Question

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in the shell side of a one-shell and two-tube heat exchanger with a surface area of $0.5 \mathrm{~m}^{2}\( and an overall heat transfer coefficient of \)2000 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}\(. If cold water \)\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\( ) flowing at \)0.5 \mathrm{~kg} / \mathrm{s}\( enters the tube side at \)15^{\circ} \mathrm{C}$, determine the outlet temperature of the cold water and the heat transfer rate for the heat exchanger.

Short answer

Answer: The outlet temperature of the cold water is 55.62°C, and the heat transfer rate is 85000 W.

Step-by-step solution

Calculate the heat transfer needed for the saturated water vapor to condense

First, we need to determine the heat transfer needed for the saturated water vapor to condense at \(100^{\circ}\mathrm{C}\). We can do this by using the heat transfer formula: $$q = U \cdot A \cdot(T_{hot} - T_{cold})$$ Where, \(q\) is the heat transfer, \(U\) is the overall heat transfer coefficient, \(A\) is the surface area of the heat exchanger, \(T_{hot}\) is the temperature of the hot fluid (saturated water vapor), and \(T_{cold}\) is the inlet temperature of the cold fluid (water). From the problem statement, we have the values: \(U = 2000 \frac{W}{m^2 K}\) \(A = 0.5 m^2\) \(T_{hot}=100^{\circ}\mathrm{C}\) \(T_{cold}=15^{\circ}\mathrm{C}\) Using these values, we can calculate the heat transfer needed for the saturated water vapor to condense: $$ q = 2000 \frac{W}{m^2 K} \cdot 0.5 m^2 \cdot (100^{\circ}\mathrm{C} - 15^{\circ}\mathrm{C})$$ $$ q = 85000 W$$

Apply the energy balance equation to determine the outlet temperature

Now that we have the heat transfer, we can apply the energy balance equation to the cold water. The energy balance equation can be expressed as: $$\dot{m}c_p (T_{out} - T_{in}) = q$$ Where, \(\dot{m}\) is the mass flow rate of the cold water, \(c_p\) is the specific heat capacity of the cold water, \(T_{out}\) is the outlet temperature of the cold water, and \(T_{in}\) is the inlet temperature of the cold water. From the problem statement, we have the values: \(\dot{m} = 0.5 \frac{kg}{s}\) \(c_p = 4179 \frac{J}{kg \cdot K}\) \(T_{in} = 15^{\circ}\mathrm{C}\) Re-arrange the equation to solve for \(T_{out}\): $$T_{out} = T_{in} + \frac{q}{\dot{m}c_p}$$ Now, plug in the values to calculate the outlet temperature: $$T_{out} = 15^{\circ}\mathrm{C} + \frac{85000 W}{0.5 \frac{kg}{s} \cdot 4179 \frac{J}{kg \cdot K}}$$ $$T_{out} = 15^{\circ}\mathrm{C} + 40.62^{\circ}\mathrm{C}$$ $$T_{out} = 55.62^{\circ}\mathrm{C}$$

Calculate the heat transfer rate

Since we have already calculated the heat transfer in step 1, the heat transfer rate for the heat exchanger is: $$q = 85000 W$$ So, the outlet temperature of cold water is \(55.62^{\circ}\mathrm{C}\), and the heat transfer rate is \(85000 W\).