Question

A heat exchanger is used to heat cold water at \(15^{\circ} \mathrm{C}\) entering at a rate of \(2 \mathrm{kg} / \mathrm{s}\) by hot air at \(85^{\circ} \mathrm{C}\) entering at a rate of \(3 \mathrm{kg} / \mathrm{s}\). The heat exchanger is not insulated and is losing heat at a rate of \(25 \mathrm{kJ} / \mathrm{s}\). If the exit temperature of hot air is \(20^{\circ} \mathrm{C},\) the exit temperature of cold water is \((a) 28^{\circ} \mathrm{C}\) (b) \(35^{\circ} \mathrm{C}\) \((c) 38^{\circ} \mathrm{C}\) \((d) 78^{\circ} \mathrm{C}\) \((e) 90^{\circ} \mathrm{C}\)

Short answer

Answer: The exit temperature of the cold water is approximately 38°C.

Step-by-step solution

Apply the conservation of energy

For the heat exchanger system, we can use the equation: \(\dot{Q}_{gain} - \dot{Q}_{loss} = \dot{m}_w \cdot c_{w} \cdot (T_{w_out} - T_{w_in}) + \dot{m}_a \cdot c_{a} \cdot (T_{a_out} - T_{a_in})\) where \(\dot{Q}_{gain}\) is the heat gained by the cold water, \(\dot{Q}_{loss}\) is the heat loss from the heat exchanger, \(\dot{m}_w\) is the mass flow rate of the cold water, \(c_w\) is the specific heat capacity of the water (approximated as \(4.186 \mathrm{kJ/(kg\cdot ^\circ C)}\)), \(T_{w_in}\) is the initial water temperature, \(T_{w_out}\) is the exit water temperature, \(\dot{m}_a\) is the mass flow rate of the hot air, \(c_a\) is the specific heat capacity of the air (approximated as \(1.005 \mathrm{kJ/(kg\cdot ^\circ C)}\)), \(T_{a_in}\) is the initial air temperature, and \(T_{a_out}\) is the exit air temperature.

Plug in the given values and solve for the unknown exit water temperature

Now, we can plug in the given values into the equation: \(\dot{Q}_{gain} - 25\mathrm{kJ/s} = (2\frac{\mathrm{kg}}{\mathrm{s}}) \cdot (4.186\mathrm{kJ/(kg\cdot ^\circ C)}) \cdot (T_{w_out} - 15\mathrm{^\circ C}) + (3\frac{\mathrm{kg}}{\mathrm{s}}) \cdot (1.005\mathrm{kJ/(kg\cdot ^\circ C)}) \cdot (20\mathrm{^\circ C} - 85\mathrm{^\circ C})\) \(\dot{Q}_{gain} = 2 \cdot 4.186 \cdot (T_{w_out} - 15) - 3 \cdot 1.005 \cdot (-65) - 25 \) Now, solve for \(\dot{Q}_{gain}\): \(\dot{Q}_{gain} = 8.372 \cdot (T_{w_out} - 15) + 195.975 - 25\) Now, solve for \(T_{w_out}\): \(T_{w_out} = \frac{\dot{Q}_{gain} + 25 - 195.975}{8.372} + 15\) \(T_{w_out} \approx 38^\circ C\) Hence, the exit temperature of the cold water is approximately \(38^{\circ} \mathrm{C}\), which corresponds to option (c).