Question

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(1.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(880 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger.

Short answer

Question: Determine the rate of heat transfer and the heat transfer surface area of the heat exchanger. Answer: The rate of heat transfer is 157.5 kW, and the heat transfer surface area of the heat exchanger is approximately 5 m².

Step-by-step solution

Find the rate of heat transfer of cold water

To find the rate of heat transfer in the cold water stream, we can use the formula \(Q_{c} = m_{c}c_{p,c}(T_{c,out} - T_{c,in})\): \(Q_{c} = (1.25\,\text{kg/s})(4180\,\text{J/kgK})(45 ^{\circ}\text{C} - 15^{\circ}\text{C})\) Calculate \(Q_{c}\): \(Q_{c} = 157500\,\text{W}\) or \(157.5\,\text{kW}\)

Calculate Temperature of Hot Water Outlet

Since the heat exchanger is a counter-flow type, the heat transfer rate for the hot water must be equal to that of the cold water. Therefore, we can use the following formula to determine the temperature of the hot water outlet: \(Q_{c} = m_{h}c_{p,h}(T_{h,in} - T_{h,out})\) Solve for \(T_{h,out}\): \(T_{h,out} = T_{h,in} - \frac{Q_{c}}{m_{h}c_{p,h}} = 100 - \frac{157500}{(3.0\,\text{kg/s})(4190\,\text{J/kgK})}\) Calculate \(T_{h,out}\): \(T_{h,out} = 80.97^{\circ}\text{C}\)

Calculate the Logarithmic Mean Temperature Difference

Now, we can calculate the logarithmic mean temperature difference (\(\Delta T_{lm}\)) using the given formula: \(\Delta T_{lm} = \frac{(100 - 45) - (80.97 - 15)}{\ln\frac{100 - 45}{80.97 - 15}}\) Calculate \(\Delta T_{lm}\): \(\Delta T_{lm} = 35.65\,\text{K}\)

Calculate the Heat Transfer Surface Area

Finally, we can use the formula \(Q = UAS\Delta T_{lm}\) to calculate the required surface area of the heat exchanger. Rearranging the formula for surface area \(S\), we have: \(S = \frac{Q}{UA\Delta T_{lm}} = \frac{157500}{(880\,\text{W/m}^{2}\text{K})(35.65\,\text{K})}\) Calculate \(S\): \(S = 4.998\,\text{m}^{2}\) So, the rate of heat transfer is \(157.5\,\text{kW}\), and the heat transfer surface area of the heat exchanger is approximately \(5\,\text{m}^{2}\).

Key concepts

Counter-Flow Heat Exchanger

A counter-flow heat exchanger is a system where two fluids move in opposite directions, exchanging heat without mixing. In this setup, the hot and cold fluids enter the heat exchanger from opposite ends. This design advantageously allows for a more uniform temperature gradient along the length of the heat exchanger, resulting in a more efficient heat transfer process compared to parallel flow exchangers.

The effectiveness of a counter-flow arrangement is largely due to the fact that it can achieve a higher temperature change for the fluids involved. This is especially important when the objective is to heat a fluid, like the cold water in our exercise, as much as possible using the cooling fluid, such as the hot water delivered to a shower system.

Logarithmic Mean Temperature Difference

The logarithmic mean temperature difference (LMTD) is a crucial concept when dealing with heat exchangers. It represents an average temperature difference between the hot and cold fluids, accounting for the fact that this difference varies over the length of the heat exchanger. The LMTD method is a key to assessing the heat transfer subject to varying temperatures and is defined mathematically considering the inlet and outlet temperatures of both fluids involved.

In the provided exercise, the LMTD calculation forms a fundamental step to understand how efficiently the heat exchanger operates, revealing the driving force behind the heat transfer from the hot water to the cold water. It’s significant to note that accurately calculating the LMTD is necessary for determining the heat exchanger's size and cost.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, denoted by 'U', is a measure of a heat exchanger's ability to transfer heat between fluids through its material. It encompasses all forms of heat loss, including conduction through the heat exchanger walls and convection on both sides of the fluid exchange. A higher 'U' value represents a more efficient heat exchanger that can facilitate more heat transfer with a smaller surface area.

In our textbook solution, knowing the value of 'U' is integral to determining both the rate of heat transfer and the necessary surface area for the heat exchanger operation. The overall heat transfer coefficient is affected by several factors, including the materials used in the construction of the heat exchanger, the nature of the fluids, and the condition of the heat exchanger surface.

Heat Transfer Rate

The heat transfer rate, often expressed in watts or kilowatts, describes the amount of heat energy moving from one fluid to another per unit time in a heat exchanger. It is indicative of the heat exchanger’s performance: the higher the rate, the more effective the exchanger is at transporting energy.

In our example, the heat transfer rate was found by multiplying the mass flow rate of the cold water, its specific heat capacity, and the difference in temperature before and after the heat exchanger. This rate must be equal for both the hot and cold fluids assuming there is no heat loss to the surroundings, which underpins the first law of thermodynamics, stating that energy cannot be created or destroyed, only transferred. The calculated heat transfer rate directly influences the design and operational parameters of the heat exchanger.