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W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of 0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in °C) is
- a)40
- b)20
- c)10
- d)5
Correct answer is option 'C'. Can you explain this answer?
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W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat excha...
°C is to be cooled to 20°C using a cooling tower. The flow rate of water is 10,000 kg/hr. The wet bulb temperature of the cooling air is 25°C and the relative humidity is 50%. Calculate the amount of water evaporated in the cooling tower per hour.
Solution:
The heat transferred from water to air can be calculated using the following formula:
Q = m*Cp*(T1-T2)
where Q is the heat transferred, m is the mass flow rate, Cp is the specific heat capacity, T1 is the inlet temperature and T2 is the outlet temperature.
Q = 10,000*4.18*(80-20)
Q = 3,344,000 kJ/hr
The heat transferred from water to air is equal to the latent heat of vaporization of water plus the sensible heat of the air. The latent heat of vaporization of water at 25°C can be obtained from the steam table:
Hv = 2441 kJ/kg
The sensible heat of the air can be calculated using the following formula:
Hs = m*Cs*(T2-Twb)
where Cs is the specific heat capacity of air and Twb is the wet bulb temperature.
Cs = 1.005 kJ/kg.K
Hs = 10,000*1.005*(20-25)
Hs = -50,250 kJ/hr
The negative sign indicates that the air is losing sensible heat.
The total heat transferred from water to air is:
Q = Hv*mev + Hs
where mev is the mass of water evaporated per hour.
mev = (Q - Hs)/Hv
mev = (3,344,000 - (-50,250))/2441
mev = 1388.6 kg/hr
Therefore, the amount of water evaporated in the cooling tower per hour is 1388.6 kg/hr.
Solution:
The heat transferred from water to air can be calculated using the following formula:
Q = m*Cp*(T1-T2)
where Q is the heat transferred, m is the mass flow rate, Cp is the specific heat capacity, T1 is the inlet temperature and T2 is the outlet temperature.
Q = 10,000*4.18*(80-20)
Q = 3,344,000 kJ/hr
The heat transferred from water to air is equal to the latent heat of vaporization of water plus the sensible heat of the air. The latent heat of vaporization of water at 25°C can be obtained from the steam table:
Hv = 2441 kJ/kg
The sensible heat of the air can be calculated using the following formula:
Hs = m*Cs*(T2-Twb)
where Cs is the specific heat capacity of air and Twb is the wet bulb temperature.
Cs = 1.005 kJ/kg.K
Hs = 10,000*1.005*(20-25)
Hs = -50,250 kJ/hr
The negative sign indicates that the air is losing sensible heat.
The total heat transferred from water to air is:
Q = Hv*mev + Hs
where mev is the mass of water evaporated per hour.
mev = (Q - Hs)/Hv
mev = (3,344,000 - (-50,250))/2441
mev = 1388.6 kg/hr
Therefore, the amount of water evaporated in the cooling tower per hour is 1388.6 kg/hr.
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W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat excha...
C
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W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer?
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W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer?.
W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for W ater (Cp = 4.18 kJ/kg.K) at 80°C enters a counterflow heat exchanger with a mass flow rate of0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectivenessof the heat exchanger is 0.8, the LMTD (in °C) isa)40b)20c)10d)5Correct answer is option 'C'. Can you explain this answer?.
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