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A double pipe counter flow heat exchanger is to be designed to cool 12000 kg/hr of an oil of specific heat 1.95 Kj/Kg-K . from 85oC to 55oC by water entering the heat exchanger at 30oC and leaving at 45oC . If the overall heat transfer coefficient of heat exchanger is 400 W/m2-. K Calculate the surface area of the heat exchanger.
(B) 15.5
Correct answer is between ','. Can you explain this answer?
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A double pipe counter flow heat exchanger is to be designed to cool 1...
For the counter flow heat exchanger
Mh = 12000/3600 = 3.33 kh/s
Cph = 1.95 kj/kh -K
Thi - 85oC; The = 55oC
Tci = 30oC; Tce = 45oC
∴ θ0 = 40oC; θ1 = 45oC
⇒ ፀm = 31.9oC
Q = mhcph(Thi - The) = 3.33 x 1.95(85-55)
⇒ Q = 194.805 kW
Again Q = UA(θm)
⇒ A = Q/u((θm)= 194.805/0.4 x 31.9
⇒ A = 15.27m2
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A double pipe counter flow heat exchanger is to be designed to cool 1...
Given data:
- Oil flow rate (m_dot_oil) = 12000 kg/hr = 3.33 kg/s
- Specific heat of oil (Cp_oil) = 1.95 kJ/kg-K
- Inlet temperature of oil (T_oil_in) = 85°C
- Outlet temperature of oil (T_oil_out) = 55°C
- Water inlet temperature (T_water_in) = 30°C
- Water outlet temperature (T_water_out) = 45°C
- Overall heat transfer coefficient (U) = 400 W/m2-K
Calculation of heat transfer:
The heat transfer rate (Q) can be calculated using the formula:
Q = m_dot_oil * Cp_oil * (T_oil_in - T_oil_out)
Substituting the given values:
Q = 3.33 kg/s * 1.95 kJ/kg-K * (85°C - 55°C)
Q = 3.33 * 1.95 * 30 kJ/s = 193.95 kW
Calculation of log mean temperature difference (LMTD):
The log mean temperature difference (LMTD) can be calculated using the formula:
LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
Where:
ΔT1 = T_oil_in - T_water_out
ΔT2 = T_oil_out - T_water_in
Substituting the given values:
ΔT1 = 85°C - 45°C = 40°C
ΔT2 = 55°C - 30°C = 25°C
LMTD = (40°C - 25°C) / ln(40°C / 25°C)
LMTD = 15°C / ln(1.6)
Using logarithmic tables or calculator, the natural logarithm of 1.6 is approximately 0.47.
LMTD = 15°C / 0.47
LMTD = 31.91°C
Calculation of surface area:
The heat transfer area (A) can be calculated using the formula:
Q = U * A * LMTD
Rearranging the formula:
A = Q / (U * LMTD)
Substituting the given values:
A = 193.95 kW / (400 W/m2-K * 31.91°C)
A = 484.875 m2
Conclusion:
The surface area of the double pipe counter flow heat exchanger required to cool 12000 kg/hr of oil from 85°C to 55°C using water entering at 30°C and leaving at 45°C, with an overall heat transfer coefficient of 400 W/m2-K, is approximately 484.875 m2.
- Oil flow rate (m_dot_oil) = 12000 kg/hr = 3.33 kg/s
- Specific heat of oil (Cp_oil) = 1.95 kJ/kg-K
- Inlet temperature of oil (T_oil_in) = 85°C
- Outlet temperature of oil (T_oil_out) = 55°C
- Water inlet temperature (T_water_in) = 30°C
- Water outlet temperature (T_water_out) = 45°C
- Overall heat transfer coefficient (U) = 400 W/m2-K
Calculation of heat transfer:
The heat transfer rate (Q) can be calculated using the formula:
Q = m_dot_oil * Cp_oil * (T_oil_in - T_oil_out)
Substituting the given values:
Q = 3.33 kg/s * 1.95 kJ/kg-K * (85°C - 55°C)
Q = 3.33 * 1.95 * 30 kJ/s = 193.95 kW
Calculation of log mean temperature difference (LMTD):
The log mean temperature difference (LMTD) can be calculated using the formula:
LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
Where:
ΔT1 = T_oil_in - T_water_out
ΔT2 = T_oil_out - T_water_in
Substituting the given values:
ΔT1 = 85°C - 45°C = 40°C
ΔT2 = 55°C - 30°C = 25°C
LMTD = (40°C - 25°C) / ln(40°C / 25°C)
LMTD = 15°C / ln(1.6)
Using logarithmic tables or calculator, the natural logarithm of 1.6 is approximately 0.47.
LMTD = 15°C / 0.47
LMTD = 31.91°C
Calculation of surface area:
The heat transfer area (A) can be calculated using the formula:
Q = U * A * LMTD
Rearranging the formula:
A = Q / (U * LMTD)
Substituting the given values:
A = 193.95 kW / (400 W/m2-K * 31.91°C)
A = 484.875 m2
Conclusion:
The surface area of the double pipe counter flow heat exchanger required to cool 12000 kg/hr of oil from 85°C to 55°C using water entering at 30°C and leaving at 45°C, with an overall heat transfer coefficient of 400 W/m2-K, is approximately 484.875 m2.
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A double pipe counter flow heat exchanger is to be designed to cool 12000 kg/hr of an oil of specific heat 1.95 Kj/Kg-K . from 85oC to 55oC by water entering the heat exchanger at 30oC and leaving at 45oC . If the overall heat transfer coefficient of heat exchanger is 400 W/m2-. K Calculate the surface area of the heat exchanger.(B) 15.5Correct answer is between ','. Can you explain this answer?
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