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Water (c = 4.18 kJ/kgK) at 80°C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/s. Air (c = 1 kJ/kgK) enter at 30°C with a mass flow rate 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in °C) is
[2012]
  • a)
    40
  • b)
    20
  • c)
    10
  • d)
    5
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Water (c = 4.18 kJ/kgK) at 80°C enters a counter flow heat exchang...

∴ 
⇒ th2 = 40°C

⇒ 

∴ LMTD = 10°C
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Most Upvoted Answer
Water (c = 4.18 kJ/kgK) at 80°C enters a counter flow heat exchang...
°C is to be cooled to 20°C. Calculate the amount of heat that must be removed from 1 kg of water.

The specific heat capacity of water is c = 4.18 kJ/kgK.

The change in temperature is ΔT = 80°C - 20°C = 60°C.

The amount of heat required to cool 1 kg of water is:

Q = m c ΔT

Q = 1 kg × 4.18 kJ/kgK × 60°C

Q = 250.8 kJ

Therefore, 250.8 kJ of heat must be removed from 1 kg of water to cool it from 80°C to 20°C.
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Water (c = 4.18 kJ/kgK) at 80°C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/s. Air (c = 1 kJ/kgK) enter at 30°C with a mass flow rate 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in °C) is[2012]a)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer?
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