91Ó°ÊÓ

The chief engineer at a university that is constructing a large number of new student dormitories decides to install a counterflow concentric tube heat exchanger on each of the dormitory shower drains. The thinwalled copper drains are of diameter \(D_{i}=50 \mathrm{~mm}\). Wastewater from the shower enters the heat exchanger at \(T_{h, i}=38^{\circ} \mathrm{C}\) while fresh water enters the dormitory at \(T_{c, l}=10^{\circ} \mathrm{C}\). The wastewater flows down the vertical wall of the drain in a thin, falling \(f\) m , providing \(h_{h}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the annular gap is \(d=10 \mathrm{~mm}\), the heat exchanger length is \(L=1 \mathrm{~m}\), and the water flow rate is \(\dot{m}=10 \mathrm{~kg} / \mathrm{min}\), determine the heat transfer rate and the outlet temperature of the warmed fresh water. (b) If a helical spring is installed in the annular gap so the fresh water is forced to follow a spiral path from the inlet to the fresh water outlet, resulting in \(h_{c}=9050 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer rate and the outlet temperature of the fresh water. (c) Based on the result for part (b), calculate the daily savings if 15,000 students each take a 10 -minute shower per day and the cost of water heating is \(\$ 0.07 / \mathrm{kW} \cdot \mathrm{h}\).

Step by step solution

01

1. Calculate the surface area for heat transfer (A)

For the annular gap, the effective surface area for heat transfer for a heat exchanger length L = 1 m is given by the formula \(A=2 \pi r L\), where r is the radius of the outer tube. With an annular gap of d = 10 mm, the outer radius would be \(r=R_o=25+5=30mm=0.03m\). Therefore, \(A=2 \pi (0.03)(1)\Longrightarrow A=0.1885\mathrm{~m}^2\)

02

2. Calculate the overall heat transfer coefficient (U)

Using the formula for the overall heat transfer coefficient in series for concentric tubes: \(\frac{1}{U A}=\frac{1}{h_{h} A_{h}}+\frac{ln \left(\frac{R_{o}}{R_{i}}\right)}{2 \pi k L}+\frac{1}{h_{c} A_{c}}\) In the given scenario, we have \(h_h = 10,000 W/m^2 \cdot K\). Since \(A_h \approx A_c\), we can assume \(A_c = A= 0.1885 m^2\). As there is no mention of any helical spring (case a), therefore, \(h_c = h_h\). The thermal conductivity of copper (k) = 385 W/m·K. Solving for U: \(\frac{1}{U (0.1885)}=\frac{1}{(10000)(0.1885)}+\frac{ln \left(\frac{0.03}{0.025}\right)}{2 \pi (385)(1)}+\frac{1}{(10000)(0.1885)}\) Solve for U: \(U = 5000 W/m^2 \cdot K \)

03

3. Calculate the heat capacity rates (C_h and C_c) of the hot and cold fluids

Given water flow rate is \(\dot{m} = \frac{10 kg}{min} = 0.1667 kg/s\). The specific heat capacity of water is \(c_p = 4190 J/kg \cdot K\). Therefore, \(C_h = C_c = \dot{m} c_p = (0.1667)(4190) = 698.3 W/K \)

04

4. Calculate the effectiveness (ε) of the heat exchanger

Using the effectiveness NTU method, we can calculate the effectiveness (ε) of the heat exchanger: \(\epsilon = \frac{1 - e^{(-NTU \cdot (1 - C_{_R}))}}{1 - C_{_R} \cdot e^{(-NTU \cdot (1 - C_{_R}))}}\) where \(NTU = \frac{U A}{C}\) and \(C_{_R} = \frac{C_{min}}{C_{max}}\). Since both heat capacity rates are equal, we have \(C_{_R} = 1\). Let's calculate the NTU: \(NTU = \frac{(5000)(0.1885)}{698.3}\Rightarrow NTU = 1.346\) Now, we can calculate the effectiveness: \(\epsilon = \frac{1 - e^{(-1.346 \cdot (1 - 1))}}{1 - 1 \cdot e^{(-1.346 \cdot (1 - 1))}} \Rightarrow \epsilon = 1 - e^0 \Rightarrow \epsilon = 1 - 1 \Rightarrow \epsilon = 0 \).

05

5. Determine the heat transfer rate (Q)

Using the effectiveness obtained, we can calculate the heat transfer rate (Q) as follows: \(Q = \epsilon C_h (T_{h,i} - T_{c,i}) \Rightarrow Q = 0\). There is no heat transfer happening in case (a). (b)

06

2b. Calculate the overall heat transfer coefficient (U) with helical spring

In this case, we are given that the fresh water flows in a spiral path due to a helical spring, and it results in a convective heat transfer coefficient, \(h_c = 9050 W/m^2 \cdot K\). With this new value of \(h_c\), we will recalculate the overall heat transfer coefficient (U): \(\frac{1}{U (0.1885)}=\frac{1}{(10000)(0.1885)}+\frac{ln \left(\frac{0.03}{0.025}\right)}{2 \pi (385)(1)}+\frac{1}{(9050)(0.1885)}\) Solve for U: \(U = 4520.3 W/m^2 \cdot K \)

07

4b. Calculate the effectiveness (ε) of the heat exchanger with helical spring

Following the same procedure as case (a), now with the new value of U (4520.3 W/m^2 · K), we calculate NTU and then effectiveness: \(NTU = \frac{(4520.3)(0.1885)}{698.3} \Rightarrow NTU = 1.2189\) Now, we can calculate the effectiveness: \(\epsilon = \frac{1 - e^{(-1.2189 \cdot (1 - 1))}}{1 - 1 \cdot e^{(-1.2189 \cdot (1 - 1))}} \Rightarrow \epsilon = 1 - e^0 \Rightarrow \epsilon = 0 \).

08

5b. Determine the heat transfer rate (Q) with helical spring

Heat transfer rate in case (b) also becomes zero following the same formula as in part (a). (c)

09

7. Calculate daily savings in water heating cost

Since there is no heat transfer happening in both cases and the effectiveness is zero, there are no savings in water heating cost in both cases. Daily savings in water heating cost: \(\$0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer Rate

In a counterflow concentric tube heat exchanger, the heat transfer rate is a critical parameter. It measures the quantity of heat exchanged between the hot and cold fluids per unit time. In our exercise, the objective was to determine this rate when wastewater from showers transfers heat to incoming fresh water. The process depends heavily on the overall difference in temperatures between these fluids, known as the driving force for heat transfer. The formula used is \[ Q = \epsilon C_h (T_{h,i} - T_{c,i}) \]where \(Q\) represents the heat transfer rate, \(\epsilon\) is the effectiveness, \(C_h\) is the heat capacity rate, and \((T_{h,i} - T_{c,i})\) is the temperature difference at the inlet.However, due to zero effectiveness discovered in the solution, the heat transfer rate resulted in zero, meaning no effective heat exchange was accomplished. This emphasizes the importance of achieving a suitable condition for heat transfer and ensuring factors like fluid flow configurations are optimally designed.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, denoted as \(U\), is a combined measure of all the resistances to heat flow encountered between the hot and cold fluids. It involves conduction through the heat exchanger material and convection on both fluid sides. In the exercise, we see calculations done for \(U\) both with normal annular flow configuration and when a helical spring is introduced to enhance flow dynamics. The final formula used was: \[ \frac{1}{U A} = \frac{1}{h_{h} A_{h}} + \frac{\ln \left( \frac{R_o}{R_i} \right)}{2 \pi k L} + \frac{1}{h_{c} A_{c}} \]where \(h_h\) and \(h_c\) are the convective heat transfer coefficients for the hot and cold sides respectively, \(A\) is the surface area, \(k\) is the thermal conductivity, and \(L\) is the length of the heat exchanger. Even small changes in these factors, as demonstrated by the influence of a helical path, can significantly alter the overall heat transfer coefficient. Yet in this scenario, calculated values of \(U\) still led to zero effectiveness, thus indicating that design modifications affecting fluid behaviors may be necessary for success.

Heat Exchanger Effectiveness

Effectiveness is a measure of how well a heat exchanger performs relative to its maximum possible performance under given conditions. It is expressed in terms of the Number of Transfer Units (NTU) and is calculated using the relationship: \[ \epsilon = \frac{1 - e^{(-NTU \cdot (1 - C_R))}}{1 - C_R \cdot e^{(-NTU \cdot (1 - C_R))}} \]where \(NTU = \frac{U A}{C}\), and \(C_R\) represents the heat capacity ratio.For the exercise case, the unfortunate result was zero effectiveness due to the NTU value and equal heat capacity rates \((C_h = C_c)\). This zero value indicates that the design wasn’t effectively transferring heat. It's a lesson in heat exchanger optimization where assessment of flow dynamics and temperature profiles need to be carefully designed to prevent similar ineffectiveness.

Water Heating Cost Savings

Economically, the efficiency of a heat exchanger translates to cost savings in terms of water heating. The goal of installing such systems in dormitory showers is to lower the overall energy expenditure by reclaiming some of the heat from wastewater. In our scenario, the savings were calculable by: - Determining the amount of heat recovered, which allows us to cut down on additional heating - Multiplying the recovered heat by the cost of water heating. However, with zero effectiveness as discovered in the exercise for both scenarios, there were no actual savings to be made. Designing effective heat exchangers is crucial because being able to reclaim even a modest amount of energy can lead to significant cost reductions, especially when scaled up to multiple installations across a large institution. Improving the effectiveness not only saves money but also contributes to environmental conservation by reducing energy consumption.