In a heat exchanger, 50000 \(\rm \frac{kg}{h}\) of hot water is cooled from 60°C to 40°C using the 40000 \(\rm \frac{kg}{h}\) of cold water entering at 5°C. If the specific heat of water is 4.18 \(\rm \frac{kJ}{kg-k}\), then the exit temperature will be:
Heat Exchanger Energy Balance Calculation
This problem asks us to determine the exit temperature of cold water in a heat exchanger given the flow rates and temperatures of both hot and cold water streams.
In a heat exchanger, energy is transferred from a hotter fluid to a colder fluid. Assuming ideal conditions with no heat loss to the surroundings, the principle of energy conservation dictates that the heat lost by the hot fluid is equal to the heat gained by the cold fluid.
The rate of heat transfer for a fluid stream is given by the formula:
\( Q = \dot{m} \cdot c_p \cdot \Delta T \)
where:
- \( Q \) is the rate of heat transfer.
- \( \dot{m} \) is the mass flow rate of the fluid.
- \( c_p \) is the specific heat capacity of the fluid.
- \( \Delta T \) is the change in temperature (\(T_{exit} - T_{inlet}\) for heat gained, \(T_{inlet} - T_{exit}\) for heat lost).
Applying Energy Balance to the Heat Exchanger
We are given the following parameters for the hot and cold water streams:
| Parameter | Hot Water | Cold Water |
|---|---|---|
| Mass flow rate (\(\dot{m}\)) | \(50000 \frac{\text{kg}}{\text{h}}\) | \(40000 \frac{\text{kg}}{\text{h}}\) |
| Inlet Temperature (\(T_{in}\)) | \(60^\circ\text{C}\) | \(5^\circ\text{C}\) |
| Outlet Temperature (\(T_{out}\)) | \(40^\circ\text{C}\) | \(T_{c,out}\) (to be found) |
| Specific Heat (\(c_p\)) | \(4.18 \frac{\text{kJ}}{\text{kg-K}}\) | \(4.18 \frac{\text{kJ}}{\text{kg-K}}\) |
The heat lost by the hot water is equal to the heat gained by the cold water:
\( Q_{hot} = Q_{cold} \)
\( \dot{m}_h \cdot c_p \cdot (T_{h,in} - T_{h,out}) = \dot{m}_c \cdot c_p \cdot (T_{c,out} - T_{c,in}) \)
Since the specific heat \(c_p\) is the same for both hot and cold water, we can cancel it out from both sides of the equation:
\( \dot{m}_h \cdot (T_{h,in} - T_{h,out}) = \dot{m}_c \cdot (T_{c,out} - T_{c,in}) \)
Calculating the Cold Water Exit Temperature
Now, let's substitute the given values into the simplified equation:
\( 50000 \frac{\text{kg}}{\text{h}} \cdot (60^\circ\text{C} - 40^\circ\text{C}) = 40000 \frac{\text{kg}}{\text{h}} \cdot (T_{c,out} - 5^\circ\text{C}) \)
\( 50000 \cdot (20^\circ\text{C}) = 40000 \cdot (T_{c,out} - 5^\circ\text{C}) \)
\( 1000000 = 40000 \cdot (T_{c,out} - 5) \)
Divide both sides by 40000:
\( \frac{1000000}{40000} = T_{c,out} - 5 \)
\( 25 = T_{c,out} - 5 \)
Now, solve for \(T_{c,out}\):
\( T_{c,out} = 25 + 5 \)
\( T_{c,out} = 30^\circ\text{C} \)
Conclusion
Based on the energy balance in the heat exchanger, the exit temperature of the cold water is 30°C.
Important Questions from Heat Exchanger
Heat exchangers are used in
A. Condensers and boilers in steam plants
B. Radiators
C. Intercoolers and preheaters
D. Condensers and evaporators in refrigerators and air conditioners
For a specified inlet and outlet temperatures, for which kind of heat exchange will the log mean temperature difference, ΔTlm will be highest?
The fouling factor in case of heat exchangers is given by ________
In heat transfe,r NTU is given by ________.
(Where U = heat transfer co-efficient, A = surface area, C = Heat capacitance, min = Minimum, max = Maximum
Which of the following heat exchangers gives parallel and linear temperature profile for both cold and hot fluid?
