# Heat Exchanger

 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).
 A 254 B 339 C 254 D 654
GATE ME 2021 SET-2   Heat Transfer
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)
 A 40 B 43.25 C 56.5 D 49.3
GATE ME 2020 SET-1   Heat Transfer
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).
 A 21.45 B 45.25 C 57.71 D 98.65
GATE ME 2019 SET-2   Heat Transfer
Question 3 Explanation:

Given data:
C_{h} = 2000 W/K Energy balance:
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).
 A 0.5 B 1.16 C 2.25 D 3.45
GATE ME 2018 SET-2   Heat Transfer
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}
 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
 A 0.104 B 0.022 C 10.4 D 21.84
GATE ME 2017 SET-2   Heat Transfer
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.
 A 30 B 80 C 64 D 84
GATE ME 2017 SET-1   Heat Transfer
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}
 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. $C_{min}$ and$C_{max}$ are the minimum and the maximum heat capacity rates. The maximum possible heat transfer ($Q_{max}$) between the two fluids is
 A $C_{min}$ LMTD B $C_{min}$ $\Delta T_{max}$ C $C_{max}$ $\Delta T_{max}$ D $C_{max}$ $\Delta T_{min}$
GATE ME 2016 SET-3   Heat Transfer
Question 7 Explanation:
Maximum heat transfer possible
$=C_{\min } \times(\Delta T)_{\max }$
 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 _________
 A 0.258 B 0.684 C 0.489 D 0.927
GATE ME 2016 SET-2   Heat Transfer
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}
 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
 A $\frac{1-exp[-NTU(1+C_{r})]}{1+C_{r}}$ B $\frac{1-exp[-NTU(1-C_{r})]}{1-C_{r} exp[-NTU(1-C_{r})]}$ C $\frac{NTU}{1+NTU}$ D $1-exp(-NTU)$
GATE ME 2015 SET-3   Heat Transfer
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
$\frac{1-\exp \left[-N T U\left(1+C_{r}\right)\right]}{1-C_{f} \exp \left[-N T U\left(1+C_{r}\right)\right]}$
$\therefore \quad c_{r}=0$
$1-\exp (-N T U)$
 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
 A 290 B 300 C 320 D 350
GATE ME 2015 SET-2   Heat Transfer
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
$320-T_{c 2}=20$
$\text{or }\quad T_{c 2}=300 \mathrm{K}$
There are 10 questions to complete.