Question 1 |
Consider a counter-flow heat exchanger with the inlet temperatures of two fluids
(1 and 2) being T_{1,in}=300K and T_{2,in}=350K. The heat capacity rates of the
two fluids are C_1=1000 W/K and C_2=400 W/K, and the effectiveness of the
heat exchanger is 0.5. The actual heat transfer rate is _____ kW. (Answer in integer)
6 | |
10 | |
12 | |
18 |
Question 1 Explanation:
Data given
\begin{aligned} \mathrm{C}_{1} & =1000 \mathrm{W} / \mathrm{K} \\ \mathrm{C}_{2} & =400 \mathrm{W} / \mathrm{K} \\ \epsilon & =\frac{\mathrm{q}_{\text {actual }}}{\mathrm{q}_{\max }} \end{aligned}
for q_{\max }, C should be minimum
So, \quad \mathrm{C}_{\min }=\mathrm{C}_{2}
So, \quad \in=\frac{\mathrm{q}_{\text {actual }}}{\mathrm{C}_{1}\left(\mathrm{T}_{\mathrm{h}_{1}}-\mathrm{T}_{\mathrm{c}_{1}}\right)}=0.5
or
\begin{aligned} q_{\text {actual }} & =0.5 \times 400(350-300) \\ & =10 \mathrm{kW} \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 |
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 3 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 4 |
A shell and tube heat exchanger is used as a steam condenser. Coolant water enters the tube at 300 K at a rate of 100 kg/s. The overall heat transfer coefficient is 1500 W/m^2.K, and total heat transfer area is 400 m^2. Steam condenses at a saturation temperature of 350 K. Assume that the specific heat of coolant water is 4000 J/kg.K. The temperature of the coolant water coming out of the condenser is _______K (round off to the nearest integer).
254 | |
339 | |
254 | |
654 |
Question 4 Explanation:
\begin{aligned} \mathrm{NTU}&=\frac{U A}{\left(\dot{m} c_{p}\right)_{\text {small }}}=\frac{1500 \times 400}{100 \times 4000}=1.5\\ \epsilon_{H E}&=1-e^{-N T U}=\frac{T_{c e}-T_{c i}}{T_{h i}-T_{c i}}=\frac{T_{C e}-300}{350-300}=0.7768\\ T_{c e}&=338.84 \mathrm{~K} \simeq 339 \mathrm{~K} \end{aligned}
Question 5 |
In a concentric tube counter-flow heat exchanger, hot oil enters at 102^{\circ}C and leaves at
65^{\circ}C. Cold water enters at 25^{\circ}C and leaves at 42^{\circ}C. The log mean temperature difference
(LMTD) is _______^{\circ}C (round off to one decimal place)
40 | |
43.25 | |
56.5 | |
49.3 |
Question 5 Explanation:
\begin{aligned} T_{h i}&= 102^{\circ} \mathrm{C}, \quad T_{h e}=65^{\circ} \mathrm{C} \\ T_{c, i}&=25^{\circ} \mathrm{C}, \quad T_{c e}=42^{\circ} \mathrm{C} \\ \mathrm{LMTD}&=\frac{\Delta T_{1}-\Delta T_{2}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{2}}\right)} \\ &\Delta T_{1}=102^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}=60^{\circ} \mathrm{C} ; \\ & \Delta T_{2}=65^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}=40^{\circ} \mathrm{C}_{7} \\ \mathrm{LMTD}&=\frac{60-40}{\ln \left(\frac{60}{40}\right)}=\frac{20}{0.4054}=49.3^{\circ} \mathrm{C} \end{aligned}
There are 5 questions to complete.