Question

A crossflow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and $20^{\circ} \mathrm{C}$, respectively, determine the exit temperature of the cold fluid.

Short answer

Answer: The exit temperature of the cold fluid is 80°C.

Step-by-step solution

Calculate the heat transfer rate

In a heat exchanger, the heat transfer rate (\(Q\)) can be calculated using the overall heat transfer coefficient (\(U\)), the heat transfer surface area (\(A\)) and the temperature difference between the hot and cold fluids (\(\Delta T\)). The formula is as follows: $$Q = UA \Delta T$$ For our problem, we have \(U = 200 \mathrm{~W/m^2\cdot K}\), \(A = 400 \mathrm{~m^2}\), and the temperature difference between the hot and cold fluids at the inlet is given as \(80^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 60^{\circ} \mathrm{C}\). Substitute these values into the formula above: $$Q = 200 \mathrm{~W/m^2\cdot K} \times 400 \mathrm{~m^2} \times (60^{\circ} \mathrm{C})$$

Calculate the exit temperature of the cold fluid

Now we will apply the energy balance equation to find the exit temperature of the cold fluid (\(T_{c,out}\)). The energy balance equation is as follows: $$Q = C_{c}(T_{c,out} - T_{c,in})$$ Where \(C_{c}\) is the heat capacity of the cold fluid, and \(T_{c,in}\) is the inlet temperature of the cold fluid. For our problem, \(C_{c} = 80,000 \mathrm{~W/K}\) and \(T_{c,in} = 20^{\circ} \mathrm{C}\). Plugging values from step 1: $$200 \mathrm{~W/m^2\cdot K} \times 400 \mathrm{~m^2} \times (60^{\circ} \mathrm{C}) = 80,000 \mathrm{~W/K}(T_{c,out} - 20^{\circ} \mathrm{C})$$ Now we will solve for \(T_{c,out}\): $$T_{c,out} = \frac{200 \mathrm{~W/m^2\cdot K} \times 400 \mathrm{~m^2} \times 60^{\circ} \mathrm{C}}{80,000 \mathrm{~W/K}} + 20^{\circ} \mathrm{C}$$ $$T_{c,out} = \frac{4800000 \mathrm{~W}}{80000 \mathrm{~W/K}} + 20^{\circ} \mathrm{C}$$ $$T_{c,out} = 60^{\circ} \mathrm{C} + 20^{\circ} \mathrm{C}$$

Conclusion

The exit temperature of the cold fluid is: $$T_{c,out} = 80^{\circ} \mathrm{C}$$