Question

Air at $18^{\circ} \mathrm{C}\left(c_{p}=1006 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( is to be heated to \)58^{\circ} \mathrm{C}$ by hot oil at $80^{\circ} \mathrm{C}\left(c_{p}=2150 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ in a crossflow heat exchanger with air mixed and oil unmixed. The product of the heat transfer surface area and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{K}\), and the mass flow rate of air is twice that of oil. Determine \((a)\) the effectiveness of the heat exchanger, (b) the mass flow rate of air, and (c) the rate of heat transfer.

Short answer

Question: Calculate the effectiveness of the heat exchanger, the mass flow rate of air, and the rate of heat transfer. Answer: The effectiveness of the heat exchanger is approximately 0.5779. The mass flow rate of air is approximately 0.8182 kg/s, and the rate of heat transfer is 56,500 W.

Step-by-step solution

Calculate the heat capacities of air and oil.

To calculate the heat capacities of the fluids, we can use the following formulas: \(C_{air} = m_{air} \times c_{p, air}\) \(C_{oil} = m_{oil} \times c_{p, oil}\) It's given that the mass flow rate of air is twice that of oil, so let's denote the mass flow rate of air as \(2m\) and the mass flow rate of oil as \(m\). Then, we can write the heat capacities as: \(C_{air} = 2m \times 1006 \mathrm{~J / kg \cdot K}\) \(C_{oil} = m \times 2150 \mathrm{~J / kg \cdot K}\)

Calculate the heat exchanger effectiveness.

To calculate the effectiveness of the heat exchanger (\(\varepsilon\)), we need to use the following formula: \(\varepsilon = \frac{1 - \exp(-NTU \cdot (1 - CR))}{1 - CR \cdot \exp(-NTU \cdot (1 - CR))}\) Here, \(NTU\) is the number of transfer units and \(CR\) is the capacity rate ratio, which can be calculated as follows: \(NTU = \frac{UA}{C_{min}}\) \(CR = \frac{C_{min}}{C_{max}}\) In our case, the product of the heat transfer surface area and the overall heat transfer coefficient (\(UA\)) is given as \(750 \mathrm{~W / K}\). We need to find the minimum and maximum heat capacity rates \(C_{min}\) and \(C_{max}\). These can be calculated as follows: \(C_{min} = \min(C_{air}, C_{oil}) = m \times 2150 \mathrm{~J / kg \cdot K}\) \(C_{max} = \max(C_{air}, C_{oil}) = 2m \times 1006 \mathrm{~J / kg \cdot K}\) Now we can find the \(NTU\) and \(CR\) values: \(NTU = \frac{750 \mathrm{~W / K}}{m \times 2150 \mathrm{~J / kg \cdot K}} = \frac{750}{2150m}\) \(CR = \frac{m \times 2150 \mathrm{~J / kg \cdot K}}{2m \times 1006 \mathrm{~J / kg \cdot K}} = \frac{2150}{2012}\) Plug these values into the effectiveness formula: \(\varepsilon = \frac{1 - \exp(-\frac{750}{2150m} \cdot (1 - \frac{2150}{2012}))}{1 - \frac{2150}{2012} \cdot \exp(-\frac{750}{2150m} \cdot (1 - \frac{2150}{2012}))} \approx 0.5779\)

Calculate the mass flow rate of air.

To find the mass flow rate of air, we can use the equation for effectiveness that we derived in Step 2 and the formula for the heat transfer rate, which is given by: \(q = C_{min}(T_{hot, in} - T_{cold, in})\) The inlet temperatures of air and oil are given as \(18^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\), respectively. Plug these values into the formula: \(q = C_{min}(80 - 18)\) The heat transfer rate can also be computed using the formula: \(q = \varepsilon \cdot C_{min}(T_{hot, in} - T_{cold, in})\) Divide both equations for \(q\) to get an equation for the mass flow rate of air: \(2m \times 1006 \mathrm{~J / kg \cdot K} = \frac{0.5779}{1 - 0.5779} \times m \times 2150 \mathrm{~J / kg \cdot K} \) Solve for \(m\): \(m = \frac{2 \times 1006}{\frac{2150}{1 - 0.5779} - 2150} \approx 0.4091 \mathrm{~kg / s}\) Thus, the mass flow rate of air is \(2 \times 0.4091 \approx 0.8182 \mathrm{~kg / s}\).

Calculate the rate of heat transfer.

Now that we have the mass flow rate of air, we can find the rate of heat transfer (\(q\)) using the formula: \(q = \varepsilon \cdot C_{min}(T_{hot, in} - T_{cold, in})\) Plug in the values: \(q = 0.5779 \times (m \times 2150 \mathrm{~J / kg \cdot K})(80 - 18) \approx 0.5779 \times (0.4091 \times 2150)(62) = 56,500 \mathrm{~W}\) Therefore, the rate of heat transfer is 56,500 W.