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A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel?
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A counter flow double pipe heat exchanges using a Super heated steam i...
Heat Transfer Rate Calculation
To calculate the heat transfer rate (\(Q\)) in the counter flow double pipe heat exchanger, we can use the formula:
\[
Q = \dot{m} \cdot C_p \cdot \Delta T
\]
Where:
- \(\dot{m}\) = mass flow rate of water (kg/s)
- \(C_p\) = specific heat capacity of water (approximately \(4.186 \, \text{kJ/kg°C}\))
- \(\Delta T\) = temperature change of water (°C)
First, convert the mass flow rate from kg/hr to kg/s:
\[
\dot{m} = \frac{10,500 \, \text{kg/hr}}{3600 \, \text{s/hr}} \approx 2.9167 \, \text{kg/s}
\]
Calculate \(\Delta T\):
\[
\Delta T = 80°C - 30°C = 50°C
\]
Now, substituting into the formula:
\[
Q = 2.9167 \, \text{kg/s} \cdot 4.186 \, \text{kJ/kg°C} \cdot 50°C \approx 609.93 \, \text{kW} \approx 610 \, \text{kW}
\]
Area Calculation for Parallel Flow
In a parallel flow configuration, the temperature difference between the two fluids is less effective compared to counter flow. The heat transfer rate can be recalculated using the logarithmic mean temperature difference (LMTD):
\[
\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln{\left(\frac{\Delta T_1}{\Delta T_2}\right)}}
\]
Where:
- \(\Delta T_1 = T_{steam, in} - T_{water, out} = 180°C - 30°C = 150°C\)
- \(\Delta T_2 = T_{steam, out} - T_{water, in} = 130°C - 80°C = 50°C\)
Calculate LMTD:
\[
\text{LMTD} = \frac{150 - 50}{\ln{\left(\frac{150}{50}\right)}} \approx 81.87°C
\]
Now, we calculate the required area (\(A\)) using:
\[
Q = U \cdot A \cdot \text{LMTD}
\]
Thus,
\[
A = \frac{Q}{U \cdot \text{LMTD}} \approx \frac{610,000 \, W}{814 \, W/m^2°C \cdot 81.87°C} \approx 9.05 \, m^2
\]
The increase in area when switching from counter flow to parallel flow generally leads to a larger area requirement, given the less effective heat transfer in the parallel configuration.
To calculate the heat transfer rate (\(Q\)) in the counter flow double pipe heat exchanger, we can use the formula:
\[
Q = \dot{m} \cdot C_p \cdot \Delta T
\]
Where:
- \(\dot{m}\) = mass flow rate of water (kg/s)
- \(C_p\) = specific heat capacity of water (approximately \(4.186 \, \text{kJ/kg°C}\))
- \(\Delta T\) = temperature change of water (°C)
First, convert the mass flow rate from kg/hr to kg/s:
\[
\dot{m} = \frac{10,500 \, \text{kg/hr}}{3600 \, \text{s/hr}} \approx 2.9167 \, \text{kg/s}
\]
Calculate \(\Delta T\):
\[
\Delta T = 80°C - 30°C = 50°C
\]
Now, substituting into the formula:
\[
Q = 2.9167 \, \text{kg/s} \cdot 4.186 \, \text{kJ/kg°C} \cdot 50°C \approx 609.93 \, \text{kW} \approx 610 \, \text{kW}
\]
Area Calculation for Parallel Flow
In a parallel flow configuration, the temperature difference between the two fluids is less effective compared to counter flow. The heat transfer rate can be recalculated using the logarithmic mean temperature difference (LMTD):
\[
\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln{\left(\frac{\Delta T_1}{\Delta T_2}\right)}}
\]
Where:
- \(\Delta T_1 = T_{steam, in} - T_{water, out} = 180°C - 30°C = 150°C\)
- \(\Delta T_2 = T_{steam, out} - T_{water, in} = 130°C - 80°C = 50°C\)
Calculate LMTD:
\[
\text{LMTD} = \frac{150 - 50}{\ln{\left(\frac{150}{50}\right)}} \approx 81.87°C
\]
Now, we calculate the required area (\(A\)) using:
\[
Q = U \cdot A \cdot \text{LMTD}
\]
Thus,
\[
A = \frac{Q}{U \cdot \text{LMTD}} \approx \frac{610,000 \, W}{814 \, W/m^2°C \cdot 81.87°C} \approx 9.05 \, m^2
\]
The increase in area when switching from counter flow to parallel flow generally leads to a larger area requirement, given the less effective heat transfer in the parallel configuration.
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A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel?
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A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel?.
A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A counter flow double pipe heat exchanges using a Super heated steam is used to heat water at the rate A 10,500 kg/hr the Steam enters the heat exchanger at 180°c and leaves at 130°c the inlet and exit temp on water are 80°c and 30°c respectively . If the overall heat transfer coefficient from steam to water is 814w/m2.°c calculate the heat transfer rate . what will be the increase in area if the fluid flow were parallel?.
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