Heat Transfer

Question 1
Consider a hydrodynamically and thermally fully-developed, steady fluid flow of 1 kg/s in a uniformly heated pipe with diameter of 0.1 m and length of 40 m. A constant heat flux of magnitude 15000 W/m^2 is imposed on the outer surface of the pipe. The bulk-mean temperature of the fluid at the entrance to the pipe is 200^{\circ}C.
The Reynolds number (R_e) of the flow is 85000, and the Prandtl number (P_r) of the fluid is 5. The thermal conductivity and the specific heat of the fluid are 0.08 W\dot m^{-1} \dot K^{-1} and 2600 J \dot kg^{-1} \dot K^{-1}, respectively. The correlation N_u = 0.023 Re^{0.8} Pr^{0.4} is applicable, where the Nusselt Number (N_u) is defined on the basis of the pipe diameter. The pipe surface temperature at the exit is ________ ^{\circ}C (round off to the nearest integer).
A
125
B
321
C
652
D
228
GATE ME 2022 SET-2      Free and Forced Convection
Question 1 Explanation: 



\begin{aligned} \dot{m}&=1kg/sec\\ Re&=85000\\ Pr&=5\\ K&=0.08 W/mK\\ C&=2600J/kgK\\ Nu&=0.023R_e^{0.8}Pr^{0.4}\\ Nu&=\frac{hD}{k}\\ T_s&=?\\ Nu&=\frac{hD}{k}=0.023R_e^{0.8}Pr^{0.4}\\ \frac{h \times 0.1}{0.08}&=0.023(85000)^{0.8}(5)^{0.4}\\ h&=307.56 W/m^2K\\ Q&=qA_s=\dot{m}c_p(T_o-T_i)\\ q \pi DL&=\dot{m}c_p(T_o-T_i)\\ 15000 \times \pi \times0.1 \times 40&=1 \times 2600 \times (T_o-200)\\ T_o&=272.5^{\circ}C\\ &\text{Newtons law of cooling}\\ q&=h(T_s-T_m)\\ T_s-T-m=\frac{q}{h}=constant \end{aligned}


At outlet, x=LT_m=T_o. T=T_s
\begin{aligned} q&=h(T_s-T_o)\\ T_s-T_o&=\frac{q}{h}\\ T_s&=T_o+\frac{q}{h}\\ T_s&=272.5+\frac{15000}{307.56}\\ T_s&=321.27^{\circ}C\\ T_s&=321^{\circ}C \end{aligned}
Question 2
Saturated vapor at 200^{\circ}C condenses to saturated liquid at the rate of 150 kg/s on the shell side of a heat exchanger (enthalpy of condensation h_{fg} = 2400 kJ/kg). A fluid with C_p = 4 kJ\dot kg^{-1} \dot K^{-1} enters at 100^{\circ}C on the tube side. If the effectiveness of the heat exchanger is 0.9, then the mass flow rate of the fluid in the tube side is ____ kg/s (in integer).
A
825
B
1126
C
1000
D
1256
GATE ME 2022 SET-2      Heat Exchanger
Question 2 Explanation: 


Hot fluid
\begin{aligned} T_{h1} &=T_{h2}=200^{\circ} C\\ \dot{m}_h &=150kg/s \\ &= h_{fg}&=2400kj/kg \end{aligned}
Cold fluid
\begin{aligned} T_{c1} &=100^{\circ} C\\ \in &=0.9\\ \dot{m}_c &= ? kg/sec \end{aligned}
If fluid is under phase change then that fluid have in finite heat capacity rate.
In this case hot fluid is under phase change so C_{max}=\infty
It means cold fluid have minimum heat capacity rate
\begin{aligned} c_{min}&=\dot{m}_c c_c\\ \in &=\frac{Q_{act}}{Q_{max}}\\ 0.9&=\frac{\dot{m}_c c_c(T_{c2}-T_{c1})}{c_{min}(T_{h1}-T_{c1})}\\ 0.9&=\frac{T_{c2}-100}{200-100}\\ T_{c2}&=190^{\circ}C\\ Q&=\dot{m}_hh_{fg}=\dot{m}_cc_c(T_{c2}-T_{c1})\\ 150 \times 2400&=\dot{m}_c \times 4 \times (190-100)\\ \dot{m}_c&=1000kg/sec \end{aligned}
Question 3
Consider steady state, one-dimensional heat conduction in an infinite slab of thickness 2L (L = 1 m) as shown in the figure. The conductivity (k) of the material varies with temperature as k = CT, where T is the temperature in K, and C is a constant equal to 2 W\dot m^{-1} \dot K^{-2}. There is a uniform heat generation of 1280 kW/m^3 in the slab. If both faces of the slab are maintained at 600 K, then the temperature at x = 0 is ________ K (in integer).

A
825
B
1126
C
1000
D
1256
GATE ME 2022 SET-2      Conduction
Question 3 Explanation: 
General Heat conduction equation in Cartesian coordinates is
\frac{\partial }{\partial x}\left ( K_x\frac{\partial T}{\partial x} \right )+\frac{\partial }{\partial y}\left ( K_y\frac{\partial T}{\partial y} \right )\frac{\partial }{\partial z}\left ( K_z\frac{\partial T}{\partial z} \right )+q_g=\rho c\frac{\partial T}{\partial t}
For one dimensional steady state heat generation with variable thermal conductivity, the above equation can be written as
\begin{aligned} \frac{\partial }{\partial x}\left ( CT\frac{\partial T}{\partial x} \right )+q_g&=0\\ C\frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+q_g&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+\frac{q_g}{C}&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\frac{q_g}{C}\\ \int \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\int \frac{q_g}{C}\\ \text{For 1-D heat flow}\\ \frac{\partial T}{\partial x}&=\frac{dT }{\partial x}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}+C_1\\ x=0\;\;\frac{dT}{dx}&=0\;\;C_1=0 \text{ (At centerline)}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}\\ \int TdT&=\int -\frac{q_gx}{C}dx\\ \frac{T^2}{2}&=\frac{-q_gx^2}{2C}+C_2 . . . (1)\\ T=T_s&=\text{Outside surface temperature}\\ \frac{T_S^2}{2}&=\frac{-q_gL^2}{2C}+C_2\\ C_2&=\frac{T-s^2}{2}+\frac{q_gL^2}{2C}\\ \text{Put }&c_2 \text{ in equation (1)}\\ \frac{T^2}{2}&=\frac{-q_ax^2}{2C}+\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \text{At }&x=0,T=T_{max}\\ \frac{T_{max}^2}{2}&=\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \frac{T_{max}^2}{2}&=\frac{600^2}{2}+\frac{128010^3 \times 1^2}{2 \times 2}\\ T_{max}&=1000K \end{aligned}
Question 4
Wien's law is stated as follows: \lambda _mT=C , where C is 2898 \mu mK and \lambda _m is the wavelength at which the emissive power of a black body is maximum for a given temperature T. The spectral hemispherical emissivity (\varepsilon _ \lambda) of a surface is shown in the figure below ( 1\mathring{A}=10^{-10}m ). The temperature at which the total hemispherical emissivity will be highest is __________ K (round off to the nearest integer).

A
4210
B
6520
C
3650
D
4830
GATE ME 2022 SET-2      Radiation
Question 4 Explanation: 
From the figure we can say that at 6000 \mathring{A} wavelength total hemispherical emissivity is maximum
\lambda _m=6000 \mathring{A} =0.6\mu m
Wein's law
\lambda _m T=C
\lambda _m T=2898 \mu mK
T=\frac{2898}{0.6}=4830K
Note: In this question, we need to find temperature at which total hemispherical emissivity will be highest.
But to calculate total hemispherical emissivity we require additional data like function chart and emissivity value at particular wavelength.
With the use of Wein's law we can find out temperature at which spectral emissivity is highest but in question it is asked temperature for total hemispherical emissivity.
Question 5
Which of the following non-dimensional terms is an estimate of Nusselt number?
A
Ratio of internal thermal resistance of a solid to the boundary layer thermal resistance
B
Ratio of the rate at which internal energy is advected to the rate of conduction heat transfer
C
Non-dimensional temperature gradient
D
Non-dimensional velocity gradient multiplied by Prandtl number
GATE ME 2022 SET-2      Free and Forced Convection
Question 5 Explanation: 


Let,
\begin{aligned} \theta ^*&=\frac{T_s-T}{T_s-T_\infty }\\ y^*&=\frac{y}{L}\\ \frac{dy^*}{dy}&=\frac{1}{L}\\ \frac{d\theta ^*}{dy}&=-\frac{dT}{dy} \end{aligned}
T_s \text{ and } T_\infty are constant
At boundary
\begin{aligned} q_{\text{conduction}}&=q_{\text{convection}}\\ -K_f\left.\begin{matrix} \frac{dT}{dy} \end{matrix}\right|_{y=0}&=h(T_s-T_\infty )\\ -K_f\frac{d}{dy}\left [ \frac{T_s-T}{T_s-T_\infty }h \right ]&=h\\ \frac{d\theta ^*}{dy}&=\frac{h}{K_f}\\ \end{aligned}
Applying chain rule
\begin{aligned} \frac{d\theta ^*}{dy^*}\frac{dy^*}{dy}&=\frac{h}{K_f}\\ \frac{d\theta ^*}{dy^*}\frac{1}{L}=\frac{h}{K_f}\\ \frac{d\theta ^*}{dy^*}=\frac{hL}{K_f}=\text{Nusselt Number}\\ Nu&=\frac{d\theta ^*}{dy^*} \end{aligned}
Nu represents non dimensional temperature gradient.
Question 6
During open-heart surgery, a patient?s blood is cooled down to 25^{\circ}C from 37^{\circ}C using a concentric tube counter-flow heat exchanger. Water enters the heat exchanger at 4^{\circ}C and leaves at 18^{\circ}C . Blood flow rate during the surgery is 5 L/minute. Use the following fluid properties:
\begin{array}{|c|c|c|} \hline \text{Fluid}& \text{Density }(kg/m^3)&\text{Specific heat (J/kg-K)}\\ \hline \text{Blood}& 1050 & 3740\\ \hline \text{Water}& 1000 & 4200 \\ \hline \end{array}
Effectiveness of the heat exchanger is _________ (round off to 2 decimal places).
A
0.42
B
0.12
C
0.36
D
0.88
GATE ME 2022 SET-1      Heat Exchanger
Question 6 Explanation: 


For preferred heat exchange between load and water.
\begin{aligned} Q_{blood}&=Q_{water}\\ \dot{m}_h c_h(T_{h_1}-T_{h_2})&=\dot{m}_c c_c(T_{c_2}-T_{c_1})\\ \dot{m}_h c_h(37-25)&=\dot{m}_c c_c(18-4)\\ 12\dot{m}_h c_h&=14\dot{m}_c c_c\\ \dot{m}_h c_h&=\frac{14}{12}\dot{m}_c c_c\\ \dot{m}_hc_h&=1.167\dot{m}_cc_c\\ \dot{m}_hc_h& \lt \dot{m}_cc_c\\ \dot{m}_cc_c&=c_{min}, \dot{m}_hc_h=c_{max}\\ \varepsilon &=\frac{Q_{act}}{Q_{max}}=\frac{\dot{m}_cc_c(T_{c_2}-T_{c_1})}{c_{min}(T_{h_1}-T_{c_1})}=\frac{T_{c_2}-T_{c_1}}{T_{h_1}-T_{c_1}}\\ \varepsilon &=\frac{T_{c_2}-T_{c_1}}{T_{h_1}-T_{c_1}}=\frac{18-4}{37-4}=0.424\\ \varepsilon &=0.42 \end{aligned}
Question 7
Consider a solid slab (thermal conductivity, k =10 W \dot m^{-1}\dot K^{-1}) with thickness 0.2 m and of infinite extent in the other two directions as shown in the figure. Surface 2, at 300 K, is exposed to a fluid flow at a free stream temperature ( T_\infty ) of 293 K, with a convective heat transfer coefficient (h) of 100 W\dot m^{-2} \dot K^{-1} . Surface 2 is opaque, diffuse and gray with an emissivity (\varepsilon ) of 0.5 and exchanges heat by radiation with very large surroundings at 0 K. Radiative heat transfer inside the solid slab is neglected. The Stefan-Boltzmann constant is 5.67 \times 10^{-8} W \dot m^{-2}\dot K^{-4} . The temperature T_1 of Surface 1 of the slab, under steady-state conditions, is _________ K (round off to the nearest integer).

A
254
B
632
C
985
D
319
GATE ME 2022 SET-1      Conduction
Question 7 Explanation: 
T_\infty =293K,T_{surrounding}=0K, h=100W/m^2K, K_{slab}=10W/mK

Under steady state
\begin{aligned} Q_{stored}&=0\\ Q_{in}&=Q_{out}\\ Q_{cond}&=Q_{conv}+Q_{rad}\\ K\left [ \frac{T_1-T_2}{L} \right ]&=h(T_2-T_\infty )+\varepsilon \sigma (T_2^4-T_{surr}^4)\\ 10\left [ \frac{T_1-300}{0.2} \right ]&=100(300-293 )+0.5 \times 5.67 \times 10^{-8} (300^4-0^4)\\ T_1-300&=\frac{929.635}{50}\\ T_1&=18.59+300\\ T_1&=318.59\\ T_1&=318.6K=319K \end{aligned}
Question 8
Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 80^{\circ}C , and the heat transfer rate per unit area to low temperature side B is 4.5 \; kW/m^2 . The thermal conductivity of the slab is 15 \; W/m.K . The rate of entropy generation per unit area during the heat transfer process is ________ W/m^2.K (round off to 2 decimal places).
A
2.25
B
3.36
C
1.18
D
1.85
GATE ME 2022 SET-1      Conduction
Question 8 Explanation: 
1D Steady heat conduction

\begin{aligned} \frac{Q}{A} &=4.5\frac{kW}{m^2}=4500 \frac{W}{m^2}\\ k&=15W/mk \\ S_{gen}&=? \\ \frac{Q}{A} &= k\left [ \frac{T_1-T_2}{dx} \right ]\\ 4500&= 15\left [ \frac{80-T_2}{0.1} \right ]\\ T_2 &=50^{\circ}C=323K \\ T_1 &=80^{\circ}C=353K \\ \therefore \;\; \frac{S_{gen}}{Area}&=\frac{-Q}{AT_1}+\frac{Q}{AT_2} \\ &=\frac{-4500}{353}+\frac{4500}{323} \\ S_{gen} &= 1.18\frac{W}{m^2K} \end{aligned}
Question 9
Consider a rod of uniform thermal conductivity whose one end (x = 0) is insulated and the other end (x = L) is exposed to flow of air at temperature T_\infty with convective heat transfer coefficient h . The cylindrical surface of the rod is insulated so that the heat transfer is strictly along the axis of the rod. The rate of internal heat generation per unit volume inside the rod is given as
\dot{q}=\cos \frac{2 \pi x}{L}
The steady state temperature at the mid-location of the rod is given as T_A. What will be the temperature at the same location, if the convective heat transfer coefficient increases to 2h?
A
T_A+\frac{\dot{q}L}{2h}
B
2T_A
C
T_A
D
T_A \left (1-\frac{\dot{q}L}{4 \pi h} \right )+\frac{\dot{q}L}{4 \pi h}T_\infty
GATE ME 2022 SET-1      Fins and Unsteady Heat Transfer
Question 9 Explanation: 


\dot{q}=\frac{\cos 2 \pi x}{L}
Total heat generation =Q_g=\int_{o}^{L}\dot{q}d(\text{Volume})
\begin{aligned} Q_g&=\int_{o}^{L}\dot{q}Adx\\ &=\int_{o}^{L}\left ( \frac{\cos 2\pi x}{L} \right )Adx\\ &=A\left [ \frac{\sin \left (\frac{2 \pi x}{L} \right ) }{\frac{ 2 \pi }{L}} \right ]_{0}^{L}\\ &=\left ( \frac{L}{2 \pi} \right )A\left [ \sin \left (\frac{2 \pi x}{L} \right )- \sin (0) \right ] \end{aligned}
Q_g=0
At steady state Q_{stored}=0
Q_{in}+Q_{gen}-Q_{out}=Q_{stored}
Q_{gen}=Q_{out}
0=hA_S(T_S-T_\infty
h \text{ and } A_S can't be zero it means T_S=T_ \infty
It indicate that T_S is independent of heat transfer coefficient.
So it means temperature of body any where is independent of heat transfer coefficient.
So by increasing heat transfer coefficient there is no change in the temperature at mid location.
Question 10
A flat plate made of cast iron is exposed to a solar flux of 600 W/m^2 at an ambient temperature of 25 ^{\circ}C. Assume that the entire solar flux is absorbed by the plate. Cast iron has a low temperature absorptivity of 0.21. Use Stefan-Boltzmann constant = 5.669 \times 10^{-8} W/m^2-K^4. Neglect all other modes of heat transfer except radiation. Under the aforementioned conditions, the radiation equilibrium temperature of the plate is __________ ^{\circ}C (round off to the nearest integer).
A
491.34
B
583.36
C
965.24
D
218.34
GATE ME 2022 SET-1      Radiation
Question 10 Explanation: 


Equilibrium Temperature =T_S=?
In this it is mentioned that entire flux is absorbed by the plate it means for solar flux absorptivity is 1. Kirchoff's law \alpha =\varepsilon =0.21
Surface of cast iron and surrounding fluid temperature difference is small due to this we can use Kirchoff's law
At equilibrium condition
Energy absorbed = Energy leaving
\begin{aligned} \alpha GA&=\varepsilon A\sigma (T_S^4-T_\infty ^4)\\ 1 \times 600 &=0.21 \times 5.669 \times 10^{-8}(T_S^4-298^4)\\ T_S&=491.34K\\ T_S&=218.34^{\circ}C\\ T_S&=218^{\circ}C \end{aligned}


There are 10 questions to complete.

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