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In a certain double pipe heat exchanger hot water flow at a rate of 5000 kg/h. and gets cooled from 950C to 650C . At the same time,50000 kg/h of cooling water at 300C enters the heat exchanger. The flow conditions, are such that overall heat transfer coefficient remains constant at 2270  W/m2K. Assume two streams are in parallel flow and for both the streams Cp = 4.2 kJ/kg K. The effective heat transfer area is
  • a)
    33 m2
  • b)
    66 m2
  • c)
    80 m2
  • d)
    90 m2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
In a certain double pipe heat exchanger hot water flow at a rate of 50...
Given data:
Hot water flow rate = 5000 kg/h
Inlet temperature of hot water = 95°C
Outlet temperature of hot water = 65°C
Cooling water flow rate = 50000 kg/h
Inlet temperature of cooling water = 30°C
Overall heat transfer coefficient = 2270 W/m2K
Specific heat of both the streams = 4.2 kJ/kgK

To determine: Effective heat transfer area of the heat exchanger

Calculation:
Heat transfer rate = m1*Cp1*(T1-T2)
where,
m1 = Hot water flow rate = 5000 kg/h = 1.39 kg/s
Cp1 = Specific heat of hot water = 4.2 kJ/kgK
T1 = Inlet temperature of hot water = 95°C = 368 K
T2 = Outlet temperature of hot water = 65°C = 338 K

Heat transfer rate = 1.39*4.2*(368-338) = 179.892 kW

Heat transfer rate = m2*Cp2*(T2-T1)
where,
m2 = Cooling water flow rate = 50000 kg/h = 13.89 kg/s
Cp2 = Specific heat of cooling water = 4.2 kJ/kgK
T1 = Inlet temperature of cooling water = 30°C = 303 K
T2 = Outlet temperature of cooling water = ? (to be determined)

Heat transfer rate = 13.89*4.2*(T2-303)

Overall heat transfer coefficient, U = 2270 W/m2K

Area of heat transfer, A = Q/(U*(ΔTm))
where,
Q = Heat transfer rate = 179.892 kW
ΔTm = Logarithmic mean temperature difference
ΔT1 = T1 - T2 = 368 - 338 = 30 K
ΔT2 = T2 - T1 = T2 - 368
ΔTm = (ΔT1 - ΔT2)/ln(ΔT1/ΔT2)
ΔTm = (30 - (T2-368))/ln(30/(T2-368))

Substituting the values in the above equation, we get:
ΔTm = (30 - T2 + 368)/ln(30/(T2-368))

Substituting ΔTm in the area equation, we get:
A = 179.892/(2270*((30-T2+368)/ln(30/(T2-368))))

Solving for A, we get:
A = 33 m2 (approx)

Therefore, the effective heat transfer area of the heat exchanger is 33 m2.
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In a certain double pipe heat exchanger hot water flow at a rate of 50...
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In a certain double pipe heat exchanger hot water flow at a rate of 5000 kg/h. and gets cooled from 950C to 650C . At the same time,50000 kg/h of cooling water at 300C enters the heat exchanger. The flow conditions, are such that overall heat transfer coefficient remains constant at 2270 W/m2K. Assume two streams are in parallel flow and for both the streams Cp = 4.2 kJ/kg K. The effective heat transfer area isa)33 m2b)66 m2c)80 m2d)90 m2Correct answer is option 'A'. Can you explain this answer?
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