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A counter flow heat exchanger is used to col 2000 kg/hr of oil (cp = 2.5 k J/kg K) from 105 degree Celsius to 30 degree Celsius by the use of water entering at 15 degree Celsius. If the overall heat transfer coefficient is expected to be 1.5 k W/m2 K, find out the water flow rate. Presume that the exit temperature of the water is not to exceed 80 degree Celsius
- a)1680.2 kg/hr
- b)1580.2 kg/hr
- c)1480.2 kg/hr
- d)1380.2 kg/hr
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A counter flow heat exchanger is used to col 2000 kg/hr of oil (cp= 2....
Mass flow rate of coolant = 2000 (2.5) (105-30)/4.18 (80-15) = 1380.2 kg/hr.
Most Upvoted Answer
A counter flow heat exchanger is used to col 2000 kg/hr of oil (cp= 2....
Given data:
Mass flow rate of oil, m1 = 2000 kg/hr
Specific heat of oil, cp1 = 2.5 kJ/kg K
Inlet temperature of oil, T1 = 105°C
Outlet temperature of oil, T2 = 30°C
Overall heat transfer coefficient, U = 1.5 kW/m²K
Inlet temperature of water, Tw1 = 15°C
Maximum outlet temperature of water, Tw2 = 80°C
To find: Water flow rate, m2
Solution:
The heat transfer rate in a counter flow heat exchanger can be calculated as:
Q = U x A x deltaT_lm
Where,
A = Heat transfer surface area
deltaT_lm = Logarithmic mean temperature difference
The heat transfer rate should be the same for both fluids in the heat exchanger, i.e., Q = m1 x cp1 x (T1 - T2) = m2 x cp2 x (Tw2 - Tw1)
The logarithmic mean temperature difference is given by:
deltaT_lm = (deltaT1 - deltaT2) / ln(deltaT1 / deltaT2)
Where,
deltaT1 = T1 - Tw2
deltaT2 = T2 - Tw1
Substituting the given values in the above equations, we get:
m2 = [m1 x cp1 x (T1 - T2)] / [cp2 x (Tw2 - Tw1)]
deltaT1 = 105 - 80 = 25°C
deltaT2 = 30 - 15 = 15°C
deltaT_lm = (25 - 15) / ln(25 / 15) = 18.394°C
A = Q / (U x deltaT_lm)
A = [m1 x cp1 x (T1 - T2)] / (U x deltaT_lm)
A = [2000 x 2.5 x (105 - 30)] / (1.5 x 18.394)
A = 420.85 m²
The water flow rate can be calculated as:
m2 = [m1 x cp1 x (T1 - T2)] / [cp2 x (Tw2 - Tw1)]
m2 = [2000 x 2.5 x (105 - 30)] / [4.18 x (80 - 15)]
m2 = 1380.2 kg/hr (approx)
Therefore, the water flow rate required is 1380.2 kg/hr.
Mass flow rate of oil, m1 = 2000 kg/hr
Specific heat of oil, cp1 = 2.5 kJ/kg K
Inlet temperature of oil, T1 = 105°C
Outlet temperature of oil, T2 = 30°C
Overall heat transfer coefficient, U = 1.5 kW/m²K
Inlet temperature of water, Tw1 = 15°C
Maximum outlet temperature of water, Tw2 = 80°C
To find: Water flow rate, m2
Solution:
The heat transfer rate in a counter flow heat exchanger can be calculated as:
Q = U x A x deltaT_lm
Where,
A = Heat transfer surface area
deltaT_lm = Logarithmic mean temperature difference
The heat transfer rate should be the same for both fluids in the heat exchanger, i.e., Q = m1 x cp1 x (T1 - T2) = m2 x cp2 x (Tw2 - Tw1)
The logarithmic mean temperature difference is given by:
deltaT_lm = (deltaT1 - deltaT2) / ln(deltaT1 / deltaT2)
Where,
deltaT1 = T1 - Tw2
deltaT2 = T2 - Tw1
Substituting the given values in the above equations, we get:
m2 = [m1 x cp1 x (T1 - T2)] / [cp2 x (Tw2 - Tw1)]
deltaT1 = 105 - 80 = 25°C
deltaT2 = 30 - 15 = 15°C
deltaT_lm = (25 - 15) / ln(25 / 15) = 18.394°C
A = Q / (U x deltaT_lm)
A = [m1 x cp1 x (T1 - T2)] / (U x deltaT_lm)
A = [2000 x 2.5 x (105 - 30)] / (1.5 x 18.394)
A = 420.85 m²
The water flow rate can be calculated as:
m2 = [m1 x cp1 x (T1 - T2)] / [cp2 x (Tw2 - Tw1)]
m2 = [2000 x 2.5 x (105 - 30)] / [4.18 x (80 - 15)]
m2 = 1380.2 kg/hr (approx)
Therefore, the water flow rate required is 1380.2 kg/hr.
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A counter flow heat exchanger is used to col 2000 kg/hr of oil (cp= 2.5 k J/kg K) from 105 degree Celsius to 30 degree Celsius by the use of water entering at 15 degree Celsius. If the overall heat transfer coefficient is expected to be 1.5 k W/m2K, find out the water flow rate. Presume that the exit temperature of the water is not to exceed 80 degree Celsiusa)1680.2 kg/hrb)1580.2 kg/hrc)1480.2 kg/hrd)1380.2 kg/hrCorrect answer is option 'D'. Can you explain this answer?
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