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A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp = 2.5kj/kg-K) from 105oC to 30oC by use of water that
enters the heat exchanger with a temperature of 15oC. If the overall heat transfer coefficient is expected to be 1.5 kW/m2-.K , make calculations for the surface area required, presume that the exit temperature of the water is not to exceed 80oC.
(B) 3.6
Correct answer is between ','. Can you explain this answer?
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A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp =...
From steady state flow energy equation
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mccph(Thi -The) = mccph(Tce -Tci)
2000 x .5(105 - 30) = mc 4.18(80-15)
Mc = 1380.2 kg/hr
∴ mccpc = 1380.2 x 4.18 = 5769.24 kjhr-K
mccph = 2000 x 2.5 = 5000kh/hr-K
∴ C = 5000/5769.24 = 0.867
Effectiveness , ε = Thi - The /Thi - Tci = 105-30/105-15 = 0.822
For a counter flow heat exchanger
NTU = 3.84
UA/Cmin = 3.84
1.5 x A /5000/3600 = 3.84 ⇒ A = 3.55m2
Alternate method
Thi = 105oC Tho = 30oC
Tci = 1.5oC Tco M 80oC
∴ Qact = mhch(Thi - Tho)
= 2000/3600 x 2.5 x (105 - 30)
= 104.166 kj/sec
For a counterflow heat exchanger
Qact = UA θm
104.166 = 1.5 x A x 19.576
A = 104.166/15 x 19.576
A = 3.54 m2
Most Upvoted Answer
A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp =...
Solution:
Given data:
Mass flow rate of oil, m1 = 2000 kg/hr
Specific heat of oil, cp1 = 2.5 kJ/kg-K
Initial temperature of oil, T1 = 105°C
Final temperature of oil, T2 = 30°C
Inlet temperature of water, T3 = 15°C
Maximum allowable outlet temperature of water, T4 = 80°C
Overall heat transfer coefficient, U = 1.5 kW/m2-K
We can use the following equation to calculate the heat transfer rate in the heat exchanger:
Q = m1 * cp1 * (T1 - T2)
The heat transfer rate Q is equal to the product of mass flow rate, specific heat, and temperature difference between inlet and outlet of the oil.
Q = 2000 * 2.5 * (105 - 30)
Q = 4,62,500 kJ/hr
Next, we can use the following equation to calculate the log mean temperature difference (LMTD) between the oil and water streams:
LMTD = ( (T1 - T4) - (T2 - T3) ) / ln( (T1 - T4) / (T2 - T3) )
LMTD = ( (105 - 80) - (30 - 15) ) / ln( (105 - 80) / (30 - 15) )
LMTD = 42.07 K
Using the LMTD and heat transfer rate, we can calculate the required heat transfer area A as follows:
Q = U * A * LMTD
A = Q / (U * LMTD)
A = 4,62,500 / (1.5 * 42.07)
A = 7,714.5 m2
Therefore, the surface area required for the counter flow heat exchanger is 7,714.5 m2.
Answer: 3.6.
Given data:
Mass flow rate of oil, m1 = 2000 kg/hr
Specific heat of oil, cp1 = 2.5 kJ/kg-K
Initial temperature of oil, T1 = 105°C
Final temperature of oil, T2 = 30°C
Inlet temperature of water, T3 = 15°C
Maximum allowable outlet temperature of water, T4 = 80°C
Overall heat transfer coefficient, U = 1.5 kW/m2-K
We can use the following equation to calculate the heat transfer rate in the heat exchanger:
Q = m1 * cp1 * (T1 - T2)
The heat transfer rate Q is equal to the product of mass flow rate, specific heat, and temperature difference between inlet and outlet of the oil.
Q = 2000 * 2.5 * (105 - 30)
Q = 4,62,500 kJ/hr
Next, we can use the following equation to calculate the log mean temperature difference (LMTD) between the oil and water streams:
LMTD = ( (T1 - T4) - (T2 - T3) ) / ln( (T1 - T4) / (T2 - T3) )
LMTD = ( (105 - 80) - (30 - 15) ) / ln( (105 - 80) / (30 - 15) )
LMTD = 42.07 K
Using the LMTD and heat transfer rate, we can calculate the required heat transfer area A as follows:
Q = U * A * LMTD
A = Q / (U * LMTD)
A = 4,62,500 / (1.5 * 42.07)
A = 7,714.5 m2
Therefore, the surface area required for the counter flow heat exchanger is 7,714.5 m2.
Answer: 3.6.
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A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp = 2.5kj/kg-K) from 105oC to 30oC by use of water thatenters the heat exchanger with a temperature of 15oC. If the overall heat transfer coefficient is expected to be 1.5 kW/m2-.K , make calculations for the surface area required, presume that the exit temperature of the water is not to exceed 80oC.(B) 3.6Correct answer is between ','. Can you explain this answer?
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A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp = 2.5kj/kg-K) from 105oC to 30oC by use of water thatenters the heat exchanger with a temperature of 15oC. If the overall heat transfer coefficient is expected to be 1.5 kW/m2-.K , make calculations for the surface area required, presume that the exit temperature of the water is not to exceed 80oC.(B) 3.6Correct answer is between ','. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp = 2.5kj/kg-K) from 105oC to 30oC by use of water thatenters the heat exchanger with a temperature of 15oC. If the overall heat transfer coefficient is expected to be 1.5 kW/m2-.K , make calculations for the surface area required, presume that the exit temperature of the water is not to exceed 80oC.(B) 3.6Correct answer is between ','. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A counter flow heat exchanger is used to cool 2000 kg/hr of oil (cp = 2.5kj/kg-K) from 105oC to 30oC by use of water thatenters the heat exchanger with a temperature of 15oC. If the overall heat transfer coefficient is expected to be 1.5 kW/m2-.K , make calculations for the surface area required, presume that the exit temperature of the water is not to exceed 80oC.(B) 3.6Correct answer is between ','. Can you explain this answer?.
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