Heat Exchanger
Exam Duration: 45 Mins Total Questions : 20
In a counter flow heat exchanger, the product of specific heat and mass flow rate is same for the hot and cold fluids. If NTU is equal to 0.5, then the effectiveness of the heat exchanger is
- (a)
1.0
- (b)
0.5
- (c)
0.33
- (d)
0.2
As Ch = Cc
\(\therefore R={C_{min}\over C_{max}}=1\)
For counter flow
\(\varepsilon ={1-e^{-NTU(1-R)}\over 1-Re^{-NTU(1-R)}}={0\over0}\)
So, apply L Hospital rule,
\(\Rightarrow lim_{R\rightarrow 1}\varepsilon =lim_{R\rightarrow 1}{1-e^{-NTU(1-R)}\over 1-Re^{-NTU(1-R)}}\)
\(={NTU\over 1+NTU}={0.5\over1.5}={1\over3}\)
Which one of the following diagrams correctly shows the temperature distribution for a gas to gas counter flow heat exchanger?
- (a)
- (b)
- (c)
- (d)
For gas to gas, heat exchange does not involve latent heat.
A condenser is designed to condense 0.76 kg/min of steam with cooling water entering at 20 oC and leaving at 65 oC. Overall heat transfer coefficient = 3400 W/m2 -K. The surface area required for this heat exchanger is (saturation temperature of steam = 95.6oC, hfgsteam = 2270 kJ/kg)
- (a)
0.17 cm2
- (b)
0.27m2
- (c)
0.17m2
- (d)
0.15m2
Applying energy balance equation
Heat lost by steam = Heat gained by cooling water
= Heat transfer across the system
\(\Rightarrow\)Q = mh hfg= mCCpc (Tc2 - Tc1)
=UA \(\triangle\)Tm ... (i)
Calculation for LMTD,
\(\triangle\)T1 = Th-1 -Tc1 = 75.6°C
\(\triangle\)T2 = Th2 - Tc2 = 30.6°C
\(\triangle T_m={\triangle T_1-\triangle T_2\over In{\triangle T_1\over \triangle T_2}}=49.75^oC\)
From Eq. (i), Q =UA \(\triangle\)Tm = mh hfg
\(A={0.76\times 2270\times10^3\over 60\times49.75\times3400}\)
=0.17m2
In a parallel flow double pipe heat exchanger water flows through the inner pipe and is heated from 20°C to 70°C. Oil flowing through the annulus is cooled from 200°C to 100 °C. It is desired to cool the a il to a lower exit temperature by increasing the length of the heat exchanger. The minimum temperature to which the oil may be cooled is
- (a)
60°C
- (b)
80°C
- (c)
70°C
- (d)
90°C
Initial case
Required case Exit temperature of oil can be lowered upto the point where it will be equal to exit temperature of cold fluid because after both fluids attain same temperature, no heat transfer will take between them as (\(\triangle\)T = 0). Applying energy balance equation in initial case
Q = Ch (Th1 - Th2 ) = Cc (Tc2 - Tc1 )
\(\Rightarrow\)Ch (200 - 100) = Cc (70 - 20)
\({C_h\over C_c}=0.5\)....(i)
Now, in required case,
Q = Ch (Th1- T)
= Cc (T- Tc1)
\({C_h\over C_c}(200-T)=(T-20)\)
\(\Rightarrow\)T = 80°C
In a counter flow heat exchanger, hot fluid enters at 60°C and cold fluid leaves at 30°C. Mass flow rate of the hot fluid is 1kg/s and that the cold fluid is 2 kg/s.Specific heat of the hot fluid is 10 kJ/kg-K and that of the cold fluid is 5 kJ/kg-K. The Log Mean Temperature Difference (LMTD) for the heat exchanger in °C is
- (a)
15
- (b)
30
- (c)
35
- (d)
45
Ch = 1 x 10 = 10 kJ/K-s
Cc = 2 x 5 = 10 kJ/K-s
As Ch = Cc
Ch . (Th1 - Th2) = Cc.(Tc1 -Tc2)
\(\triangle\)T1 =\(\triangle\)T2
\(\triangle T_m={\triangle T_1-\triangle T_2\over In{\triangle T_1\over \triangle T_2}}={0\over0}\)
(when \(\triangle\)T1 =\(\triangle\)T2)
\(\therefore\) Applying L Hospital rule,
\(lim_{\triangle T_1\rightarrow \triangle T_2}\triangle T_m=lim_{\triangle T_1\rightarrow \triangle T_2}\)\({\triangle T_1-\triangle T_2\over In \triangle T_1/ \triangle T_2}\)
(differentiate w.r.t to \(\triangle\)T1)
\(=lim_{\triangle T_1\rightarrow \triangle T_2}[{1-0\over {1\over \triangle T_1}}]\)
\(=lim_{\triangle T_1\rightarrow \triangle T_2}(\triangle T_1)=\triangle T_2\)
and we know \(\triangle T_1=\triangle T_2=60-30=30^oC\)
AMTD (Arithmetic Mean Temperature Difference) will be 5% higher than LMTD (Log Mean Temperature Difference) when \({\triangle T_1\over \triangle T_2}\) is equal to
- (a)
3.2
- (b)
2.2
- (c)
2.0
- (d)
3.0
AMTD = \({\triangle T_1+\triangle T_2\over 2}\)
LMTD =\({\triangle T_1+\triangle T_2\over in {\triangle T_1\over \triangle T_2}}\)
Given, AMTD=LMTD+LMTD x \({5\over 100}\)
\({\triangle T_1+\triangle T_2\over 2}={\triangle T_2({\triangle T_1\over \triangle T_2}-1)\over in {\triangle T_1\over \triangle T_2}}\times\)\({5\over 100}\)
\(\Rightarrow \triangle T_1 / \triangle T_2=2.2\)
For a given counter flow heat exchanger, given m1 = m2 = 1 kg/s.
Th1 = 420oC, TC1 = 20oC, C1 = 1 kJ/kg-K, C2 = 4 kJ/kg-K
Exit temperature To2 of fluid 2 is equal to
- (a)
85°C
- (b)
95°C
- (c)
190°C
- (d)
105°C
By energy balance,
Ch(Th1-Th2) = Cc (Tc2 - Tc1) =Q
Tc2=\({300\over4}+20\)
Tc2=95oC
An oil flows into a double pipe counter flow heat exchanger at the rate of 0.4 kg/s. The inner tube has internal diameter of 2.0 cm and the inner diameter of the outer tube is 3.0 cm. The water flows in the inner tube at the rate of 0.6 kg/s with its inlet and exit temperatures as 200C and 50oC respectively. The convective heat transfer coefficient on water side is 8000 W/m2-K and on oil side of tube is 80 W/m2-K. The oil temperature at inlet is 160°C. Neglecting the thickness of both the inner and outer tubes, then
Overall heat transfer coefficient, if Cpo = 2.25 kJ/kg-K
- (a)
81.07 W/m2-K
- (b)
91.73W/m2-K
- (c)
85.89W/m2-K
- (d)
79.208W/m2-K
As we know,
\(U_oA_o={1\over {1\over h_iA_i}+{In(r_2/r_1)\over 2\pi KL}+{1\over h_oA_o}}\)
\(U_o={1\over {r_2\over r_1}\times {1\over h_1}+{r_2In(r_2/r_1)\over K}+{1\over h_o}}\)
Since, the thickness of the tube have been neglected, the overall heat transfer coefficient can be written as
\(U=U_1=U_o={1\over{1\over h_1}+{1\over h_o}}={1\over {1\over8000}+{1\over80}}\)
= 79.208 W/m2 -K
An oil flows into a double pipe counter flow heat exchanger at the rate of 0.4 kg/s. The inner tube has internal diameter of 2.0 cm and the inner diameter of the outer tube is 3.0 cm. The water flows in the inner tube at the rate of 0.6 kg/s with its inlet and exit temperatures as 200C and 50oC respectively. The convective heat transfer coefficient on water side is 8000 W/m2-K and on oil side of tube is 80 W/m2-K. The oil temperature at inlet is 160°C. Neglecting the thickness of both the inner and outer tubes, then
Heat transfer rate
- (a)
77.36 kW
- (b)
80.09 kW
- (c)
75.36 kW
- (d)
70.73 kW
Heat transfer rate Q -
Assuming, Cpw= 4.187 kJ/kg-K
As Q = mwCpw (Tc1 -Tc2)
=mo Cpo (Th1 - Th2 )
=Uo Ao \(\triangle\)Tm
= 0.6 x 4.187 x (60 - 30)
= 75.366 kJ/s or kW
An oil flows into a double pipe counter flow heat exchanger at the rate of 0.4 kg/s. The inner tube has internal diameter of 2.0 cm and the inner diameter of the outer tube is 3.0 cm. The water flows in the inner tube at the rate of 0.6 kg/s with its inlet and exit temperatures as 200C and 50oC respectively. The convective heat transfer coefficient on water side is 8000 W/m2-K and on oil side of tube is 80 W/m2-K. The oil temperature at inlet is 160°C. Neglecting the thickness of both the inner and outer tubes, then
Exit temperature of oil Th2 is
- (a)
70oC
- (b)
81oC
- (c)
76.26oC
- (d)
93oC
Exit temperature of oil Th2
Again, Q = mo.Cpo(Th1-Th2)
75.366 = 0.4 x 2.25 x (160 - Th2) Th2 = 76.26°C
If, Th1 = 500oC, Th2 = 300oC, TC1 = 50oC, TC2 = 140oC,di = 5 cm,do = 6 cm, hi = 250 W/m2-K,ho = 400 W/m2-K.Assume thermal conductivity of tube is very large.
Length of tube, if Q = 100 kW
- (a)
130 m
- (b)
145 m
- (c)
138 m
- (d)
149 m
As K= \(\infty\)
\(\therefore R_{tube}={Inr_2 / r_1\over 2\pi KL}=0\)
By energy balance, Q = Ch (Th1 - Th2) = Cc(TC2 - Tc1)
= U· A· \(\triangle \)Tm
As Ch or Cc are not given so,
Q = U.A.\(\triangle \)Tm
Here, if U = Uo, then A = Ao = \(\pi d_o\)L
We get, L = 138.088 m
If, Th1 = 500oC, Th2 = 300oC, TC1 = 50oC, TC2 = 140oC,di = 5 cm,do = 6 cm, hi = 250 W/m2-K,ho = 400 W/m2-K.Assume thermal conductivity of tube is very large.
Number of tubes, if tube length is 3 m is
- (a)
50
- (b)
45
- (c)
47
- (d)
42
As K= \(\infty\)
\(\therefore R_{tube}={Inr_2 / r_1\over 2\pi KL}=0\)
If length of each tube = 3 m
Then, number of tubes =\({138.088\over3}=46.03=47\)
In a counter flow heat exchanger, a hot fluid enters at 80°C having heat capacity Ch = 2 kJ/kg-K. Cold water enters at 20oC having heat capacity Cc = 4 kJ/kg-K. If NTU = 0.25, then exit temperature of hot fluid is
- (a)
\(T_{h_2}\cong 58^oC\)
- (b)
\(T_{h_2}\cong 50^oC\)
- (c)
\(T_{h_2}\cong 64^oC\)
- (d)
\(T_{h_2}\cong 68^oC\)
Given,
T = 80oC,
Ch = 2 kJ/kg-K
Tc1 = 20oC,
Cc= 4 kJ/kg-K
NTU = 0.25, TC2 =?
As Ch < Cc,
\(\therefore\)Th2 = Tc1 (for maximum rate of heat transfer)
Hence,
\(\varepsilon ={C_h(T_{h1}-T_{h2})\over C_h(T_{h1}-T_{C_1})}=\)\({C_c(T_{c2}-T_{c1})\over C_h(T_{h1}-T_{C_1})}\)......(i)
and for \(\varepsilon _{counter}={1-e^{-NTU(1-R)}\over 1-Re^{-NTU(1-R)}}\)......(ii)
\(R={2\over4}=0.5\)
From Eqs. (i) and (ii)
\({2(80-T_{h_2})\over2(80-20)}={1-e^{-0.5(1-0.5)}\over 1-0.5e^{-0.5(1-0.5)}}={0.221\over 0.610}\)
Th2 = 80 - 21.73 = 58.2°C
The temperature distribution Curve for a heat exchanger as shown in the figure below (with usual notations) refers to which one of the following?
- (a)
Tubular parallel flow heat exchanger
- (b)
Tube in tube counter flow heat exchanger
- (c)
Boiler
- (d)
Condenser
In a double pipe counter flow heat exchanger, if Ch = Cc, then temperature profiles of two fluids along the length will be
- (a)
- (b)
- (c)
- (d)
A liquid having specific heat of 3.3 kJ/kg-K flowing at the rate of 20000 kg/h enters a parallel flow heat exchanger at 100oC. The flow rate of cooling water is 50000 kg/h with an inlet temperature of 30oC.The heat transfer area is 10m2 and the overall heat transfer coefficient is 1050 W/m2-K. Take for water, specific heat = 4.186 kJ/kg-K.
Effectiveness of the heat exchanger will be
- (a)
0.2
- (b)
0.3
- (c)
0.4
- (d)
0.6
Heat capacity,
Ch = mh Cph
= 556 x 3.3
= 18.35 kJ/K
For cold fluid,
Cc=mc Cpc
= 13.89 x 4.186
= 58.14 kJ/K
\(\because\)Ch < Cc
\(\therefore\)Cmin = Ch = 18.35
\(NTC={UA\over C_{min}}\)
\(={1050\times 10\over 108.35\times 1000}=0.572\)
and \(R={C_{min}\over C_{max}}={18.35\over 58.4}=0.316\)
Effectiveness (for parallel flow),|
\(\varepsilon ={1-e^{-NTU}(1+R)\over 1+R}\)
\(={1-e^{-0.572(1+0316)}\over 1+0.316}=0.402\)
A liquid having specific heat of 3.3 kJ/kg-K flowing at the rate of 20000 kg/h enters a parallel flow heat exchanger at 100oC. The flow rate of cooling water is 50000 kg/h with an inlet temperature of 30oC.The heat transfer area is 10m2 and the overall heat transfer coefficient is 1050 W/m2-K. Take for water, specific heat = 4.186 kJ/kg-K.
The outlet temperature of chemical will be
- (a)
72oC
- (b)
76oC
- (c)
80oC
- (d)
82oC
As Ch < C
\(\therefore T_{h_o}\rightarrow T_{ci}\)
\(\varepsilon ={C_h(T_{hi}-T_{ho})\over C_{min}(T_{hi}-T_{ci})}\)
\(\therefore 0.402={100-T_{ho}\over 100-30}\)
or Tho = 71.8°C
A one ton window air-conditioner removes 3.5 kJ/s from a room and in the process rejects 4.2 kJ/s in the air-cooled condenser. The ambient temperature is 30oC whereas condensing temperature of the refrigerant is 45oC. For the condenser, the product of overall heat transfer coefficient and corresponding area is 350 W/K. The temperature rise of the air as it flows over the condenser tubes is
- (a)
35oC
- (b)
5oC
- (c)
40oC
- (d)
None of these
For condenser, the temperature distribution along the length of tubes can be represented as
By energy balance 45oC
Heat lost by the refrigerant in condenser = Heat transferred to air
\(Q=m\times h_{fg}\)
\(=C_c(T_{c2}-T_{c1})\)
\(=UA\triangle T_m\)
Given, Q=4.2 kJ/s and U/A=350 W/K
\(\therefore 4.2 \times 1000=UA\theta_m=UA{\theta_1-\theta_2\over In(\theta_1/\theta_2)}\)
\(=UA \times {(T_{h1}-T_{c1})-(T_{h_2}-T_{c2})\over In({T_{h_1}-T_{c1}\over T_{h_2}-T_{c2}})}\)
\(=UA{T_{c2}-T_{c1}\over In({T_{h_1}-T_{c1}\over T_{h_2}-T_{c2}})}\)
Here Th1=Th2
\(\Rightarrow 4200=350\times{(T_{c2}-30)\over In({45-30\over 45-T_{c2}})}\)
\(\therefore {T_{c2}-30\over In({15\over 45-T_{c2}})}={4200\over 350}=12\)
We get, Tc2= 35°C
\(\therefore\)Temperature rise of air = 35 - 30 = 5°C
If Ch = Cc and NTU = 0.5, then which will be true?
- (a)
\(\varepsilon \)parallel > \(\varepsilon \)counter
- (b)
\(\varepsilon \)parallel = \(\varepsilon \)counter
- (c)
\(\varepsilon \)parallel < \(\varepsilon \)counter
- (d)
None of these
We know for Ch = Cc,
\(\varepsilon _{parallel}={1-e^{-2NTU}\over 2}\)
\(={1-e^{-2\times 0.5}\over 2}=0.315\)
\(\varepsilon _{counter}={NTU\over 1+NTU}={0.5\over 1.5}=0.33\)
\(\Rightarrow \varepsilon_c>\varepsilon _p\)
For evaporators, what will be effectiveness, if NTU = 0.5?
- (a)
\(\varepsilon_p>\varepsilon _c,\varepsilon _p=0.39\)
- (b)
\(\varepsilon_p<\varepsilon _c,\varepsilon _c=0.19\)
- (c)
\(\varepsilon_p=\varepsilon _c=0.39\)
- (d)
\(\varepsilon_p=\varepsilon _c=0.19\)
For evaporators, Cmax = \(\infty\)
(Because, Q = C \(\triangle\)T and \(\triangle\)T = 0 \(\Rightarrow\) C =\(\infty\))
\(\therefore R={C_{min}\over C_{max}}=0\)
\(\varepsilon _{parallel}={1-e^{-NTU(1+R)}\over 1+R}\)
\(={1-e^{-0.5}\over 1}=0.39\)
\(\varepsilon _{counter}={1-e^{-NTU(1-R)}\over 1-Re^{-NTU(1-R)}}\)
= 1 - e-0·5 = 0.39
\(\therefore \varepsilon_P=\varepsilon _c\)