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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?
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Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both...
Problem Statement: Determine the NTU value for a cross flow exchanger, given the specific heat of water, air, their flow rates, inlet temperatures, overall heat transfer coefficient, and surface area.
Solution:
Step 1: Calculate the mass flow rate of air (ma) and water (mw)
ma = 10 kg/s
mw = 7.5 kg/s
Step 2: Determine the inlet temperature difference (ΔT1) and outlet temperature difference (ΔT2)
ΔT1 = Tai - Twi = 120 - 15 = 105°C
ΔT2 = Tao - Two
Step 3: Calculate the heat capacity rate of air (Ca) and water (Cw)
Ca = ma x CPa = 10 x 1 = 10 kJ/s K
Cw = mw x CPw = 7.5 x 4 = 30 kJ/s K
Step 4: Calculate the overall heat transfer coefficient (U), and the surface area (A)
U = 780 kJ/m2 hr K
A = 240 m2
Step 5: Calculate the effectiveness (ε) of the heat exchanger using the NTU method
NTU = U x A / Cmin
Cmin = min(Ca, Cw) = 10 kJ/s K
NTU = 780 x 240 / 10 = 18,720
ε = 1 - exp(-NTU x (Cmin/Cmax))
Cmax = max(Ca, Cw) = 30 kJ/s K
ε = 1 - exp(-18,720 x (10/30)) = 0.999
Step 6: Calculate the outlet temperature of air (Tao)
ε = (Tao - Twi) / ΔT1
Tao = ε x ΔT1 + Twi = 0.999 x 105 + 15 = 119.5°C
Step 7: Calculate the outlet temperature of water (Two)
ε = (Tao - Twi)
Solution:
Step 1: Calculate the mass flow rate of air (ma) and water (mw)
ma = 10 kg/s
mw = 7.5 kg/s
Step 2: Determine the inlet temperature difference (ΔT1) and outlet temperature difference (ΔT2)
ΔT1 = Tai - Twi = 120 - 15 = 105°C
ΔT2 = Tao - Two
Step 3: Calculate the heat capacity rate of air (Ca) and water (Cw)
Ca = ma x CPa = 10 x 1 = 10 kJ/s K
Cw = mw x CPw = 7.5 x 4 = 30 kJ/s K
Step 4: Calculate the overall heat transfer coefficient (U), and the surface area (A)
U = 780 kJ/m2 hr K
A = 240 m2
Step 5: Calculate the effectiveness (ε) of the heat exchanger using the NTU method
NTU = U x A / Cmin
Cmin = min(Ca, Cw) = 10 kJ/s K
NTU = 780 x 240 / 10 = 18,720
ε = 1 - exp(-NTU x (Cmin/Cmax))
Cmax = max(Ca, Cw) = 30 kJ/s K
ε = 1 - exp(-18,720 x (10/30)) = 0.999
Step 6: Calculate the outlet temperature of air (Tao)
ε = (Tao - Twi) / ΔT1
Tao = ε x ΔT1 + Twi = 0.999 x 105 + 15 = 119.5°C
Step 7: Calculate the outlet temperature of water (Two)
ε = (Tao - Twi)
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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?
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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? 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 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? covers all topics & solutions for Mechanical 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 (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?.
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? 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 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? covers all topics & solutions for Mechanical 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 (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?.
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