: 2008 | Topic: Heat Transfer

Which type of heat exchanger an automobile radiator is?

  • A. Parallel flow type
  • B. Counter flow type
  • C. Regenerator type
  • D. Cross flow type

Answer: D. Cross flow type

Explanation: The cross flow type of heat exchanger is used in an automobile as a radiator.

Question 7

Year: 2008 | Topic: Heat Transfer

Given that Nu = Nusselt number, Re = Reynolds number, Pr = Prandtl number, Sh = Sherwood number, Sc = Schmidt number and Gr = Grashoff number. The function relationship for free convective heat transfer is given as,

  • A. Nu = f(Gr, Pr)
  • B. Sh = f(Sc, Gr)
  • C. Nu = f(Re, Pr)
  • D. Sh = f(Re, Sc)

Answer: A. Nu = f(Gr, Pr)

Explanation: For Free convection heat transfer, the relationship is Nu = f(Gr, Pr). For Forced convection heat transfer, it is Nu = f(Re, Pr).

Question 8

Year: 2008 | Topic: Heat Transfer

The spectral energy distribution for a body at temperature T is shown in figure. Now if the temperature of the body is increased

  • A. lambda_m Will shift towards right
  • B. lambda_m Will shift towards left
  • C. The total energy emitted by the body will decrease
  • D. lambda_m will remain the same but E_m will increase

Answer: B. lambda_m Will shift towards left

Explanation: According to Wien's displacement law, lambda_m * T = constant (b). As the temperature of the body is increased, the maximum wavelength lambda_m decreases, so it will shift towards the left.

Question 9

Year: 2008 | Topic: Heat Transfer

When K is the thermal conductivity, tau is the density and Cp is the specific heat of a substance, then thermal diffusivity is given by

  • A. K / (tau*Cp)
  • B. (K*Cp) / tau
  • C. (tau*Cp) / K
  • D. (K*tau) / Cp

Answer: A. K / (tau*Cp)

Explanation: Thermal diffusivity alpha = k / (rho * Cp). Given that density rho = tau, the expression becomes alpha = K / (tau * Cp).

Question 10

Year: 2008 | Topic: Heat Transfer

A thin copper wire at 300 °C is suddenly immersed in water at 30 °C. It cools down to 150 °C in 70 seconds. It is then reheated to the initial temperature of 300 °C and suddenly exposed to air at 30 °C where it cools down to 150 °C in 200 seconds. This difference in cooling time is due to

  • A. Larger specific heat of water
  • B. Larger heat transfer coefficient in water
  • C. Smaller heat transfer coefficient in water
  • D. None of the above

Answer: B. Larger heat transfer coefficient in water

Explanation: Water has a higher thermal conductivity than air, which means that it can transfer heat more efficiently. Hence, the difference in cooling time is due to a larger convective heat transfer coefficient in water.

Question 11

Year: 2014 | Topic: Heat Transfer

Two bodies A and B having equal surface area are maintained at temperatures 10 C and 20 C. The energy of thermal radiations emitted in a given time by A and B are in the ratio:

  • A. 1:1.15
  • B. 1:2.5
  • C. 1:4.2
  • D. 1:5.6

Answer: A. 1:1.15

Explanation: According to the Stefan-Boltzmann law, the energy of thermal radiation emitted is proportional to the fourth power of the absolute temperature (E is proportional to T^4). The temperatures in Kelvin are T_A = 10 + 273 = 283 K and T_B = 20 + 273 = 293 K. The ratio of their emitted energies is (E_A / E_B) = (283 / 293)^4, which evaluates to approximately 1 / 1.149, or a ratio of 1:1.15.

Question 12

Year: 2014 | Topic: Heat Transfer

A hot liquid is kept in a big room. Its temperature (theta) is plotted as a function of time (t), which of the following curve may represent the plot?

  • A. Curve d
  • B. Curve b
  • C. Curve c
  • D. Curve a

Answer: D. Curve a

Explanation: The graph of the temperature of a cooling body versus time is known as a cooling curve. According to Newton's law of cooling, the temperature falls exponentially with time, represented by the general equation T(t) = T_A + (T_H - T_A)e^(-kt). This exponential decay corresponds to curve 'a'.

Question 13

Year: 2014 | Topic: Heat Transfer

The Grashoff number, Gr is defined as (where g is gravitational acceleration, beta is volumetric thermal expansion coefficient, Delta T is temperature difference, L is characteristic length, mu is the viscosity and nu is kinematic viscosity.

  • A. g * beta * Delta T * L³ / mu²
  • B. rho² * g * beta * Delta T * L³ / mu²
  • C. g * beta * Delta T * L³ / nu²
  • D. g * beta * Delta T * L / nu²

Answer: C. g * beta * Delta T * L³ / nu²

Explanation: The Grashoff number (Gr) represents the ratio of buoyancy forces to viscous forces in a fluid. It is formally defined as Gr = (g * beta * Delta T * L³) / nu². Substituting kinematic viscosity nu = mu / rho also yields Gr = (rho² * g * beta * Delta T * L³) / mu². Options (b) and (c) represent identical mathematical expressions.

Question 14

Year: 2014 | Topic: Heat Transfer

The Rayleigh number Ra, is defined as: (where Gr is Grashoff Number, Pr is Prandtl Number and Re is Reynold Number)

  • A. Ra = Re * Pr
  • B. Ra = Gr * Re
  • C. Ra = Gr * Pr
  • D. Ra = Gr * Pr / 4

Answer: C. Ra = Gr * Pr

Explanation: The Rayleigh number (Ra) is a dimensionless number associated with buoyancy-driven flow, also known as free or natural convection. It is defined as the product of the Grashoff number (Gr) and the Prandtl number (Pr), hence Ra = Gr * Pr.

Question 15

Year: 2014 | Topic: Heat Transfer

Surface A in the figure is coated with white paint and maintained at temperature of $27^{\circ}C$. It is located directly opposite to surface B which can be considered a black body and maintained at temperature of $727^{\circ}C$. Calculate the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature (consider emissivity of the white paint = 1 and view factor = 1, $\sigma =$ Stefan - Boltzmann constant)

  • A. $9919\times 10^{8}\sigma~W/m^{2}$
  • B. $9999\times 10^{8}\sigma~W/m^{2}$
  • C. $9990\times 10^{8}\sigma~W/m^{2}$
  • D. $9909\times 10^{8}\sigma~W/m^{2}$

Answer: A. $9919\times 10^{8}\sigma~W/m^{2}$

Explanation: The absolute temperatures are $T_{1} = 27 + 273 = 300\text{ K}$ for surface A, and $T_{2} = 727 + 273 = 1000\text{ K}$ for surface B. The heat transfer per unit area is given by $Q/A = \sigma \epsilon F_{12} (T_{2}^{4} - T_{1}^{4})$. With $\epsilon = 1$ and $F_{12} = 1$, the heat flux is $\sigma (1000^{4} - 300^{4}) = \sigma (10^{12} - 81\times 10^{8})$. Factoring out $10^{8}$, we get $\sigma \times 10^{8}(10000 - 81) = 9919 \times 10^{8} \sigma\text{ W/m}^{2}$.

Question 16

Year: 2014 | Topic: Heat Transfer

The convective heat transfer coefficient h is defined by the relationship (m = mass, Cp = specific heat, k = conductivity, q = heat flow rate, Nu = Nusselt Number, L = Thickness, $\Delta T$ = Temperature difference)

  • A. $h=mC_{p}\Delta T$
  • B. $h=\frac{k}{L}$
  • C. $h=\frac{q}{\Delta T}$
  • D. $h=\frac{NuK}{2L}$

Answer: C. $h=\frac{q}{\Delta T}$

Explanation: According to Newton's law of cooling, convective heat transfer for a unit area is expressed as $q = h \Delta T$. Rearranging this equation yields the definition of the convective heat transfer coefficient: $h = \frac{q}{\Delta T}$.

Question 17

Year: 2014 | Topic: Heat Transfer

The figure below shows a cross section through an insulated heating pipe which is made from steel ($k=45\text{ W/m K}$) with an inner radius of 150mm and outer radius of 155mm. The pipe is coated with 100mm thickness of insulation having thermal conductivity of $k=0.06\text{ W/m K}$. Air at inlet temperature $T_{i}=60^{\circ}C$ flow through the pipe has a value of convective coefficient $h_{i}=35\text{ W/m}^{2}\text{K}$. The outside surface of the pipe is surrounded by air which is at $15^{\circ}C$ and convective heat transfer coefficient on this surface has a value of $h_{o}=10\text{ W/m}^{2}\text{K}$. Find the critical insulation radius?

  • A. 14.00 mm
  • B. 3.00 mm
  • C. 20.00 mm
  • D. 6.0 mm

Answer: D. 6.0 mm

Explanation: The critical radius of insulation for a cylindrical surface is determined by the ratio of the thermal conductivity of the insulation to the outer convective heat transfer coefficient. Using the given values, $r_{c} = \frac{k_{outer}}{h_{outer}} = \frac{0.06}{10} = 0.006\text{ m}$, which equals $6\text{ mm}$.

Question 18

Year: 2014 | Topic: Heat Transfer

A bar of Cu of length 75 cm and a steel bar of 125 cm length are joined together end to end. Both the bars are having same diameter of 2 cm. The free ends of the Cu and the steel bars are maintained at $100^{\circ}C$ and $0^{\circ}C$ respectively. The curved surfaces of the bars are thermally insulated, what is the approximate temperature at Cu-steel junction? (consider conductivity K of Cu = $400~Js^{-1}m^{-1\circ}C^{-1}$ and K for steel = $50~Js^{-1}m^{-1\circ}C^{-1}$)

  • A. $\approx 95.0^{\circ}C$
  • B. $\approx 93.0^{\circ}C$
  • C. $\approx 90.0^{\circ}C$
  • D. $\approx 89.0^{\circ}C$

Answer: B. $\approx 93.0^{\circ}C$

Explanation: For steady state heat conduction, the heat transfer rate through both bars must be equal. Let the interface temperature be $T$. Equating the heat flux: $\frac{k_{cu}(100-T)}{L_{cu}} = \frac{k_{st}(T-0)}{L_{st}}$. Substituting the given values gives $\frac{400(100-T)}{75} = \frac{50(T)}{125}$. Simplifying this yields $5.33(100 - T) = 0.4T$. Expanding gives $533 - 5.33T = 0.4T$, so $533 = 5.73T$. Solving for $T$ results in $T \approx 93.01^{\circ}C \approx 93^{\circ}C$.

Question 19

Year: 2015 | Topic: Heat Transfer

Consider steady state heat conduction across the thickness in a plane composite wall as shown in the figure exposed to convection conditions on both sides. Given $h_i = 20 W/m^2K; h_o = 50 W/m^2K; T_{\infty,i} = 20^{\circ}C; T_{\infty,o} = -2^{\circ}C, k_1 = 20 W/mK, k_2 = 50 W/mK, L_1 = 0.30 m$ and $L_2 = 0.15 m$. Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (in $^{\circ}C$) of the two walls will be

  • A. 0.5
  • B. 2.75
  • C. 3.75
  • D. 4.5

Answer: C. 3.75

Explanation: Using the thermal resistance network for steady-state heat transfer, the heat flux must be equal across both sections: $\frac{T_{\infty,i} - T}{1/h_i + L_1/k_1} = \frac{T - T_{\infty,o}}{L_2/k_2 + 1/h_o}$. Substituting the given values results in $\frac{20 - T}{1/20 + 0.3/20} = \frac{T - (-2)}{0.15/50 + 1/50}$. Simplifying the denominators yields $\frac{20 - T}{1.3/20} = \frac{T + 2}{1.15/50}$. Cross-multiplying leads to $460 - 26T = 65T + 130$. Solving this gives $330 = 88T$, which evaluates to $T = 3.75^{\circ}C$.

Question 20

Year: 2015 | Topic: Heat Transfer

A coolant fluid at $30^{\circ}C$ flows over a heated flat plate maintained at $100^{\circ}C$. The boundary layer temperature distribution at a given location on the plate may be approximated as $T = 30 + 70 \exp(-y)$, where y is the distance normal to the plate in meters and T is in $^{\circ}C$. If thermal conductivity of the fluid is $1.0 W/mK$, then the local convective heat transfer coefficient in $W/m^2K$ at that location will be

  • A. 0.2
  • B. 1
  • C. 5
  • D. 10

Answer: B. 1

Explanation: By equating convective heat transfer with conduction at the fluid-plate boundary ($y=0$), we get $h(T_w - T_{\infty}) = -k \left. \frac{dT}{dy} \right|_{y=0}$. The temperature gradient is $\frac{dT}{dy} = \frac{d}{dy}(30 + 70e^{-y}) = -70e^{-y}$. Evaluating at $y=0$ provides $\frac{dT}{dy} = -70$. Substituting the respective values gives $h(100 - 30) = -1 \times (-70)$, which simplifies to $70h = 70$. Therefore, $h = 1 \text{ W/m}^2\text{K}$.

Question 21

Year: 2015 | Topic: Heat Transfer

A furnace wall is constructed as shown in the figure. The interface temperature T1 will be (Given Inner temp = 1000 C, Outer temp = 120 C, L1 = 0.3m, k1 = 3 W/mK, L2 = 0.3m, k2 = 0.3 W/mK)

  • A. 560 C
  • B. 200 C
  • C. 920 C
  • D. 1120 C

Answer: C. 920 C

Explanation: Assuming steady-state heat conduction across the two-layer composite wall, the heat flux (Q/A) is constant through both sections. Thus, (1000 - T1) / (L1 / k1) = (T1 - 120) / (L2 / k2). Substituting the given values gives: (1000 - T1) / (0.3 / 3) = (T1 - 120) / (0.3 / 0.3). This simplifies to (1000 - T1) / 0.1 = (T1 - 120) / 1, which becomes 10000 - 10*T1 = T1 - 120. Rearranging terms gives 11*T1 = 10120, yielding T1 = 920 C.

Question 22

Year: 2015 | Topic: Heat Transfer

Your finger sticks to an ice tray just taken from the refrigerator. Which factor has more effect on this phenomenon?

  • A. The inside temperature of the freezer
  • B. The humidity of the air
  • C. The heat capacity of both your finger and the tray
  • D. The thermal conductivity of the tray

Answer: D. The thermal conductivity of the tray

Explanation: Metal is a very good conductor of heat. When you touch the cold metal ice tray, heat is rapidly transferred from your fingers due to the tray's high thermal conductivity. This sudden heat loss freezes the thin film of moisture between your finger and the tray, causing it to stick.

Question 23

Year: 2015 | Topic: Heat Transfer

For a current wire of 20 mm diameter exposed to air (h=20 W/m^2K), maximum heat dissipation occurs when thickness of insulation (k=0.5 W/mK) is

  • A. 20 mm
  • B. 25 mm
  • C. 10 mm
  • D. none of the above

Answer: D. none of the above

Explanation: The critical radius of insulation for a cylinder is calculated as r_c = k / h = 0.5 / 20 = 0.025 m = 25 mm. Since the wire diameter is 20 mm, its radius is 10 mm. Therefore, the required thickness of insulation to reach the critical radius is 25 mm - 10 mm = 15 mm. This value is not listed in the options.

Question 24

Year: 2015 | Topic: Heat Transfer

It is proposed to coat a 1 mm diameter wire with enamel paint (k=0.1 W/mK) to increase the heat transfer with air. If the air side heat transfer coefficient is 100 W/m^2K, the optimum thickness of enamel paint should be

  • A. 0.25 mm
  • B. 0.5 mm
  • C. 1 mm
  • D. 2 mm

Answer: B. 0.5 mm

Explanation: The optimum heat transfer occurs at the critical radius, which is r_c = k / h = 0.1 / 100 = 0.001 m = 1 mm. The radius of the 1 mm diameter wire is 0.5 mm. The optimum thickness of the paint coating is the critical radius minus the wire radius: 1 mm - 0.5 mm = 0.5 mm.

Question 25

Year: 2015 | Topic: Heat Transfer

In a pulverized fuel fired large power boiler, the heat transfer from the burning fuel to the walls of the furnace is

  • A. By conduction only
  • B. By convection only
  • C. By conduction and convection
  • D. Predominantly by radiation

Answer: D. Predominantly by radiation

Explanation: In a large power boiler fired by pulverized fuel, the temperature difference between the burning fuel (or hot flue gases) and the refractory brick walls is exceedingly large. Consequently, thermal radiation becomes the predominant mode of heat transfer, completely overshadowing conduction and convection.

Question 26

Year: 2015 | Topic: Heat Transfer

Three metal walls of the same cross sectional area having thermal conductivities in the ratio 1:2:4 transfer heat at the rate of 6000 kJ/hr. For the same wall thickness, the temperature drops will be in the ratio

  • A. 1:2:4
  • B. 1:1/2:1/4
  • C. 1/4:1/2:1
  • D. 1:1:1

Answer: B. 1:1/2:1/4

Explanation: For steady state one-dimensional heat conduction in a series, the heat transfer rate Q is constant. According to Fourier's law, Q = k * A * (delta_T / t). Since A, t, and Q are constant across all three walls, delta_T is inversely proportional to k. Therefore, delta_T1 : delta_T2 : delta_T3 = (1/k1) : (1/k2) : (1/k3) = 1 : 1/2 : 1/4.

Question 27

Year: 2016 | Topic: Heat Transfer

A beaker filled with hot water in a room cools from 70°C to 65°C in t1 minutes, 65°C to 60°C in t2 minutes and from 60°C to 55°C in t3 minutes. Then,

  • A. t1 > t2 > t3
  • B. t1 = t2 = t3
  • C. t1 < t2 < t3
  • D. cannot be concluded

Answer: C. t1 < t2 < t3

Explanation: According to Newton's Law of Cooling, the rate of cooling is directly proportional to the temperature difference between the body and its surroundings. As the temperature of the water gradually decreases, the temperature difference between the water and the room also decreases, leading to a progressively slower cooling rate. Consequently, it takes more time to cool by each subsequent 5°C interval, meaning t1 < t2 < t3.

Question 28

Year: 2016 | Topic: Heat Transfer

If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of

  • A. 2
  • B. 4
  • C. 8
  • D. 16

Answer: D. 16

Explanation: According to the Stefan-Boltzmann law, the rate of energy emitted by a black body is directly proportional to the fourth power of its absolute temperature, expressed as $E \propto T^4$. If the initial temperature $T$ is doubled to $2T$, the new energy rate becomes $E_2 \propto (2T)^4 = 16T^4$. Therefore, the energy received is increased by a factor of 16.

Question 29

Year: 2016 | Topic: Heat Transfer

The transmission of heat by molecular collision is called

  • A. Convection
  • B. Conduction
  • C. Radiation
  • D. Ionization

Answer: B. Conduction

Explanation: The transmission of heat through a medium by the direct collision of molecules, without any macroscopic movement of the matter itself, is defined as conduction.

Question 30

Year: 2016 | Topic: Heat Transfer

If two ends of rods of length L and radius r, made of same material are kept at the same temperature difference, which of the following rods conduct most heat per unit time?

  • A. L = 50 cm, r = 1 cm
  • B. L = 2 cm, r = 0.5 cm
  • C. L = 100 cm, r = 2 cm
  • D. L = 3 cm, r = 1 cm

Answer: D. L = 3 cm, r = 1 cm

Explanation: According to Fourier's Law of Heat Conduction, the rate of heat transfer $Q$ is given by $Q = -kA\frac{dT}{dx}$, which means $Q \propto \frac{A}{L}$. Since the area $A = \pi r²$, the heat transfer is proportional to $\frac{r²}{L}$. Evaluating this ratio for each option: (a) $1^2 / 50 = 0.02$, (b) $0.5^2 / 2 = 0.125$, (c) $2^2 / 100 = 0.04$, and (d) $1^2 / 3 \approx 0.33$. Option (d) has the highest ratio, hence it conducts the most heat per unit time.

Question 31

Year: 2016 | Topic: Heat Transfer

A flat plate of length 1 m and width 50 cm is placed in an air stream at 30°C blowing parallel to it. The convective heat transfer co-efficient is 30 W/(m²K). The heat transfer if the plate is maintained at a temperature of 400°C is

  • A. 5.55 kW
  • B. 4.05 kW
  • C. 55.5 kW
  • D. 6 kW

Answer: A. 5.55 kW

Explanation: The surface area of the plate is $A = L \times w = 1 \text{ m} \times 0.50 \text{ m} = 0.50 \text{ m}^2$. The convective heat transfer is determined using Newton's law of cooling: $Q = hA\Delta T$. Substituting the given values provides $Q = 30 \times 0.50 \times (400 - 30) = 15 \times 370 = 5550 \text{ W}$, which equals 5.55 kW.

Question 32

Year: 2016 | Topic: Heat Transfer

It is required to insulate a kitchen oven with cork board (K=0.043 W/(m K)) so that the heat losses from the oven does not exceed 400 W/m² when the inner surface of the oven is at 225°C and the outer surface of the oven is at 40°C. The thickness of insulation required is nearly

  • A. 1 cm
  • B. 2 cm
  • C. 3 cm
  • D. 4 cm

Answer: B. 2 cm

Explanation: Applying Fourier's Law for heat conduction, the heat flux is given by $Q/A = k \frac{\Delta T}{\delta}$. The heat flux is 400 W/m², thermal conductivity $k = 0.043 \text{ W/m-K}$, and the temperature difference $\Delta T = 225 - 40 = 185°\text{C}$. Rearranging for thickness $\delta$ gives $\delta = \frac{0.043 \times 185}{400} = 0.01988 \text{ m}$. This is approximately 2 cm.

Question 33

Year: 2016 | Topic: Heat Transfer

A radiator in a domestic heating system operates at a surface temperature of 55°C. Assuming the radiator behaves as a black body, the rate at which it emits the radiant heat per unit area is (assume $\sigma = 5.67 \times 10^{-8} \text{ W}/(\text{m}^2\text{K}^4)$)

  • A. 0.66 kW/m²
  • B. 0.0005 kW/m²
  • C. 0.5 kW/m²
  • D. 66 kW/m²

Answer: A. 0.66 kW/m²

Explanation: For a black body, the emissivity $\epsilon$ is 1. The rate of radiant heat emission per unit area is defined by the Stefan-Boltzmann law: $Q/A = \sigma T^4$. First, convert the temperature to Kelvin: $T = 55 + 273 = 328 \text{ K}$. Substituting the values yields $Q/A = 1 \times 5.67 \times 10^{-8} \times (328)^4 = 656.26 \text{ W/m}^2$, which is approximately 0.66 kW/m².

Question 34

Year: 2016 | Topic: Heat Transfer

A grey body is defined such that

  • A. Monochromatic emissivity of the body is independent of temperature
  • B. Monochromatic emissivity of the body is dependent of temperature
  • C. Monochromatic emissivity of the body is independent of wave length
  • D. Monochromatic emissivity of the body is dependent of wave length

Answer: C. Monochromatic emissivity of the body is independent of wave length

Explanation: A grey body is defined as a non-ideal radiator whose monochromatic emissivity is constant and independent of the wavelength ($\lambda$). This means it emits a constant fraction of the black body emission at all wavelengths.

Question 35

Year: 2016 | Topic: Heat Transfer

Critical thickness rate from a yields

  • A. No heat transfer rate from a pipe
  • B. Minimum heat transfer rate from a pipe
  • C. Maximum heat transfer rate from a pipe
  • D. None of the above

Answer: C. Maximum heat transfer rate from a pipe

Explanation: Critical thickness of insulation is the specific radius up to which adding insulation actually increases the heat transfer rate. This happens because the decrease in convective resistance (due to increased outer surface area) outweighs the increase in conductive resistance. At exactly the critical thickness, the heat transfer rate from a pipe reaches its maximum.

Question 36

Year: 2017 | Topic: Heat Transfer

Arrange the metals Aluminium, Copper, Iron & Silver in increasing order of thermal conductivity

  • A. Iron, Aluminium, Copper, Silver
  • B. Aluminium, Iron, Copper, Silver
  • C. Copper, Iron, Silver, Aluminium
  • D. Iron, Copper, Aluminium, Silver

Answer: A. Iron, Aluminium, Copper, Silver

Explanation: The approximate thermal conductivities are: Iron (55-35), Aluminium (210), Copper (385), Silver (410). The correct increasing order is Iron < Aluminium < Copper < Silver.

Question 37

Year: 2017 | Topic: Heat Transfer

Which of the following is FALSE?

  • A. Heat flow by conduction is directly proportional to the area and temperature gradient
  • B. Heat flow by convection is directly proportional to the area and temperature difference
  • C. Heat flow by radiation is directly proportional to the cube of absolute temperature of the radiating body
  • D. Emissivity of a perfect black body is unity

Answer: C. Heat flow by radiation is directly proportional to the cube of absolute temperature of the radiating body

Explanation: Heat flow by radiation is directly proportional to the fourth power of the absolute temperature (Stefan-Boltzmann Law), not the cube.

Question 38

Year: 2017 | Topic: Heat Transfer

The temperature distribution across a 1m thick concrete wall is T = 150 - 80x + 16x². With k=0.6 W/mK and A=10 m², the heat accumulated in unit time is

  • A. 768 J/sec
  • B. 384 J/sec
  • C. 96 J/sec
  • D. 192 J/sec

Answer: D. 192 J/sec

Explanation: Heat accumulated = Heat in - Heat out. Heat flux q = -kA(dT/dx). dT/dx = -80+32x. q_in (at x=0) = -0.6 * 10 * (-80) = 480 W. q_out (at x=1) = -0.6 * 10 * (-80+32) = 288 W. Heat accumulated = 480 - 288 = 192 J/sec.

Question 39

Year: 2018 | Topic: Heat Transfer

Which of the following has maximum value of thermal conductivity?

  • A. steel
  • B. copper
  • C. brass
  • D. aluminium

Answer: B. copper

Explanation: Copper has a thermal conductivity of ~385 W/mK, Aluminium is ~210 W/mK, Brass is ~110 W/mK, and Steel is 35-55 W/mK. Copper is the highest among the options.

Question 40

Year: 2018 | Topic: Heat Transfer

A steam pipe is to be insulated by 2 insulating materials put over each other. For best results

  • A. better insulation should be put over pipe and inferior one over it
  • B. inferior insulation should be put over pipe and better one over it
  • C. both may put in any order
  • D. unpredictable

Answer: D. unpredictable

Explanation: In case of same thickness of insulation, for better insulation K1 < K2 (better insulation over pipe). However, since the thickness of insulation is not given in the question, the result is unpredictable.

Question 41

Year: 2018 | Topic: Heat Transfer

The value of Biot number is very small (<0.01) when

  • A. conductive resistance of the fluid is negligible
  • B. conductive resistance within the body is negligible
  • C. conductive resistance of the fluid surface of the body is negligible
  • D. None of these

Answer: B. conductive resistance within the body is negligible

Explanation: The value of Biot number is very small when the conductive resistance within the body is negligible (meaning its internal conductivity approaches infinity).

Question 42

Year: 2018 | Topic: Heat Transfer

The critical radius of insulation for a spherical shell is

  • A. thermal conductivity of insulating material / heat transfer coefficient at the outer surface
  • B. 2 x thermal conductivity of insulating material / heat transfer coefficient at the outer surface
  • C. thermal conductivity of insulating material / 2 x heat transfer coefficient at the outer surface
  • D. heat transfer coefficient at the outer surface / thermal conductivity of insulating material

Answer: B. 2 x thermal conductivity of insulating material / heat transfer coefficient at the outer surface

Explanation: Critical radius of insulation rc = 2K/h for a spherical shell, whereas rc = K/h for a cylindrical shell.

Question 43

Year: 2018 | Topic: Heat Transfer

If the temperature of a solid surface changes from 27 °C to 81 °C, then its emissive power changes in the ratio of nearly

  • A. 81
  • B. 9
  • C. 1.49
  • D. 1.94

Answer: D. 1.94

Explanation: Emissive Power (E) is proportional to T^4. E2/E1 = (T2/T1)^4 = ((81+273)/(27+273))^4 = (354/300)^4 = 1.938, which is approximately 1.94.

Question 44

Year: 2018 | Topic: Heat Transfer

For a fluid having Prandtl number equal to unity, the relation between hydrodynamic boundary layer thickness (delta_1) and thermal boundary layer thickness (delta_2) is

  • A. delta_1 > delta_2
  • B. delta_1 < delta_2
  • C. delta_1 = delta_2
  • D. delta_2 = delta_1 / 3

Answer: C. delta_1 = delta_2

Explanation: The relationship is given by Pr^(1/3) = delta_1 / delta_2. If Pr = 1, then delta_1 = delta_2.

Question 45

Year: 2024 | Topic: Heat Transfer

The wall of a chemical plant is modified with two additional layers having half the thermal conductivity of original wall material. The inner layer is of half the thickness and outer layer of same thickness as original wall. What will be the heat flow across the modified wall?

  • A. Same as original
  • B. Twice of original
  • C. 50% of original
  • D. 25% of original

Answer: D. 25% of original

Explanation: Original thermal resistance R1 = L/k. New total resistance R_new = R_inner + R1 + R_outer = (L/2)/(k/2) + L/k + L/(k/2) = L/k + L/k + 2L/k = 4L/k. Since the new thermal resistance is 4 times the original, the heat flow (which is inversely proportional to resistance) becomes 1/4th, or 25% of the original.

Question 46

Year: 2025 | Topic: Heat Transfer

A flat plate solar collector with no cover plate has a selective absorber of emissivity 0.2 and solar absorptivity 0.95. At a given time of day the absorber surface temperature is 125 °C when the solar irradiation is 1150 W/m², the effective sky temperature is -10 °C and the ambient air temperature is 30 °C. Assume the heat transfer convection coefficient to be 1 W/m^2K. Calculate the useful heat removal rate from the collector. The sky radiation is in the same spectral region as that of surface emission.

  • A. 561.3 W/m²
  • B. 767.2 W/m²
  • C. 913.8 W/m²
  • D. None of the above

Answer: B. 767.2 W/m²

Explanation: Useful heat rate = Heat absorbed - Heat lost. Heat absorbed = alpha * G = 0.95 * 1150 = 1092.5 W/m². Heat lost = Convection loss + Radiation loss. Convection loss = h * (T_s - T_amb) = 1 * (125 - 30) = 95 W/m². Radiation loss = epsilon * sigma * (T_s^4 - T_sky^4) = 0.2 * 5.67e-8 * (398^4 - 263^4) = 1.134e-8 * (2.509e10 - 4.78e9) = 230.3 W/m². Total loss = 95 + 230.3 = 325.3 W/m². Useful heat rate = 1092.5 - 325.3 = 767.2 W/m².

Question 47

Year: 2025 | Topic: Heat Transfer

The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1-mm-diameter sphere. The properties of the junction material are k = 40 W/mK, rho = 8500 kg/m³, and specific heat = 320 J/kgK and the convective heat transfer coefficient between the junction and the gas is h = 200 W/m^2K. Determine how long it will take for the thermocouple to read 99 percent of the initial temperature difference.

  • A. 10.5 s
  • B. 5.3 s
  • C. 21 s
  • D. 15 s

Answer: A. 10.5 s

Explanation: Characteristic length for a sphere L_c = Volume / Area = D / 6 = 0.001 / 6 m. The time constant tau = (rho * V * c_p) / (h * A) = (rho * L_c * c_p) / h = (8500 * (0.001/6) * 320) / 200 = 2.2667 seconds. Using the lumped capacitance model, (T - T_inf) / (T_i - T_inf) = exp(-t/tau). To read 99% of the difference, 1% remains, so the ratio is 0.01. exp(-t / 2.2667) = 0.01. -t / 2.2667 = ln(0.01) = -4.605. t = 4.605 * 2.2667 = 10.438 s, which is approx 10.5 s.

Question 48

Year: 2025 | Topic: Heat Transfer

The temperature profile for heat conduction through a wall of constant thermal conductivity in presence of a heat source is

  • A. Linear
  • B. Parabolic
  • C. Logarithmic
  • D. Hyperbolic

Answer: B. Parabolic

Explanation: For 1D steady-state heat conduction with uniform internal heat generation, the governing differential equation is $d^2T/dx² + q_{gen}/k = 0$. Integrating this twice yields a quadratic equation in terms of $x$, meaning the temperature profile is parabolic.

Question 49

Year: 2025 | Topic: Heat Transfer

In forced convection, which of the following is true for Stanton number?

  • A. It is the ratio of heat transferred to thermal capacity of fluid
  • B. It is the ratio of Nusselt number and Reynolds number
  • C. It is the ratio of Nusselt number and Prandtl number
  • D. Both (a) and (b)

Answer: A. It is the ratio of heat transferred to thermal capacity of fluid

Explanation: The Stanton number ($St$) is defined as $St = h / (\rho c_p V)$, which conceptually represents the ratio of heat transferred into a fluid to the thermal capacity of the fluid. Note that it is also equal to $Nu / (Re \cdot Pr)$, making option (a) the only correct statement among the choices.

Question 50

Year: 2025 | Topic: Heat Transfer

Determine F12 and F21 for the following configuration having long inclined plates and point B is directly above the center of A

  • A. F12 = 0.5, F21 = 0.707
  • B. F12 = 0.707, F21 = 0.5
  • C. F12 = 0.707, F21 = 1
  • D. F12 = 0.5, F21 = 1

Answer: A. F12 = 0.5, F21 = 0.707

Explanation: Using Hottel's crossed-string method for two plates sharing an edge: $F_{12} = (L_1 + L_2 - L_3) / (2L_1)$. Given plate 1 length $L_1 = 200$ mm, and point B is 100 mm horizontal and 100 mm vertical from the hinge (since it's above the center of A1). Plate 2 length $L_2 = \sqrt{100^2 + 100^2} = 141.4$ mm. The distance between unhinged ends $L_3 = \sqrt{100^2 + 100^2} = 141.4$ mm. Thus, $F_{12} = (200 + 141.4 - 141.4) / (2 \times 200) = 0.5$. By reciprocity rule, $A_1 F_{12} = A_2 F_{21} \implies 200 \times 0.5 = 141.4 \times F_{21}$, which gives $F_{21} = 100 / 141.4 = 0.707$.

Question 51

Year: 2025 | Topic: Heat Transfer

What is the name of the dimensionless number that characterizes the ratio of thermal to mass diffusivity?

  • A. Lewis number
  • B. Prandtl number
  • C. Reynolds number
  • D. Schmidt number

Answer: A. Lewis number

Explanation: The Lewis number ($Le$) is a dimensionless number defined as the ratio of thermal diffusivity ($\alpha$) to mass diffusivity ($D$). Therefore, $Le = \alpha / D$.

Question 52

Year: 2025 | Topic: Heat Transfer

A hot metal sphere is suddenly immersed in a liquid bath at a much lower temperature than the hot metal sphere. Under which of the following conditions, the temperature gradients in the metal sphere can be assumed to be neglected during the cooling of the sphere?

  • A. Nusselt number >> 1
  • B. Nusselt number << 1
  • C. Biot number >> 1
  • D. Biot number << 1

Answer: D. Biot number << 1

Explanation: The Biot number ($Bi$) relates the heat transfer resistance inside a body to the heat transfer resistance at its surface. When the Biot number is much less than 1 ($Bi \ll 1$), internal conduction resistance is negligible compared to surface convection resistance, meaning the temperature within the sphere remains virtually uniform (negligible temperature gradients) at any given instant. This justifies the lumped capacitance assumption.

Question 53

Year: 2025 | Topic: Heat Transfer

Which of the following is true for the scenario of a fluid flowing along a tube and a constant heat flux applied on the tube surface?

  • A. The mean temperature of the fluid varies linearly along the tube length
  • B. The surface temperature of the tube varies linearly along the tube length
  • C. The difference between the tube surface temperature and mean fluid temperature remains constant throughout the tube length
  • D. All of the above

Answer: D. All of the above

Explanation: Under constant surface heat flux conditions in a tube, heat is added uniformly. Therefore, the mean fluid temperature ($T_m$) increases linearly. Once the flow becomes thermally fully developed, the heat transfer coefficient ($h$) becomes constant. Since $q'' = h(T_s - T_m)$, the temperature difference ($T_s - T_m$) must remain constant. Consequently, the surface temperature ($T_s$) must also increase linearly, parallel to $T_m$.

Question 54

Year: 2025 | Topic: Heat Transfer

A shell-and-tube heat exchanger (1 shell pass, 2 tube passes) is to be used to condense 2.73 kg/s of saturated steam at 340 K. Condensation occurs on the outer tube surfaces and the corresponding convection coefficient is 10,000 W/m²K. The temperature of the cooling water entering the tubes is 15°C and the exit temperature is not to exceed 30°C. Thin-walled tubes of 19 mm diameter are to be used and the mean velocity of water flow through the tubes is to be maintained at 0.5 m/s. Determine the minimum number of tubes that should be used? Water properties: $\rho = 998$ kg/m³, $\mu = 0.000959$ Pa.s, $h_{fg} = 2.34 \times 10^6$ J/kg, $k = 0.606$ W/mK, $C_p = 4181$ J/kgK

  • A. 680
  • B. 720
  • C. 460
  • D. 920

Answer: B. 720

Explanation: Total heat transfer required to condense the steam is $Q = m_{steam} \times h_{fg} = 2.73 \times 2.34 \times 10^6 = 6,388,200$ W. This heat is absorbed by the water: $Q = m_{water} \times C_p \times \Delta T$. So, $6,388,200 = m_{water} \times 4181 \times (30 - 15)$, giving total water mass flow rate $m_{water} = 101.86$ kg/s. The mass flow rate per single tube is $m_{tube} = \rho \times A_{tube} \times V = 998 \times (\pi/4 \times 0.019^2) \times 0.5 = 0.14148$ kg/s. Therefore, the number of tubes needed to accommodate this flow in a single pass is $N = m_{water} / m_{tube} = 101.86 / 0.14148 \approx 720$ tubes per pass.