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A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :
Mass flow rate of the hot fluid = 0.25 kg/s
Mass flow rate of the cold water = 0.4 kg/s
Specific heat of hot fluid = 6000 J/kg.K
Specific heat of water = 4184 J/kg.K
The inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.
Inlet temperature of cold water = 20°C
The heat transfer area is 1.82 m2
The overall heat transfer coefficient (in W/m2 K) is .
Correct answer is '500'. Can you explain this answer?
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A hot fluid entering a well stirred vessel is cooled by feeding cold ...
Energy released by hot fluid = energy received by cold water
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Given Data:
- Mass flow rate of hot fluid (mh) = 0.25 kg/s
- Mass flow rate of cold water (mcw) = 0.4 kg/s
- Specific heat of hot fluid (Cph) = 6000 J/kg.K
- Specific heat of water (Cw) = 4184 J/kg.K
- Inlet temperature of hot fluid (Th1) = 150°C
- Exit temperature of hot fluid (Th2) = 100°C
- Inlet temperature of cold water (Tcw) = 20°C
- Heat transfer area (A) = 1.82 m2
- Overall heat transfer coefficient (U) = ?
Approach:
To determine the overall heat transfer coefficient, we can use the equation for heat transfer rate (Q) in a heat exchanger:
Q = U * A * ΔTlm
Where:
- Q is the heat transfer rate
- U is the overall heat transfer coefficient
- A is the heat transfer area
- ΔTlm is the logarithmic mean temperature difference
Calculating ΔTlm:
To calculate ΔTlm, we need to determine the temperature differences at each end of the heat exchanger.
ΔT1 = Th1 - Tcw
ΔT2 = Th2 - Tcw
The logarithmic mean temperature difference (ΔTlm) can be calculated using the formula:
ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
Calculating Heat Transfer Rate:
The heat transfer rate (Q) can be calculated using the formula:
Q = mh * Cph * (Th1 - Th2) = mcw * Cw * (Tcw - Th2)
Substituting values:
Substituting the given values into the equation for Q, we get:
0.25 * 6000 * (150 - 100) = 0.4 * 4184 * (20 - 100)
Calculating Overall Heat Transfer Coefficient:
Now we can rearrange the equation for Q to solve for U:
U = Q / (A * ΔTlm)
Substituting the calculated values for Q, A, and ΔTlm, we can solve for U:
U = (0.25 * 6000 * (150 - 100)) / (1.82 * ΔTlm)
Calculating ΔTlm:
Substituting the calculated values for ΔT1 and ΔT2 into the equation for ΔTlm, we get:
ΔTlm = (150 - 20 - (100 - 20)) / ln((150 - 20) / (100 - 20))
Calculating U:
Now we can substitute the calculated values for ΔTlm into the equation for U:
U = (0.25 * 6000 * (150 - 100)) / (1.82 * ΔTlm)
Calculating this expression gives us the final answer of U = 500 W/m2 K.
- Mass flow rate of hot fluid (mh) = 0.25 kg/s
- Mass flow rate of cold water (mcw) = 0.4 kg/s
- Specific heat of hot fluid (Cph) = 6000 J/kg.K
- Specific heat of water (Cw) = 4184 J/kg.K
- Inlet temperature of hot fluid (Th1) = 150°C
- Exit temperature of hot fluid (Th2) = 100°C
- Inlet temperature of cold water (Tcw) = 20°C
- Heat transfer area (A) = 1.82 m2
- Overall heat transfer coefficient (U) = ?
Approach:
To determine the overall heat transfer coefficient, we can use the equation for heat transfer rate (Q) in a heat exchanger:
Q = U * A * ΔTlm
Where:
- Q is the heat transfer rate
- U is the overall heat transfer coefficient
- A is the heat transfer area
- ΔTlm is the logarithmic mean temperature difference
Calculating ΔTlm:
To calculate ΔTlm, we need to determine the temperature differences at each end of the heat exchanger.
ΔT1 = Th1 - Tcw
ΔT2 = Th2 - Tcw
The logarithmic mean temperature difference (ΔTlm) can be calculated using the formula:
ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
Calculating Heat Transfer Rate:
The heat transfer rate (Q) can be calculated using the formula:
Q = mh * Cph * (Th1 - Th2) = mcw * Cw * (Tcw - Th2)
Substituting values:
Substituting the given values into the equation for Q, we get:
0.25 * 6000 * (150 - 100) = 0.4 * 4184 * (20 - 100)
Calculating Overall Heat Transfer Coefficient:
Now we can rearrange the equation for Q to solve for U:
U = Q / (A * ΔTlm)
Substituting the calculated values for Q, A, and ΔTlm, we can solve for U:
U = (0.25 * 6000 * (150 - 100)) / (1.82 * ΔTlm)
Calculating ΔTlm:
Substituting the calculated values for ΔT1 and ΔT2 into the equation for ΔTlm, we get:
ΔTlm = (150 - 20 - (100 - 20)) / ln((150 - 20) / (100 - 20))
Calculating U:
Now we can substitute the calculated values for ΔTlm into the equation for U:
U = (0.25 * 6000 * (150 - 100)) / (1.82 * ΔTlm)
Calculating this expression gives us the final answer of U = 500 W/m2 K.
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A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer?
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A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer?.
A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer?.
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A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer?, a detailed solution for A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? has been provided alongside types of A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A hot fluid entering a well stirred vessel is cooled by feeding cold water through a jacket around the vessel in opposite direction. Consider the following data :Mass flow rate of the hot fluid = 0.25 kg/sMass flow rate of the cold water = 0.4 kg/sSpecific heat of hot fluid = 6000 J/kg.KSpecific heat of water = 4184 J/kg.KThe inlet aid exit temperatures of hot fluid are 150°C and 100°C, respectively.Inlet temperature of cold water = 20°CThe heat transfer area is 1.82 m2The overall heat transfer coefficient (in W/m2 K) is .Correct answer is '500'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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