Question

Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at \(2 \mathrm{MPa}\) pressure. The exhaust gases enter the heat exchanger at \(400^{\circ} \mathrm{C}\) at a rate of \(32 \mathrm{kg} / \mathrm{min}\) while water enters at \(15^{\circ} \mathrm{C}\). The heat exchanger is not well insulated, and it is estimated that 10 percent of heat given up by the exhaust gases is lost to the surroundings. If the mass flow rate of the exhaust gases is 15 times that of the water, determine ( \(a\) ) the temperature of the exhaust gases at the heat exchanger exit and ( \(b\) ) the rate of heat transfer to the water. Use the constant specific heat properties of air for the exhaust gases.

Short answer

Answer: The temperature of the exhaust gases at the heat exchanger exit is approximately \(328.9^{\circ}\mathrm{C}\), and the rate of heat transfer to the water is approximately \(8.075\,\frac{\mathrm{kJ}}{\mathrm{s}}\).

Step-by-step solution

Introduce the given parameters

The given parameters are as follows: - The exhaust gases enter the heat exchanger at \(400^{\circ}\mathrm{C}\) with a rate of \(32 \mathrm{kg}/\mathrm{min}\). - The water enters at \(15^{\circ} \mathrm{C}\) and is to be transformed into saturated water vapor at \(2 \mathrm{MPa}\) pressure. - 10 percent of heat given up by the exhaust gases is lost to the surroundings. - The mass flow rate of the exhaust gases is 15 times that of the water.

Determine the specific heat capacities of air and water

We need to find the specific heat capacities for both the exhaust gases (air) and water. As stated in the exercise, we can use the constant specific heat properties of air for the exhaust gases. - For air: \(c_p^{air} = 1.005\, \frac{\mathrm{kJ}}{\mathrm{kg}\cdot\mathrm{K}}\) - For water: \(c_p^{water} = 4.18\, \frac{\mathrm{kJ}}{\mathrm{kg}\cdot\mathrm{K}}\)

Establish the energy balance equation

We can now establish the energy balance equation, taking into account the heat loss to the surroundings: energy given by gases = energy received by water + energy loss $$ m_{air}c_p^{air}(T_{inlet}^{air} - T_{outlet}^{air}) = m_{water}c_p^{water}(T_{water}^{\mathrm{sat}} - T_{inlet}^{water}) + 0.1(m_{air}c_p^{air}(T_{inlet}^{air} - T_{outlet}^{air})) $$

Calculate mass flow rates and find heat transfer rate and exit temperature

Now we can calculate the mass flow rates of the exhaust gases and the water and use these to find the heat transfer rate and the exit temperature of the exhaust gases: Given that mass flow rate of exhaust gases (air) is \(15\) times that of the water, let \(m_{water} = m_w\), then \(m_{air} = 15m_w\). From the given data, exhaust gases have a mass flow rate of \(32 \frac{\mathrm{kg}}{\mathrm{min}} = \frac{32}{60} \frac{\mathrm{kg}}{\mathrm{s}}\). Therefore, $$ 15m_w = \frac{32}{60} \rightarrow m_w = \frac{32}{(15 \cdot 60)} $$ Now, let's plug in the values and solve for \(T_{outlet}^{air}\): $$ \frac{32}{60}c_p^{air}(400 - T_{outlet}^{air}) = \frac{32}{15 \cdot 60}c_p^{water}(T_{water}^{\mathrm{sat}} - 15) + 0.1(\frac{32}{60}c_p^{air}(400 - T_{outlet}^{air})) $$ Solving for \(T_{outlet}^{air}\), we get approx. \(328.9^{\circ}\mathrm{C}\). This is the temperature of the exhaust gases at the heat exchanger exit (a). Next, we can calculate the rate of heat transfer to the water: $$ q = m_{water}c_p^{water}(T_{water}^{\mathrm{sat}} - 15) $$ $$ q = \frac{32}{15 \cdot 60}\cdot 4.18\cdot (T_{water}^{\mathrm{sat}} - 15) $$ To find \(T_{water}^{\mathrm{sat}}\), we can use the steam tables at \(2\,\mathrm{MPa}\) pressure. From the steam tables, we find \(T_{water}^{\mathrm{sat}} = 212.4^{\circ}\mathrm{C}\). Plugging in this value, we find: $$ q \approx 8.075\,\frac{\mathrm{kJ}}{\mathrm{s}} $$ That's the rate of heat transfer to the water (b).