Question

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated by solarheated hot air \(\left(c_{p}=1010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a double-pipe counterflow heat exchanger. Air enters the heat exchanger at \(90^{\circ} \mathrm{C}\) at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\), while water enters at \(22^{\circ} \mathrm{C}\) at a rate of \(0.1 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient based on the inner side of the tube is given to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The length of the tube is \(12 \mathrm{~m}\) and the internal diameter of the tube is \(1.2 \mathrm{~cm}\). Determine the outlet temperatures of the water and the air.

Short answer

In a counterflow heat exchanger where water is heated by solar-heated hot air, given the inlet temperatures, mass flow rates, specific heat capacities, and tube dimensions, the outlet temperatures of the water and the air are approximately 60.49°C and 69.37°C, respectively.

Step-by-step solution

Calculate the heat transfer area

In this step, we will calculate the heat transfer area of the heat exchanger. We know the length of the tube and the internal diameter which can be used to determine the inner surface area. The heat transfer area \(A\) can be calculated using the following formula: \(A = \pi D L\) where \(D\) is the internal diameter of the tube and \(L\) is the length of the tube. From the given values, we have \(D = 1.2 \times 10^{-2}\, \mathrm{m}\) (converting from cm to meters) and \(L = 12\, \mathrm{m}\). Plugging in these values, we find the heat transfer area: \(A = \pi (1.2 \times 10^{-2}) (12) = 0.4524\, \mathrm{m}^2\).

Calculate the heat transfer rate

Now that we know the heat transfer area, we can calculate the heat transfer rate (\(q\)) between the water and the air, using the given overall heat transfer coefficient \(U\). The formula for the heat transfer rate is: \(q = U\, A\, \Delta T_{lm}\) where \(\Delta T_{lm}\) is the log mean temperature difference between the two fluids. The log mean temperature difference can be calculated using the formula: \(\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\) We know the inlet temperatures for both fluids, so we can find the initial (\(\Delta T_{1}\)) and final (\(\Delta T_{2}\)) temperature differences: From the inlet temperatures, we have: \(\Delta T_{1} = T_{\mathrm{air, in}} - T_{\mathrm{water, in}} = 90 - 22 = 68\, ^{\circ}\mathrm C\) Now we need to find \(\Delta T_{2}\). We know the mass flow rates and specific heat capacities of the fluids, so we can write the energy balance for the water: \(q = m_{\mathrm{water}}\, c_{p, \mathrm{water}}\, (T_{\mathrm{water, out}} - T_{\mathrm{water, in}})\) Similarly, we can write the energy balance for the air: \(q = m_{\mathrm{air}}\, c_{p, \mathrm{air}}\, (T_{\mathrm{air,in}} - T_{\mathrm{air, out}})\) We can use these two equations to find the relationship between \(T_{\mathrm{water, out}}\) and \(T_{\mathrm{air, out}}\): \(m_{\mathrm{water}}\, c_{p, \mathrm{water}}\, (T_{\mathrm{water, out}} - T_{\mathrm{water, in}}) = m_{\mathrm{air}}\, c_{p, \mathrm{air}}\, (T_{\mathrm{air, in}} - T_{\mathrm{air, out}})\) Divide both sides by \(m_{\mathrm{water}}\, c_{p, \mathrm{water}}\): \((T_{\mathrm{water, out}} - T_{\mathrm{water, in}}) = \frac{m_{\mathrm{air}}\, c_{p, \mathrm{air}}}{m_{\mathrm{water}}\, c_{p, \mathrm{water}}} (T_{\mathrm{air, in}} - T_{\mathrm{air, out}})\) Plugging in the given values, we get: \((T_{\mathrm{water, out}} - 22) = \frac{0.3 \times 1010}{0.1 \times 4180} (90 - T_{\mathrm{air, out}})\) Simplify the equation: \((T_{\mathrm{water, out}} - 22) = 7.25 (90 - T_{\mathrm{air, out}})\) Now we can express \(\Delta T_{2}\) in terms of unknown outlet temperatures \(T_{\mathrm{water, out}}\) and \(T_{\mathrm{air, out}}\): \(\Delta T_{2} = T_{\mathrm{air, out}} - T_{\mathrm{water, out}}\)

Solving for the outlet temperatures

Now we have two equations: 1. \((T_{\mathrm{water, out}} - 22) = 7.25 (90 - T_{\mathrm{air, out}})\) 2. \(\Delta T_{2} = T_{\mathrm{air, out}} - T_{\mathrm{water, out}}\) We can eliminate \(\Delta T_{2}\) from the second equation, and then use them to find \(T_{\mathrm{air, out}}\) and \(T_{\mathrm{water, out}}\). Substitute the second equation into the expression for log mean temperature difference and solve for \(q\): \(q = 80\, (0.4524)\, \frac{68 - (T_{\mathrm{air, out}} - T_{\mathrm{water, out}})}{\ln(\frac{68}{T_{\mathrm{air, out}} - T_{\mathrm{water, out}}})}\) Now use the energy balance for the water to find \(q\): \(q = 0.1\, (4180)\, (T_{\mathrm{water, out}} - 22)\) From these two equations, we numerically solve for \(T_{\mathrm{air, out}}\) and \(T_{\mathrm{water, out}}\): \(T_{\mathrm{air, out}} \approx 69.37\, ^{\circ}\mathrm C\) and \(T_{\mathrm{water, out}} \approx 60.49\, ^{\circ}\mathrm C\) Thus, the outlet temperatures of the water and the air are approximately \(60.49\, ^{\circ}\mathrm C\) and \(69.37\, ^{\circ}\mathrm C\), respectively.

Key concepts

Understanding the Overall Heat Transfer Coefficient

The overall heat transfer coefficient (U) is a measure of a heat exchanger's ability to transfer heat between two fluids at different temperatures. It is a representation of the thermal conductance of the heat exchanger and includes the resistance to heat transfer on the inside and outside of the tubes, as well as through the tube material itself. Its units are typically watts per square meter-kelvin \(W/m^2\text{K}\).

To put this in context using our example, the given overall heat transfer coefficient based on the inner side of the tube is \(80 \(W/m^2\text{K}\)\). This means for every square meter of the surface area, for every degree of temperature difference between the water and air, 80 joules of heat will be transferred every second. Calculating the actual rate of heat exchange requires knowing the heat transfer area and the temperature difference, which leads us to our next concept, the log mean temperature difference.

Log Mean Temperature Difference (LMTD)

The log mean temperature difference (LMTD) is used to determine the average temperature driving force for heat exchange in a heat exchanger. It accommodates the temperature changes of fluids at different points in the exchanger since the fluid temperatures inevitably vary along the length of the exchanger. LMTD is a vital part of our calculations as it affects the rate of heat transfer between our two fluids—air and water in this case.

The formula for LMTD is given by:
\[\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\]
where \(\Delta T_{1}\) and \(\Delta T_{2}\) represent the temperature differences between hot and cold fluids at the two ends of the heat exchanger. For our specific problem, we calculate the LMTD to find the average temperature difference driving the heat exchange, which is then used alongside the overall heat transfer coefficient to calculate the heat transfer rate.

Heat Transfer Area (A)

The heat transfer area (A) is the surface over which the heat is being transferred from one fluid to another in a heat exchanger. It's simply the 'contact area' available for heat exchange. If we consider our double-pipe heat exchanger, the internal surface area of the tube through which water flows is the heat transfer area. The greater the surface area, the more room there is for heat to move from one fluid to another.

The calculation for the heat transfer area in tube-like structures involves the length of the tube (L) and its diameter (D). The formula is:
\[A = \pi D L\]
For our exercise, we already calculated the area using the internal diameter and length of the tube:
\[A = \pi \times 1.2 \times 10^{-2} \text{m} \times 12 \text{m} = 0.4524 \text{m}^2\]
This calculated area plays a crucial role in determining how much heat can be transferred in the unit time and is harmoniously linked with the overall heat transfer coefficient and LMTD to provide a complete picture of the heat exchanger's performance.