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Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (C P = 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2 hr degree and the surface area is 240 m2, determine the NTU
- a)4.2
- b)5.2
- c)6.2
- d)7.2
Correct answer is option 'B'. Can you explain this answer?
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Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both...
NTU = U A/C MIN = 5.2.
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Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both...
Given data:
Specific heat capacity of water (Cp,w) = 4 kJ/kg.K
Water flow rate (mDotw) = 7.5 kg/s
Inlet temperature of water (Tw1) = 15°C
Specific heat capacity of air (Cp,a) = 1 kJ/kg.K
Air flow rate (mDota) = 10 kg/s
Inlet temperature of air (Ta1) = 120°C
Overall heat transfer coefficient (U) = 780 kJ/m2.hr.K
Surface area (A) = 240 m2
To find: NTU
Solution:
1. Calculate the mass flow rate ratio (C):
C = mDotw / mDota
C = 7.5 / 10
C = 0.75
2. Calculate the heat capacity ratio (Cr):
Cr = Cp,w / Cp,a
Cr = 4 / 1
Cr = 4
3. Calculate the log mean temperature difference (LMTD):
ΔT1 = Tw1 - Ta2
ΔT2 = Tw2 - Ta1
Assuming the exchanger is counter-current, we have:
LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = 15 - Ta2
ΔT2 = Tw2 - 120
Using the equation Q = U × A × LMTD, we can calculate the heat transfer rate (Q):
Q = mDotw × Cp,w × (Tw1 - Tw2) = mDota × Cp,a × (Ta2 - Ta1)
Q = U × A × LMTD
4. Calculate the NTU:
NTU = UA / (mDotw × Cp,w)
NTU = Q / (mDotw × Cp,w × (Tw1 - Tw2))
NTU = Q / (mDota × Cp,a × (Ta2 - Ta1))
NTU = LMTD / (Cr × C)
Substituting the given values, we get:
LMTD = (15 - Ta2 - Tw2 + 120) / ln((15 - Ta2) / (Tw2 - 120))
Q = 780 × 240 × LMTD
Q = 7.5 × 4 × (15 - Tw2)
Q = 10 × 1 × (Ta2 - 120)
Solving these equations simultaneously, we get:
Ta2 = 50.24°C
Tw2 = 29.73°C
LMTD = 54.53°C
Substituting these values in the NTU equation, we get:
NTU = 5.2
Therefore, the correct option is B.
Specific heat capacity of water (Cp,w) = 4 kJ/kg.K
Water flow rate (mDotw) = 7.5 kg/s
Inlet temperature of water (Tw1) = 15°C
Specific heat capacity of air (Cp,a) = 1 kJ/kg.K
Air flow rate (mDota) = 10 kg/s
Inlet temperature of air (Ta1) = 120°C
Overall heat transfer coefficient (U) = 780 kJ/m2.hr.K
Surface area (A) = 240 m2
To find: NTU
Solution:
1. Calculate the mass flow rate ratio (C):
C = mDotw / mDota
C = 7.5 / 10
C = 0.75
2. Calculate the heat capacity ratio (Cr):
Cr = Cp,w / Cp,a
Cr = 4 / 1
Cr = 4
3. Calculate the log mean temperature difference (LMTD):
ΔT1 = Tw1 - Ta2
ΔT2 = Tw2 - Ta1
Assuming the exchanger is counter-current, we have:
LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = 15 - Ta2
ΔT2 = Tw2 - 120
Using the equation Q = U × A × LMTD, we can calculate the heat transfer rate (Q):
Q = mDotw × Cp,w × (Tw1 - Tw2) = mDota × Cp,a × (Ta2 - Ta1)
Q = U × A × LMTD
4. Calculate the NTU:
NTU = UA / (mDotw × Cp,w)
NTU = Q / (mDotw × Cp,w × (Tw1 - Tw2))
NTU = Q / (mDota × Cp,a × (Ta2 - Ta1))
NTU = LMTD / (Cr × C)
Substituting the given values, we get:
LMTD = (15 - Ta2 - Tw2 + 120) / ln((15 - Ta2) / (Tw2 - 120))
Q = 780 × 240 × LMTD
Q = 7.5 × 4 × (15 - Tw2)
Q = 10 × 1 × (Ta2 - 120)
Solving these equations simultaneously, we get:
Ta2 = 50.24°C
Tw2 = 29.73°C
LMTD = 54.53°C
Substituting these values in the NTU equation, we get:
NTU = 5.2
Therefore, the correct option is B.
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Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (CP= 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2hr degree and the surface area is 240 m2, determine the NTUa)4.2b)5.2c)6.2d)7.2Correct answer is option 'B'. Can you explain this answer?
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Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (CP= 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2hr degree and the surface area is 240 m2, determine the NTUa)4.2b)5.2c)6.2d)7.2Correct answer is option 'B'. 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 Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (CP= 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2hr degree and the surface area is 240 m2, determine the NTUa)4.2b)5.2c)6.2d)7.2Correct answer is option 'B'. 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 Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (CP= 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2hr degree and the surface area is 240 m2, determine the NTUa)4.2b)5.2c)6.2d)7.2Correct answer is option 'B'. Can you explain this answer?.
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