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).
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).
125 | |
321 | |
652 | |
228 |
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).
825 | |
1126 | |
1000 | |
1256 |
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).
825 | |
1126 | |
1000 | |
1256 |
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}
\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).
4210 | |
6520 | |
3650 | |
4830 |
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.
\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?
Ratio of internal thermal resistance of a solid to the boundary layer thermal resistance | |
Ratio of the rate at which internal energy is advected to the rate of conduction heat transfer | |
Non-dimensional temperature gradient | |
Non-dimensional velocity gradient multiplied by Prandtl number |
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).
\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).
0.42 | |
0.12 | |
0.36 | |
0.88 |
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).
254 | |
632 | |
985 | |
319 |
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}
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).
2.25 | |
3.36 | |
1.18 | |
1.85 |
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}
\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?
\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?
T_A+\frac{\dot{q}L}{2h} | |
2T_A | |
T_A | |
T_A \left (1-\frac{\dot{q}L}{4 \pi h} \right )+\frac{\dot{q}L}{4 \pi h}T_\infty |
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).
491.34 | |
583.36 | |
965.24 | |
218.34 |
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.