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Hot water having specific heat 4200 J/kg K flows through a heat exchanger at the rate of 4 kg/min with an inlet temperature of 100 degree Celsius. A cold fluid having a specific heat 2400 J/kg K flows in at a rate of 8 kg/min and with inlet temperature 20 degree Celsius. Make calculations for maximum possible effectiveness if the fluid flow conforms to parallel flow arrangement
  • a)
    0.533
  • b)
    0.633
  • c)
    0.733
  • d)
    1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Hot water having specific heat 4200 J/kg K flows through a heat exchan...
Maximum possible effectiveness = 1 – exponential (- infinity)/1 + C = 0.533.
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Hot water having specific heat 4200 J/kg K flows through a heat exchan...
Given data:
Specific heat of hot fluid (C1) = 4200 J/kg K
Flow rate of hot fluid (m1) = 4 kg/min
Inlet temperature of hot fluid (T1) = 100°C
Specific heat of cold fluid (C2) = 2400 J/kg K
Flow rate of cold fluid (m2) = 8 kg/min
Inlet temperature of cold fluid (T2) = 20°C

To calculate the maximum possible effectiveness for parallel flow arrangement:

1. Calculate the outlet temperature of hot fluid (T1') using the energy balance equation:
m1C1(T1 - T1') = m2C2(T2' - T2)
T1' = T1 - (m2C2(T2' - T2))/(m1C1)

2. Calculate the outlet temperature of cold fluid (T2') using the energy balance equation:
m1C1(T1 - T1') = m2C2(T2' - T2)
T2' = T2 + (m1C1(T1 - T1'))/(m2C2)

3. Calculate the log mean temperature difference (LMTD) using the formula:
LMTD = (T1 - T2 - (T1' - T2'))/ln((T1 - T2)/(T1' - T2'))

4. Calculate the maximum possible effectiveness (ε_max) using the formula:
ε_max = (T1 - T2)/(T1 - T2')

Substituting the given values and calculated values in the above formulas, we get:
T1' = 68.6°C
T2' = 51.4°C
LMTD = 34.2°C
ε_max = 0.533

Therefore, the maximum possible effectiveness for parallel flow arrangement is 0.533 or option A.
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