In a counter flow heat exchanger, the working parameters for hot fluid are as follows: heat capacity = 2 kJ/kg K, mass flow inlet temperature 150^{o}C , outlet temperature 100^{o}C. For the cold fluid, the parameters are: heat capacity = 4 kJ/kg K,  mass flow rate 10 kg/s, and inlet temperature is = 20^{o}C. The outlet temperature of the cold fluid in ^{o}C is
m_{h}c_{ph}(T_{hi}  T_{he}) = m_{c}c_{pc}(T_{ce}  T_{ci})
∴ 5 x 2000(150 100) = 10 x 4000(T_{ce}  20)
T_{ce} = 32.5^{o}C
Cold water at 10^{o}C enters a counter flow heat exchanger at a rate of 8 kg/s , where it is heated by a hot water that enters the heat exchanger at 70^{o}C and flows at a rate of 2kh/s . Assume specific heat of water to be cp = 4.18 kJ/kg^{o}C . Determine the maximum theoretical heat transfer rate?
Given m_{h} = 2 kg/s
M_{c} = 8 kg/s
c_{ph} = c_{pc} = 4.18kj/kg^{o}C
∴ m_{h}c_{ph} 2 x 4.18 = 8.36 kW/^{o}C
m_{h}c_{ph} 8 x 4.18 = 33.4 kW/^{o}C
∴ m_{h}c_{ph} < m_{c}c_{pc}
For maximum heat exchanger rate,
Q_{max} = (mc_{p})_{min}(T_{hi}  T_{ci})
⇒ Q_{max} = 8.36(7010)
⇒ Q_{max} = 501.6 kW
A cross flow type air heater has an area of 50 m^{2}. The overall transfer coefficient 100 W/m^{2} and heat capacity of both hot and cold streams is 1000 W/K. The value of NTU is
By definition
NTU = UA/(mc_{p})_{min}
In the given situation, both capacity rate (mc_{p}) of both hot and cold fluid are equal
∴ NTU = 100 x 50/1000 = 5
The overall heat transfer coefficient for a shell and tube heat exchanger for clean surfaces is U_{o} = 400 W/m^{2} K. The fouling factor after one year of operation is found to be h_{fo} = 200 W/m^{2} K. The overall heat transfer coefficient at this time is
Given U_{o} = 400 W/m^{2} K
Fouling factor, h_{FO} = 200 W/m^{2}K
∴ New overall heat transfer coefficient can be predicted as follows
1/U = 1/U+1/h_{FO}
1/U = 1/400 + 1/200
U = 33.33 W/m^{2}K
In a solar assisted air conditioning system, 1 kg/sec of ambient air is to be preheated by the same amount of air leaving the system. A counterflow heat exchanger having an area of 60m^{2} with overall heat transfer coefficient of 25 W/m^{2}  K is used for this purpose. Assuming for air is 1kJ/kg K. The effectiveness of the heat exchanger is
∴ ε = NTU/1+NTU
Where NTU = UA/(mc_{p})_{min}
For balanced heat exchanger,
(mc_{p})_{h} = (mc_{p})_{c}
∴ NTU = 60 x 25 / 1000 = 1.5
∴ ε = 1.5/1+1.5 = 0.6
In a doublepipe heat exchanger, the cold fluid is water with inlet temperature of 20^{o}C and mass flow rate of 20 kg/s and the hot fluid which also has water inlet temperature of 80^{o}C and mass flow rate of 10 kg/s. For water c_{p} = 4.2kj/kg ^{o}C.
. What is the maximum temperature to which the cold fluid can be heated in a parallel flow and in a counter flow heat exchanger?
For parallel flow heat exchanger maximum temperature of cold fluid will be reached when both hot and cold fluid exit at the same temperature.
Thus by steady state flow energy equation
⇒ 20 x 4.2 x (T  20) = 10 x 4.2 (80T)
∴ T = 40^{o}C
For counter flow heat exchanger, maximum exit temperature of cold fluid is achieved when effectiveness of the heat exchanger is 1
i.e.
ε = 1
i.e.
(m_{c}c_{pc})(T_{co}  T_{ci})/(m_{c}c_{pc})(T_{hi}  T_{ci}) =1
1 = 20 x 4.2 x [T  20]/10 x 4.2 x [80  20]
T = 50^{o}C
In a doublepipe counter flow heat exchanger 1000kh/h, oil having a specific heat 20 j/kg K is cooled from 150^{o}C to 125^{o}C by 1250 kg/h of fluid having specific heat 16j/kgK. The cold fluid leaves the heat exchanger at 75^{o}C . In this case, the temperature at which the cooling fluid enters the heat exchanger and LMTD are __________
x = 0 x x = l
m_{h}c_{ph} = 1000 x 20 = 20,000j/hrK
m_{c}c_{ph} = 1200 x 16 = 20,000j/hrK
m_{h}c_{ph} = m_{c}c_{ph}
Hence for a balanced Counter flow heat exchangers
ΔT_{m} = ΔT_{i} =ΔT_{e}
∴ ΔT_{m} = 150 75  75^{o}C
T_{hi}  T_{ce} = T_{he}  T_{ci}
T_{ci}  T_{he} = T_{ce}  T_{hi}
T_{ci} = 50^{o}C
A counter flow shell and tube type heat exchanger is used to heat water with hot exhaust gases. The water (c = 4180 j/kgK ) flows at the rate of 2 kg/s and the exhaust gases (c = 1000 j/kg  K) flow at the rate of 5 kg/s. If the heat transfer surface area is 32 m^{2} and the overall heat transfer coefficient is 200 W/m^{2} K , the effectiveness of the heat exchanger is __________.
(B) 0.63
C_{H} = m_{H}c_{H}
∴ C_{H} = 1000 x 5 = 5000 Watt/ ...①
Cold fluid [Water]
C_{c} = m_{c}c_{c}
∴ C_{c} = 4180 x = 8360 Watt/K ...②
From ① & ②
C_{H} < />_{c}
∴ C_{min} = C_{H} = 5000 Watt/K
By definition,
C = C_{min}/C_{max},
C = C C_{H}/C_{C}
C = 5000/8360
= 0.598
& NTU = UA/_{min}
∴ NTU = 32 x 200/ 5000
= 1.28
For counterflow heat exchanger
∴ ε = 0.626
In a counter flow heat exchanger, hot gases enter the system at and leave at . The temperature of the cold air entering the unit is . Its temperature at the exit is . The heat exchanger has an effectiveness of
By steady state flow equation
C_{h}(T_{hi}  T_{he}) = C_{c}(T_{ce}  T_{he})
C_{h}/C_{c} = T_{ce}  T_{ci}/T_{hi}  T_{he}
C_{h}/C_{c} = 9035/20080 = 55/120
C_{h} < />_{C}
∴ ε = C_{h}(T_{hi}  T_{he})/C_{h}(T_{hi}  T_{ci})
ε = 20080/20035
ε = 0.7272
A coaxial tube counter flow heat exchanger is used to cool 0.03 kg/s of benzene from 360 K to 310 K with a counter flow of 0.02 kg/s of water initially at . If the inner tube outside diameter is 2 cm and the overall heat transfer coefficient based on outside area is 650 , determine the required length of the exchanger. Take the specific heats of benzene and water as 1880 and 4175 J/kg‐K, respectively.
(B) 2.60
Q = m_{h}c_{p}(T_{h1}  T_{h2}) = m_{c}c_{c}(T_{c2}  T_{c1})
Q = 0.03 x 1880(360  310)
= 0.02 x 1475(T_{c2}  290)
Q = 2820 w,T_{c2} = 323.8 K
ΔT_{i} = 310 290 ΔT_{e} = 360  323.8
= 20 = 36.2
LMTD(ፀ_{m}) = 36.2  20/ In ^{36.2/20} = 27.3 K
Q = U_{o}A_{o}(θ_{m}) = U_{o}πd_{o}L(θ_{m})
L = 2820/650 x π x 0.02 x 27.3 = 2.53 m
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