Question 1 |
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 1 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 2 |
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 2 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}
Question 3 |
Hot and cold fluids enter a parallel flow double tube heat exchanger at 100^{\circ}C \; and \; 15^{\circ}C, respectively. The heat capacity rates of hot and cold fluids are C_h=2000W/K and C_c=1200W/K, respectively. If the outlet temperature of the cold fluid is 45^{\circ}C, the log mean temperature difference (LMTD) of the heat exchanger is _________K (round off to two decimal places).
21.45 | |
45.25 | |
57.71 | |
98.65 |
Question 3 Explanation:
Given data:
C_{h} = 2000 W/K
Energy released by hot fluid = Energy received by cold fluid
\begin{aligned} \mathrm{C}_{\mathrm{h}}\left(\mathrm{T}_{\mathrm{hi}}-\mathrm{T}_{\mathrm{he}}\right) &=\mathrm{C}_{\mathrm{c}}\left(\mathrm{T}_{\mathrm{cc}}-\mathrm{T}_{\mathrm{ci}}\right) \\ 2000\left(100-\mathrm{T}_{\mathrm{hc}}\right)&=1200(45-15) \\ 2000 \times\left(100-\mathrm{T}_{\mathrm{he}}\right)& =1200 \times 30 \mathrm{T}_{\mathrm{he}}=82^{\circ} \mathrm{C} \\ \mathrm{LMTD}&=\frac{\theta_{1}-\theta_{2}}{\ln \left(\frac{\theta_{1}}{\theta_{2}}\right)} \\ &=\frac{85-37}{\ln \left(\frac{85}{37}\right)}=\frac{48}{\ln \left(\frac{85}{37}\right)} \\ \mathrm{LMTD}&=57.71^{\circ} \mathrm{C} \text{ or } \mathrm{K}\end{aligned}
This is logarithmic mean temperature difference either we take in ^{o}C or we take in K the final answer will be 57.71 K (or)^{o}C because it is a difference.
Question 4 |
Steam in the condenser of a thermal power plant is to be condensed at a temperature of
30^{\circ}C with cooling water which enters the tubes of the condenser at 14^{\circ}C and exits at 22^{\circ}C. The total surface area of the tubes is 50 m^{2}, and the overall heat transfer coefficient
is 2000 W/m^{2} K. The heat transfer (in MW) to the condenser is ______ (correct to two
decimal places).
0.5 | |
1.16 | |
2.25 | |
3.45 |
Question 4 Explanation:
Since steam is condensing, the temperature of hot fluid remains constant.
\therefore Temperature profiles of hot and cold fluids are
\therefore (\mathrm{LMTD}) of H E=\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}}=\frac{16-8}{\ln \left(\frac{16}{8}\right)}=11.54^{\circ} \mathrm{C}
Total heat transfer rate between steam
\begin{aligned} \mathrm{C.W.} &=Q=U A \mathrm{LMTD} \mathrm{W} \\ &=2000 \times 50 \times 11.54 \\ &=11.54 \times 10^{5} \mathrm{W} \\ &=1.154 \mathrm{MW} \end{aligned}
\therefore Temperature profiles of hot and cold fluids are
\therefore (\mathrm{LMTD}) of H E=\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}}=\frac{16-8}{\ln \left(\frac{16}{8}\right)}=11.54^{\circ} \mathrm{C}
Total heat transfer rate between steam
\begin{aligned} \mathrm{C.W.} &=Q=U A \mathrm{LMTD} \mathrm{W} \\ &=2000 \times 50 \times 11.54 \\ &=11.54 \times 10^{5} \mathrm{W} \\ &=1.154 \mathrm{MW} \end{aligned}
Question 5 |
In a counter-flow heat exchanger, water is heated at the rate of 1.5kg/s from 40\,^{\circ} C to 80\,^{\circ} C by an oil entering at 120\,^{\circ} C and leaving at 60\,^{\circ} C. The specific heat of water and oil are 4.2\:kJ/kg \! \cdot\!K and 2\:kJ/kg \! \cdot\!K, respectively. The overall heat transfer coefficient is 400\, W/m^{2} \cdot K The required heat transfer surface area (in m^{2} ) is
0.104 | |
0.022 | |
10.4 | |
21.84 |
Question 5 Explanation:
\begin{aligned} \Delta T_{i} &=120-80=40^{\circ} \mathrm{C} \\ \Delta T_{e} &=60-40=20^{\circ} \mathrm{C} \\ (L M T D) \text { Counterflow }&=\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}} =28.86^{\circ} \mathrm{C} \\ &=\frac{40-20}{\ln \frac{40}{20}} = 28.86^{o}\mathrm{C}\\ Q &=\dot{m}_{C} c_{p_{c}}\left(T_{c e}-T_{c i}\right) \\ &=1.5 \times 4.2 \times 10^{3}(80-40) \mathrm{watt} \\ Q &=252000 \mathrm{W} \\ Q &=U A \Delta T_{m} \\ A &=\frac{Q}{U \Delta T_{m}}=\frac{1.5 \times 4.2 \times 10^{3} \times 40}{400 \times 28.86} \\ &=21.83 \mathrm{m}^{2} \end{aligned}
Question 6 |
Saturated steam at 100^{\circ}C condenses on the outside of a tube.cold fluid enters the tube at 20^{\circ}C and exits at 50^{\circ}C the value of the log mean Temperature Difference (LMTD) is _________^{\circ}C.
30 | |
80 | |
64 | |
84 |
Question 6 Explanation:
Temperature profiles are
\begin{aligned} \Delta T_{i}=100-20=80^{\circ} \mathrm{C} \\ \Delta T_{e}=100-50=50^{\circ} \mathrm{C} \end{aligned} (LMTD) (parallel or counter HE)
\begin{aligned} =\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}}=\frac{80-50}{\ln \frac{80}{50}} \\ =63.82^{\circ} \mathrm{C} \end{aligned}
\begin{aligned} \Delta T_{i}=100-20=80^{\circ} \mathrm{C} \\ \Delta T_{e}=100-50=50^{\circ} \mathrm{C} \end{aligned} (LMTD) (parallel or counter HE)
\begin{aligned} =\frac{\Delta T_{i}-\Delta T_{e}}{\ln \frac{\Delta T_{i}}{\Delta T_{e}}}=\frac{80-50}{\ln \frac{80}{50}} \\ =63.82^{\circ} \mathrm{C} \end{aligned}
Question 7 |
For a heat exchanger, \Delta T_{max} is the maximum temperature difference and \Delta T_{min} is the minimum temperature difference between the two fluids. LMTD is the log mean temperature difference. C_{min} andC_{max} are the minimum and the maximum heat capacity rates. The maximum possible heat transfer (Q_{max}) between the two fluids is
C_{min} LMTD | |
C_{min} \Delta T_{max} | |
C_{max} \Delta T_{max} | |
C_{max} \Delta T_{min} |
Question 7 Explanation:
Maximum heat transfer possible
=C_{\min } \times(\Delta T)_{\max }
=C_{\min } \times(\Delta T)_{\max }
Question 8 |
Consider a parallel-flow heat exchanger with area A_{p} and a counter-flow heat exchanger with area A_{c}. In both the heat exchangers, the hot stream flowing at 1 kg/s cools from 80^{\circ}C to 50^{\circ}C . For the
cold stream in both the heat exchangers, the flow rate and the inlet temperature are 2 kg/s and 10^{\circ}C , respectively. The hot and cold streams in both the heat exchangers are of the same fluid.
Also, both the heat exchangers have the same overall heat transfer coefficient. The ratio A_{c}/ A_{p} is _________
0.258 | |
0.684 | |
0.489 | |
0.927 |
Question 8 Explanation:
Parallel flow
A_{p} = Area of parallel flow heat exchanger
Counter flow
A_{c}= Area of counter flow heat exchanger
Heat lost by hot stream in both heat exchanger
\begin{aligned} &=m c\left[T_{h i}-T_{h o}\right] \\ &=1 \times c \times\left[T_{h i}-T_{h o}\right] \\ &=1 \times c \times[80-50]=30 \mathrm{ckW} \end{aligned}
Heat gained by cold stream in both exchange
\begin{aligned} &=m c\left[T_{co}-T_{c i}\right] \\ &=2 \times c\left[T_{c o}-10\right] \end{aligned}
Heat lost by hot stream = Heat gain by cool stream
\begin{aligned} 30c &=2c(T_\infty -10) \\ T_\infty &= \frac{30}{2}+10\\ &=15+10=25^{\circ} C\\ A_P, LMTD&= \frac{\Delta T_i-\Delta T_e}{ln(\Delta T_i/\Delta T_e)}\\ &= \frac{70-25}{ln(70/25)}=43.705^{\circ} C\\ Q_H&=UA_P 43.705 \\ A_C, LMTD&= \frac{\Delta T_i-\Delta T_e}{ln(\Delta T_i/\Delta T_e)}\\ &= \frac{55-40}{ln(55/40)}=47.103^{\circ} C\\ Q_H&=UA_C 47.103 \\ \therefore \; UA_P 43.705 &=UA_C 47.103\\ \frac{43.705}{47.103}&=\frac{A_C}{A_P}=0.927 \end{aligned}
A_{p} = Area of parallel flow heat exchanger
Counter flow
A_{c}= Area of counter flow heat exchanger
Heat lost by hot stream in both heat exchanger
\begin{aligned} &=m c\left[T_{h i}-T_{h o}\right] \\ &=1 \times c \times\left[T_{h i}-T_{h o}\right] \\ &=1 \times c \times[80-50]=30 \mathrm{ckW} \end{aligned}
Heat gained by cold stream in both exchange
\begin{aligned} &=m c\left[T_{co}-T_{c i}\right] \\ &=2 \times c\left[T_{c o}-10\right] \end{aligned}
Heat lost by hot stream = Heat gain by cool stream
\begin{aligned} 30c &=2c(T_\infty -10) \\ T_\infty &= \frac{30}{2}+10\\ &=15+10=25^{\circ} C\\ A_P, LMTD&= \frac{\Delta T_i-\Delta T_e}{ln(\Delta T_i/\Delta T_e)}\\ &= \frac{70-25}{ln(70/25)}=43.705^{\circ} C\\ Q_H&=UA_P 43.705 \\ A_C, LMTD&= \frac{\Delta T_i-\Delta T_e}{ln(\Delta T_i/\Delta T_e)}\\ &= \frac{55-40}{ln(55/40)}=47.103^{\circ} C\\ Q_H&=UA_C 47.103 \\ \therefore \; UA_P 43.705 &=UA_C 47.103\\ \frac{43.705}{47.103}&=\frac{A_C}{A_P}=0.927 \end{aligned}
Question 9 |
Saturated vapor is condensed to saturated liquid in a condenser. The heat capacity ratio is C_{r}=\frac{C_{min}}{C_{max}}. The effectiveness (\varepsilon ) of the condenser is
\frac{1-exp[-NTU(1+C_{r})]}{1+C_{r}} | |
\frac{1-exp[-NTU(1-C_{r})]}{1-C_{r} exp[-NTU(1-C_{r})]} | |
\frac{NTU}{1+NTU} | |
1-exp(-NTU) |
Question 9 Explanation:
For parallel
\varepsilon=\frac{1-\exp \left[-N T U\left(1+C_{f}\right)\right]}{1+C_{r}}
For condenser,
\begin{aligned} C_{\max } &=\infty \\ C_{r} &=0 \\ \varepsilon &=\frac{1-\exp [-\mathrm{NTU}]}{1} \end{aligned}
Therefore (D) is correct option.
Considering counter-flow heat exchanger
\frac{1-\exp \left[-N T U\left(1+C_{r}\right)\right]}{1-C_{f} \exp \left[-N T U\left(1+C_{r}\right)\right]}
\therefore \quad c_{r}=0
1-\exp (-N T U)
\varepsilon=\frac{1-\exp \left[-N T U\left(1+C_{f}\right)\right]}{1+C_{r}}
For condenser,
\begin{aligned} C_{\max } &=\infty \\ C_{r} &=0 \\ \varepsilon &=\frac{1-\exp [-\mathrm{NTU}]}{1} \end{aligned}
Therefore (D) is correct option.
Considering counter-flow heat exchanger
\frac{1-\exp \left[-N T U\left(1+C_{r}\right)\right]}{1-C_{f} \exp \left[-N T U\left(1+C_{r}\right)\right]}
\therefore \quad c_{r}=0
1-\exp (-N T U)
Question 10 |
A balanced counterflow heat exchanger has a surface area of 20 m^{2} and overall heat transfer coefficient of 20 W/m^{2} -K. Air (C_{p} =1000 J/kg-K) entering at 0.4 kg/s and 280 K is to be preheated by the air leaving the system at 0.4 kg/s and 300 K. The outlet temperature (in K) of the preheated air is
290 | |
300 | |
320 | |
350 |
Question 10 Explanation:
Given data:
\begin{aligned} A&=20 \mathrm{m}^{3} ; \quad U=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad c_{p h}=1000 \mathrm{J} / \mathrm{kg} \mathrm{K} \\ m_{c}&=0.4 \mathrm{kg} / \mathrm{s} ; \quad T_{c 1}=280 \mathrm{K} ; \quad m_{h}=0.4 \mathrm{kg} / \mathrm{s} \\ T_{h 2}&=300 \mathrm{K} ; \quad T_{c 2}=? \end{aligned}
By energy balance,
Heat lost by hot fluid = Heat gained by cold fluid
m_{h} c_{p h}\left(T_{h 1}-T_{h 2}\right)=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right)
\text{or }\quad T_{h 1}-T_{h 2}=T_{c 2}-T_{c 1} \quad \because m_{h} c_{p h}=m_{c} c_{p c}
\begin{aligned} T_{h 1}-T_{c 2} &=T_{h 2}-T_{c 1} \\ \theta_{1} &=\theta_{2} \\ \text { If } \theta_{1}&=\theta_{2} \text { , then } L M T D=\theta_{1}=\theta_{2} \\ Q &=A U L M T D \\ Q &=A U \times \theta_{1} \\ &=20 \times 20 \times\left(T_{h 2}-T_{c 1}\right) \\ &=400(300-280)=8000 \mathrm{kW} \\ \text{also }\quad Q&=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right) \\ \therefore \quad 8000 &=0.4 \times 1000\left(T_{c 2}-280\right) \\ 20&=T_{c 2}-280 \\ \text{or}\quad T_{c 2}&=300^{\circ} \mathrm{C} \end{aligned}
Alternatively:
Given data:
A=20 \mathrm{m}^{3} ; U=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad
c_{\mathrm{ph}}=1000 \mathrm{J} / \mathrm{kgK}
m_{c}=0.4 \mathrm{kg} / \mathrm{s} ; \quad T_{c 1}=280 \mathrm{K} ; \quad m_{h}=0.4 \mathrm{kg} / \mathrm{s}
T_{h 2}=300 \mathrm{K} ; \quad T_{c 2}=?
By energy balance,
Heat lost by hot fluid = Heat gained by cold fluid
m_{h} c_{p h}\left(T_{h 1}-T_{h 2}\right)=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right)
\begin{aligned} \text { or } T_{h 1}-T_{h 2} &=T_{c 2}-T_{c 1} \qquad \because m_{h} c_{p h}=m_{c} c_{p c} \\ T_{h 1}-300 &=T_{c 2}-280 \\ T_{h 1}-T_{c 2} &=300-280 \qquad \ldots(1) \\ T_{h 1}-T_{c 2} &=20 \\ \epsilon&=\frac{N T U}{N T U+1} \quad[ \text{For balancing heat exchanger}]\\ \frac{T_{h 1}-T_{h 2}}{T_{h 1}-T_{c 1}} &=\frac{1}{1+1} \\ \frac{T_{h 1}-300}{T_{h 1}-280} &=\frac{1}{2} \\ 2 T_{h 1}-600 &=T_{h 1}-280 \\ T_{h 1} &=600-280=320 \mathrm{K} \end{aligned}
Substitute the value of T_{h 1}=320 \mathrm{K} in Eq. (1), we get
320-T_{c 2}=20
\text{or }\quad T_{c 2}=300 \mathrm{K}
\begin{aligned} A&=20 \mathrm{m}^{3} ; \quad U=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad c_{p h}=1000 \mathrm{J} / \mathrm{kg} \mathrm{K} \\ m_{c}&=0.4 \mathrm{kg} / \mathrm{s} ; \quad T_{c 1}=280 \mathrm{K} ; \quad m_{h}=0.4 \mathrm{kg} / \mathrm{s} \\ T_{h 2}&=300 \mathrm{K} ; \quad T_{c 2}=? \end{aligned}
By energy balance,
Heat lost by hot fluid = Heat gained by cold fluid
m_{h} c_{p h}\left(T_{h 1}-T_{h 2}\right)=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right)
\text{or }\quad T_{h 1}-T_{h 2}=T_{c 2}-T_{c 1} \quad \because m_{h} c_{p h}=m_{c} c_{p c}
\begin{aligned} T_{h 1}-T_{c 2} &=T_{h 2}-T_{c 1} \\ \theta_{1} &=\theta_{2} \\ \text { If } \theta_{1}&=\theta_{2} \text { , then } L M T D=\theta_{1}=\theta_{2} \\ Q &=A U L M T D \\ Q &=A U \times \theta_{1} \\ &=20 \times 20 \times\left(T_{h 2}-T_{c 1}\right) \\ &=400(300-280)=8000 \mathrm{kW} \\ \text{also }\quad Q&=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right) \\ \therefore \quad 8000 &=0.4 \times 1000\left(T_{c 2}-280\right) \\ 20&=T_{c 2}-280 \\ \text{or}\quad T_{c 2}&=300^{\circ} \mathrm{C} \end{aligned}
Alternatively:
Given data:
A=20 \mathrm{m}^{3} ; U=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad
c_{\mathrm{ph}}=1000 \mathrm{J} / \mathrm{kgK}
m_{c}=0.4 \mathrm{kg} / \mathrm{s} ; \quad T_{c 1}=280 \mathrm{K} ; \quad m_{h}=0.4 \mathrm{kg} / \mathrm{s}
T_{h 2}=300 \mathrm{K} ; \quad T_{c 2}=?
By energy balance,
Heat lost by hot fluid = Heat gained by cold fluid
m_{h} c_{p h}\left(T_{h 1}-T_{h 2}\right)=m_{c} c_{p c}\left(T_{c 2}-T_{c 1}\right)
\begin{aligned} \text { or } T_{h 1}-T_{h 2} &=T_{c 2}-T_{c 1} \qquad \because m_{h} c_{p h}=m_{c} c_{p c} \\ T_{h 1}-300 &=T_{c 2}-280 \\ T_{h 1}-T_{c 2} &=300-280 \qquad \ldots(1) \\ T_{h 1}-T_{c 2} &=20 \\ \epsilon&=\frac{N T U}{N T U+1} \quad[ \text{For balancing heat exchanger}]\\ \frac{T_{h 1}-T_{h 2}}{T_{h 1}-T_{c 1}} &=\frac{1}{1+1} \\ \frac{T_{h 1}-300}{T_{h 1}-280} &=\frac{1}{2} \\ 2 T_{h 1}-600 &=T_{h 1}-280 \\ T_{h 1} &=600-280=320 \mathrm{K} \end{aligned}
Substitute the value of T_{h 1}=320 \mathrm{K} in Eq. (1), we get
320-T_{c 2}=20
\text{or }\quad T_{c 2}=300 \mathrm{K}
There are 10 questions to complete.
