1. In shell and tube heat exchanger, baffles are mainly used to
(a) Increase the mixing of fluid
(b) Increase the heat transfer area
(c) Deflect the flow in desired direction
(d) Reduce fouling of the tube surface
(1 Mark, 1991)
2. The practice to use steam on the shell side and cooling water on the tube side in condensers of steam power plant is because
(a) To increase overall heat transfer coefficient water side velocity can be increased if water is at the tube side
(b) Condenser can act as a storage unit for condensed steam
(c) Rate of condensation of steam is invariably smaller than the mass flow rate of cooling water
(d) It is easier to maintain vacuum on the shell side than on the tube side
(1 Mark, 1994)

Ans: d
3. In a certain heat exchanger, both the fluids have identical mass flow ratespecific heat product. The hot fluid enters at 76^{0}C and leaves at 47^{0}C and the cold fluid entering at 26^{0}C leaves at 55^{0}C. The effectiveness of the heat exchanger is
(a) 0.16
(b) 0.60
(c) 0.72
(d) 1.0
(2 Mark, 1997)

Ans: b
Explanation:
{ \dot { m } }_{ c }{ c }_{ pc } = { \dot { m } }_{ h }{ c }_{ ph } = { \left( \dot { m } { c }_{ p } \right) }_{ min }
Now, Effectiveness, ε=\frac { { \dot { m } }_{ h }{ c }_{ ph }\left( { T }_{ h1 }{ T }_{ h2 } \right) }{ { \left( \dot { m } { c }_{ p } \right) }_{ min }\left( { T }_{ h1 }{ T }_{ c1 } \right) }
=> ε=\frac { \left( { T }_{ h1 }{ T }_{ h2 } \right) }{ \left( { T }_{ h1 }{ T }_{ c1 } \right) }
=> ε=\frac { 7647 }{ 7626 } = 0.6
4. Air enters a counterflow heat exchanger at 70^{0}C and leaves at 40^{0}C. Water enters at 30^{0}C and leaves at 50^{0}C. The LMTD in deg. C is
(a) 5.65
(b) 14.43
(c) 19.52
(d) 20.17
(2 Mark, 2000)

Ans: b
Explanation:
\Delta { T }_{ 1 } = 70^{0}C – 50^{0}C = 20^{0}C
\Delta { T }_{ 2 } = 40^{0}C – 30^{0}C = 10^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 2010 }{ ln\left( \frac { 20 }{ 10 } \right) } = 14.43
5. For the same inlet and outlet temperatures of hot and cold fluids, the Log Mean Temperature Difference (LMTD) is
(a) Greater for parallel flow heat exchanger than for counter flow heat exchanger.
(b) Greater for counter flow heat exchanger than for parallel flow heat exchanger.
(c) Same for both parallel and counter flow heat exchangers.
(d) Dependent on the properties of the fluids.
(1 Mark, 2002)

Ans:b
6. In a counter flow heat exchanger, for the hot fluid the heat capacity = 2 kJ/kgK, mass flow rate = 5 kg/s, inlet temperature =150^{0}C, outlet temperature =100^{0}C. for the cold fluid, heat capacity = 4 kJ/kgK, mass flow rate = 10 kg/s, inlet temperature = 20^{0}C. Neglecting heat transfer to the surroundings, the outlet temperature of the cold fluid in ^{0}C is
(a) 7.5
(b) 32.5
(c) 45.5
(d) 70.0
(2 Mark, 2003)

Ans:b
Explanation:
We know: (mC)_{h}(T_{h1} – T_{h2}) = (mC)_{c}(T_{c2} – T_{c1})
=> 2 * 5(150 – 100) = 4 * 10(T_{c2} – 20)
=> T_{c2} = 32.5^{0}C
7. In a condenser, water enters a 30^{0}C and flows at the rate 1500 kg/hr. The condensing steam is at a temperature of 120^{0}C and cooling water leaves the condenser at 80^{0}C. Specific heat of water is 4.187 kJ/kgK. If the overall heat transfer coefficient is 2000 W/m^{2} K, the heat transfer area is
(a) 0.707 m^{2}
(b) 7.07 m^{2}
(c) 70.7 m^{2}
(d) 141.4 m^{2}
(2 Mark, 2004)

Ans: a
Explanation:
\Delta { T }_{ 1 } = 120^{0}C – 30^{0}C = 90^{0}C
\Delta { T }_{ 2 } = 120^{0}C – 80^{0}C = 40^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 9040 }{ ln\left( \frac { 90 }{ 40 } \right) } = 61.657^{0}C
Again, (mc)_{c}(T_{c2} – T_{c1}) = UA\Delta { T }_{ m }
=> \frac { 1500 }{ 3600 } * 4187 * (80 – 30) = 2000 * A * 61.657
=> A = 0.707 m^{2}
8. Hot oil is cooled from 80 to 50°C in an oil cooler which uses air as the coolant. The air temperature rises from 30 to 40°C. the designer uses a LMTD value of 26°C. the type of heat exchanger is
(a) Parallel flow
(b) Double pipe
(c) Counter flow
(d) Cross flow
(2 Mark, 2005)

Ans: d
Explanation:
For parallel flow:
\Delta { T }_{ 1 } = 80^{0}C – 30^{0}C = 50^{0}C
\Delta { T }_{ 2 } = 50^{0}C – 40^{0}C = 10^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 5010 }{ ln\left( \frac { 50 }{ 10 } \right) } = 24.85^{0}C
For counter flow:
\Delta { T }_{ 1 } = 80^{0}C – 40^{0}C = 40^{0}C
\Delta { T }_{ 2 } = 50^{0}C – 30^{0}C = 20^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 4020 }{ ln\left( \frac { 40 }{ 20 } \right) } = 28.85^{0}C
We know: { \left( LMTD \right) }_{ PF }\le { \left( LMTD \right) }_{ CrossFlow }\le { \left( LMTD \right) }_{ CF }
As, 24.85^{0}C < 26^{0}C < 28.85^{0}C.
So, the Heat exchanger is crossflow.
9. In a counter flow heat exchanger, hot fluid enters at 60^{0}C and cold fluid leaves at 30^{0}C. Mass flow rate of the hot fluid is 1 kg/s and that of the cold fluid is 2 kg/s. Specific heat of the hot fluid is 10 kJ/kgK and that of the cold fluid is 5 kJ/kgK. The Log Mean Temperature Difference (LMTD) for the heat exchanger in ^{0}C is
(a) 15
(b) 30
(c) 35
(d) 45
(2 Mark, 2007)
10. The logarithmic mean temperature difference (LMTD) of a counterflow heat exchanger is 20^{0}C. The cold fluid enters at 20^{0}C and the hot fluid enters at 100^{0}C. Mass flow rate of the cold fluid is twice that of the hot fluid. Specific heat at constant pressure of the hot fluid is twice that of the cold fluid. The exit temperature of the cold fluid
(a) Is 40^{0}C
(b) Is 60^{0}C
(c) Is 80^{0}C
(d) Cannot be determined
(2 Mark, 2008)
11. In a parallel flow heat exchanger operating under steady state, the heat capacity rates (product of specific heat at constant pressure and mass flow rate) of the hot and cold fluid are equal. The hot fluid, flowing at 1kg/s with C_{p} = 4kJ/kgK, enters the heat exchanger at 102°C while the cold fluid has an inlet temperature of 15°C. The overall heat transfer coefficient for the heat exchanger is estimated to be 1kW/m^{2}K and the corresponding heat transfer surface area is 5m^{2}. Neglect heat transfer between the heat exchanger and the ambient. The heat exchanger is characterized by the following relation: 2ε = 1 − exp (−2NTU). The exit temperature (in °C) for the cold fluid is
(a) 45
(b) 55
(c) 65
(d) 75
(2 Mark, 2009)

Ans: b
Explanation:
Here, { \dot { m } }_{ c }{ c }_{ pc }={ \dot { m } }_{ h }{ c }_{ ph }_{ }= 1 * 4 = 4 kJ/K = { \left( \dot { m } { c }_{ p } \right) }_{ min }
We know: m_{h}c_{ph}(T_{h1} – T_{h2}) = m_{c}c_{pc}(T_{c2} – T_{c1})
=> (T_{h1} – T_{h2}) = (T_{c2} – T_{c1})
Again, NTU = \frac { UA }{ { C }_{ min } } =\frac { 1\times 5 }{ 4 } = 1.25
Effectiveness \left( ε \right) =\frac { 1−exp(−2NTU) }{ 2 }
=> ε=\frac { 1−exp(−2\times 1.25) }{ 2 } = 0.46
Again, ε=\frac { { Q }_{ actual } }{ { Q }_{ max } } =\frac { { \dot { m } }_{ c }{ c }_{ pc }\left( { T }_{ c2 }{ T }_{ c1 } \right) }{ { \left( \dot { m } { c }_{ p } \right) }_{ min }\left( { T }_{ h1 }{ T }_{ c1 } \right) }
=> ε=\frac { \left( { T }_{ c2 }{ T }_{ c1 } \right) }{ \left( { T }_{ h1 }{ T }_{ c1 } \right) } =\frac { { T }_{ c2 }15 }{ 10215 }
=> T_{c2 }= 55^{0}C
12. An industrial gas (C_{p} = 1 kJ/kgK) enters a parallel flow heat exchanger at 250^{0}C with a flow rate of 2 kg/s to heat a water stream. The water stream (C_{p} = 4 kJ/kgK) enters the heat exchanger at 50^{0}C with a flow rate of 1kg/s. The heat exchanger has an effectiveness of 0.75. The gas stream exit temperature will be
(a) 75^{0}C
(b) 100^{0}C
(c) 125^{0}C
(d) 150^{0}C
(2 Mark, 2010)

Ans: b
Explanation:
{ \dot { m } }_{ c }{ c }_{ pc } = 4 * 1 = 4 kJ/K
{ \dot { m } }_{ h }{ c }_{ ph } = 2 * 1 = 2 kJ/K = { \left( \dot { m } { c }_{ p } \right) }_{ min }
Now, Effectiveness, ε=\frac { { \dot { m } }_{ h }{ c }_{ ph }\left( { T }_{ h1 }{ T }_{ h2 } \right) }{ { \left( \dot { m } { c }_{ p } \right) }_{ min }\left( { T }_{ h1 }{ T }_{ c1 } \right) }
=> .75=\frac { \left( 250{ T }_{ h2 } \right) }{ \left( 25050 \right) }
=> T_{h2 }= 100^{0}C
13. Cold water flowing at 0.1 kg/s is heated from 20^{0}C and 70^{0}C in a counter flow type heat exchanger by a hot water stream flowing at 0.1kg/s and entering at 90^{0}C. The specific heat of water is 4200 J/kgK and density is 1000 kg/m^{3}. If the overall heat transfer coefficient (U) for the heat exchanger is 2000 W/(m^{2}K), the required heat exchange area (in m^{2}) is
(a) 0.052
(b) 0.525
(c) 0.151
(d) 0.202
(2 Mark, 2011(PI))

Ans: b
Explanation:
Here, (mc_{p})_{h} = (mc_{p})_{c}
So, (T_{h1} – T_{h2}) = (T_{c2} – T_{c1})
=> (90 – T_{h2}) = (70 – 20) => T_{h2 }= 40^{0}C
\Delta { T }_{ 1 } = 90^{0}C – 70^{0}C = 20^{0}C
\Delta { T }_{ 2 } = 40^{0}C – 20^{0}C = 20^{0}C
\Delta { T }_{ m } = ΔT_{1 }= ΔT_{2 }= 20^{0}C
=> (mc_{p})_{h}(90 – 40) = UA\Delta { T }_{ m }
=> 0.1 * 4200 * (90 – 40) = 2000 * A * 20
=> A = 0.525 m^{2}
14. In a condenser of a power plant, the steam condenses at a temperature of 60^{0}C. The cooling water enters at 30^{0}C and leaves at 45^{0}C. The logarithmic mean temperature difference (LMTD) of the condenser is
(a) 16.2^{0}C
(b) 21.6^{0}C
(c) 30^{0}C
(d) 37.5^{0}C
(1 Mark, 2011)

Ans: b
Explanation:
\Delta { T }_{ 1 } = 60^{0}C – 30^{0}C = 30^{0}C
\Delta { T }_{ 2 } = 60^{0}C – 45^{0}C = 15^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 3015 }{ ln\left( \frac { 30 }{ 15 } \right) } = 21.6
15. Water (C_{p} = 4.18 kJ/kgK) at 80^{0}C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/s. Air (C_{p} = 1 kJ/kgK) enter at 30^{0}C with a mass flow rate 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in ^{0}C) is
(a) 40
(b) 20
(c) 10
(d) 5
(2 Mark, 2012)

Ans: c
Explanation:
Here, { \dot { m } }_{ c }{ c }_{ pc }={ \dot { m } }_{ h }{ c }_{ ph }_{ }= 2.09 kJ/K = { \left( \dot { m } { c }_{ p } \right) }_{ min }
Now, Effectiveness, ε=\frac { { \dot { m } }_{ h }{ c }_{ ph }\left( { T }_{ h1 }{ T }_{ h2 } \right) }{ { \left( \dot { m } { c }_{ p } \right) }_{ min }\left( { T }_{ h1 }{ T }_{ c1 } \right) }
=> ε=\frac { \left( { T }_{ h1 }{ T }_{ h2 } \right) }{ \left( { T }_{ h1 }{ T }_{ c1 } \right) }
=> 0.8=\frac { 80{ T }_{ h2 } }{ 8030 }
=> T_{h2 }= 40^{0}C
Now LMTD = ΔT1 = ΔT_{2 }= 40 – 30 = 10^{0}C
16. In a heat exchanger, it is observed that \Delta { T }_{ 1 }=\Delta { T }_{ 2 }, where \Delta { T }_{ 1 } is the temperature difference between the two single phase fluid streams at one end and \Delta { T }_{ 2 } is the temperature difference at the other end. This heat exchanger is
(a) A condenser
(b) An evaporator
(c) A counter flow heat exchanger
(d) A parallel flow heat exchanger
(1 Mark, 2014[2])

Ans: c
Explanation:
It is only possible in CFHE.
17. In a concentric counter flow heat exchanger, water flows through the inner tube at 25^{0}C and leaves at 42^{0}C. The engine oil enters at 100°C and flows in the annular flow passage. The exit temperature of the engine oil is 50^{0}C. Mass flow rate of water and the engine oil are 1.5 kg/s and 1 kg/s, respectively. The specific heat of water and oil are 4178 J/kg.K and 2130 J/kg.K, respectively. The effectiveness of this heat exchanger is _______.
(2 Mark, 2014[2])

Ans: 0.667
Explanation:
{ \dot { m } }_{ c }{ c }_{ pc } = 1.5 * 4178 = 6267 W/K
{ \dot { m } }_{ h }{ c }_{ ph } = 1 * 2130 = 2130 W/K = { \left( \dot { m } { c }_{ p } \right) }_{ min }
ε=\frac { { \dot { m } }_{ h }{ c }_{ ph }\left( { T }_{ h1 }{ T }_{ h2 } \right) }{ { \left( \dot { m } { c }_{ p } \right) }_{ min }\left( { T }_{ h1 }{ T }_{ c1 } \right) }=> ε=\frac { 10050 }{ 10025 } = 0.667
18. A doublepipe counterflow heat exchanger transfers heat between two water streams. Tube side water at 19 litre/s is heated from 10^{0}C to 38^{0}C. Shell side water at 25 litre/s is entering at 46^{0}C. Assume constant properties of water, density is 1000 kg/m^{3} and specific heat is 4186 J/kg K. The LMTD (in ^{0}C) is ______.
(2 Mark, 2014[3])

Ans: 11
Explanation:
(mc)_{h}(T_{h1} – T_{h2}) = (mc)_{c}(T_{c2} – T_{c1})
=> 25(46 – T_{h2}) = 19(38 – 10)
=> T_{h2 }= 24.72^{0}C
\Delta { T }_{ 1 } = 46^{0}C – 38^{0}C = 8^{0}C
\Delta { T }_{ 2 } = 24.72^{0}C – 10^{0}C = 14.72^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac {814.72 }{ ln\left( \frac { 8 }{ 14.72 } \right) } = 11
19. 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/kgK) 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
(2 Mark, 2015[2])

Ans: a
Explanation:
(mc_{p})_{h} = (mc_{p})_{c} = 0.4 * 1000 = 400 W/K
From energy balance:
(mc_{p})_{h} (T_{h1} – T_{h2})= (mc_{p})_{c }(T_{c2} – T_{c1})
=> T_{h1} – T_{h2 }= T_{c2} – T_{c1}
=> 300 – T_{h2} = T_{c2 }– 280
=> 300 – T_{c2} = T_{h2}_{ }– 280
Since the temperature difference is same at both side, So LMTD = 300 – T_{c2}
Again, U * A * LMTD = (mc_{p})_{c }(T_{c2} – T_{c1})
=> 20 * 20 * (300 – T_{c2}) = 400 * (T_{c2} – 280)
=> 300 – T_{c2 }= T_{c2} – 280 => T_{c2} = 290 K
20. Saturated vapour is condensed to saturated liquid in a condenser. The heat capasity ratio is C_{r} = \frac { { C }_{ min } }{ { C }_{ max } } . The effectiveness (ε) of the condenser is
(a) \frac { 1exp[NTU\left( 1+{ C }_{ r } \right) ] }{ 1{ C }_{ r } }
(b) \frac { 1exp[NTU\left( 1{ C }_{ r } \right) ] }{ 1{ +C }_{ r }[NTU\left( 1{ C }_{ r } \right) ] }
(c) \frac { NTU }{ 1+NTU }
(d) 1 exp( NTU)
(1 Mark, 2015[3])

Ans: d
21. Consider a parallelflow heat exchanger with area A_{p} and a counterflow heat exchanger with area A_{c} . In both the heat exchangers, the hot stream flowing at 1 kg/s cools from 80^{o}C to 50^{o}C. For the cold stream in both the heat exchangers, the flow rate and the inlet temperature are 2 kg/s and 10^{o}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 ___.
(2 Mark, 2016[2])

Ans: 0.927
Explanation:
We know: (mC)_{h}(T_{h1} – T_{h2}) = (mC)_{c}(T_{c2} – T_{c1})
=> C_{h}(T_{h1} – T_{h2}) = C_{c}(T_{c2} – T_{c1})
=> 1(80 – 50) = 2(T_{c2} – 10) => T_{c2} = 25^{0}C
Parallel flow heat exchanger:
\Delta { T }_{ 1 } = 80^{0}C – 10^{0}C = 70^{0}C
\Delta { T }_{ 2 } = 50^{0}C – 25^{0}C = 25^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 7025 }{ ln\left( \frac { 70 }{ 25 } \right) } = 43.705
Counter flow heat exchanger:
\Delta { T }_{ 1 } = 80^{0}C – 25^{0}C = 55^{0}C
\Delta { T }_{ 2 } = 50^{0}C – 10^{0}C = 40^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 5540 }{ ln\left( \frac { 55 }{ 40 } \right) } = 47.1
Again, Q = { \left( UA\Delta { T }_{ m } \right) }_{ C }={ \left( UA\Delta { T }_{ m } \right) }_{ P }
=> \frac { { A }_{ C } }{ { A }_{ P } } =\frac { { \left( \Delta { T }_{ m } \right) }_{ P } }{ { \left( \Delta { T }_{ m } \right) }_{ C } } =\frac { 43.705 }{ 47.1 } = 0.927
22. For a heat exchanger, ΔT_{max} is the maximum temperature difference and ΔT_{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} ΔT_{max}
(c) C_{max} ΔT_{max}
(d) C_{max }ΔT_{min}
(2 Mark, 2016[3])

Ans: b
23. Saturated steam at 100^{0}C condenses on the outside of a tube. Cold enters the tube at 20^{0}C and exists at 50^{0}C. The value of the Log Mean Temperature Difference (LMTD) is ____^{0}C.
(1 Mark, 2017[1])

Ans: 63.82
Explanation:
\Delta { T }_{ 1 } = 100^{0}C – 20^{0}C = 80^{0}C
\Delta { T }_{ 2 } = 100^{0}C – 50^{0}C = 50^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 8050 }{ ln\left( \frac { 80 }{ 50 } \right) } = 63.82
24. In a counterflow heat exchanger, water is heated at the rate of 1.5 kg/s from 40^{0}C to 80^{0}C by an oil entering at 120^{0}C and leaving 60^{0}C.The specific heats of water and oil are 4.2 kJ/kgK and 2 kJ/kgK, respectively. The overall heat transfer coefficient is 400 W/m^{2}K. The required heat transfer surface area (in m^{2}) is _____.
(a) 0.104
(b) 0.022
(c) 10.4
(d) 21.84
(2 Mark, 2017[2])

Ans: d
Explanation:
\Delta { T }_{ 1 } = 120^{0}C – 80^{0}C = 40^{0}C
\Delta { T }_{ 2 } = 60^{0}C – 40^{0}C = 20^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 4020 }{ ln\left( \frac { 40 }{ 20 } \right) } = 28.86
We know: Q = UA\Delta { T }_{ m }
=> mC_{w}(80 – 40) = UA\Delta { T }_{ m }
=> 1.5 * 4.2 * (80 – 40) = 0.4 * A * 28.86
=> A = 21.83 m^{2}
25. Steam in the condenser of a thermal power plant is to be condensed at a temperature of 30^{0}C with cooling water which enters the tubes of the condenser at 14^{0}C and exits at 22^{0}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).
(2 Mark, 2018[2])

Ans: 1.15
Explanation:
\Delta { T }_{ 1 } = 30^{0}C – 14^{0}C = 16^{0}C
\Delta { T }_{ 2 } = 30^{0}C – 22^{0}C = 8^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 168 }{ ln\left( \frac { 16 }{ 8 } \right) } = 11.54
Heat transfer (Q) = UA\Delta { T }_{ m }
=> Q = 2000 * 50 * 11.54 = 1.15 MW
26. Hot and cold fluids enter a parallel flow double tube heat exchanger at 100^{0}C and 15^{0}C, respectively. The heat capacity rates of hot and cold fluids are C_{h} = 200 W/k and C_{c} = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45^{0}C, the log mean temperature difference (LMTD) of the heat exchanger is ______ K.(round of to two decimal places).
(2 Mark, 2019[2])

Ans: 57.76
Explanation:
From energy balance:
C_{h}(T_{hi} – T_{ho}) = C_{c}(T_{co} – T_{ci})
=> 2000(100 – T_{ho}) = 1200(45 – 15)
=> T_{ho }= 82^{0}C
\Delta { T }_{ 1 } = 100^{0}C – 15^{0}C = 85^{0}C
\Delta { T }_{ 2 } = 82^{0}C – 45^{0}C = 37^{0}C
\Delta { T }_{ m }=\frac { \Delta { T }_{ 1 }\Delta { T }_{ 2 } }{ ln\left( \frac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } \right) } =\frac { 8537 }{ ln\left( \frac { 85 }{ 37 } \right) } = 57.76