Question

In a textile manufacturing plant, the waste dyeing water $\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)80^{\circ} \mathrm{C}$ is to be used to preheat fresh water $\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)10^{\circ} \mathrm{C}$ at the same flow rate in a double-pipe counterflow heat exchanger. The heat transfer surface area of the heat exchanger is \(1.65 \mathrm{~m}^{2}\), and the overall heat transfer coefficient is $625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the rate of heat transfer in the heat exchanger is \)35 \mathrm{~kW}$, determine the outlet temperature and the mass flow rate of each fluid stream.

Short answer

Based on the given information about a double-pipe counterflow heat exchanger, the outlet temperature of the hot fluid stream is approximately 46.06°C, and the mass flow rate of each fluid stream is approximately 1.44 kg/s.

Step-by-step solution

Calculate the heat transfer rate

Given the heat transfer rate (\(q = 35 \mathrm{~kW}\)), we first convert it to Watts to have consistent units for our calculations: \(q = 35,000 \mathrm{~W}\).

Calculate the temperature difference

The temperature difference between the streams can be found using the equation for the heat transfer rate: \(q = U \cdot A (T_{hot, in} - T_{cold, in})\). With the given information, we can rewrite the equation to find the temperature difference: $$T_{hot, in} - T_{cold, in} = \frac{q}{U \cdot A} = \frac{35,000}{625 * 1.65} \approx 33.94^{\circ}C$$.

Determine the outlet temperature of the hot fluid stream

Since hot and cold fluid streams are flowing at the same rate, the heat transfer to the cold stream is equal to the heat transfer from the hot stream. Thus, we use the heat transfer rate equation to find the outlet temperature of the hot fluid stream: $$q = m_{hot} \cdot c_{p_{hot}} \cdot \left(T_{hot, in} - T_{hot, out}\right)$$, where \(q\) is the heat transfer rate, \(m_{hot}\) is the mass flow rate of the hot fluid, and \(T_{hot, in}\) and \(T_{hot, out}\) are the inlet and outlet temperatures of the hot fluid stream, respectively. Rearranging this equation for the outlet temperature \(T_{hot, out}\), we get $$T_{hot, out} = T_{hot, in} - \frac{q}{m_{hot} \cdot c_{p_{hot}}}$$. Now we have \(T_{hot, in} = 80^{\circ}C\), \(c_{p_{hot}} = 4295 \mathrm{~J / kg \cdot K}\) and \(q = 35,000 \mathrm{~W}\). The only unknown is \(m_{hot}\), which we will determine in the next step.

Determine the mass flow rate of the hot fluid stream

We know that the heat transfer rate is equal for both the hot and cold fluid streams, so we can write the equation for the heat transfer rate of the cold fluid stream as $$q = m_{cold} \cdot c_{p_{cold}} \cdot \left(T_{cold, out} - T_{cold, in}\right)$$. Since the flow rates are the same for both streams, \(m_{hot} = m_{cold} = m\). With the given information, we have \(c_{p_{cold}} = 4180 \mathrm{~J / kg \cdot K}\), \(T_{cold, in} = 10^{\circ}C\), and \(T_{cold, out} = T_{cold, in} + 33.94^{\circ}C = 43.94^{\circ}C\). Plug in the values into the equation for the cold fluid stream and solve for \(m\): $$35,000 = m \cdot 4180 \cdot (43.94 - 10)$$ $$m \approx 1.44 \mathrm{~kg/s}$$.

Determine the outlet temperature of the hot fluid stream

Now we can use the mass flow rate and substituted it in step 3 equation and find the outlet temperature of the hot fluid: $$T_{hot, out} = 80 - \frac{35,000}{1.44 \cdot 4295} \approx 46.06^{\circ}C$$. The outlet temperature of the hot fluid stream is \(46.06^{\circ}C\), and the mass flow rate of each fluid stream is approximately \(1.44 \mathrm{~kg/s}\).