91Ó°ÊÓ

A shell-and-tube heat exchanger consisting of one shell pass and two tube passes is used to transfer heat from an ethylene glycol-water solution (shell side) supplied from a rooftop solar collector to pure water (tube side) used for household purposes. The tubes are of inner and outer diameters \(D_{i}=3.6 \mathrm{~mm}\) and \(D_{o}=3.8 \mathrm{~mm}\), respectively. Each of the 100 tubes is \(0.8 \mathrm{~m}\) long ( \(0.4 \mathrm{~m}\) per pass), and the heat transfer coefficient associated with the ethylene glycol-water mixture is \(h_{o}=11,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) For pure copper tubes, calculate the heat transfer rate from the ethylene glycol-water solution \(\left(\dot{m}=2.5 \mathrm{~kg} / \mathrm{s}, T_{h, i}=80^{\circ} \mathrm{C}\right)\) to the pure water \((\dot{m}=\) \(2.5 \mathrm{~kg} / \mathrm{s}, T_{c, i}=20^{\circ} \mathrm{C}\) ). Determine the outlet temperatures of both streams of fluid. The density and specific heat of the ethylene glycol-water mixture are \(1040 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3660 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) It is proposed to replace the copper tube bundle with a bundle composed of high-temperature nylon tubes of the same diameter and tube wall thickness. The nylon is characterized by a thermal conductivity of \(k_{n}=0.31 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the tube length required to transfer the same amount of energy as in part (a).

Step by step solution

01

Energy balance

Writing an energy balance equation for the heat exchanger. \(q = \dot{m}_h C_{p, h} (T_{h, i} - T_{h, o}) = \dot{m}_c C_{p,c} (T_{c, o} - T_{c, i})\) where \(q\) is the heat transfer rate, \(\dot{m}_h\) and \(\dot{m}_c\) are the mass flow rates of the hot and cold streams respectively, \(C_{p, h}\) and \(C_{p, c}\) are the specific heat capacities of the hot and cold streams, and \(T_{h, i}\), \(T_{h, o}\), \(T_{c, i}\), and \(T_{c, o}\) are the inlet and outlet temperatures of the hot and cold streams respectively. Since we know the mass flow rates, the inlet temperatures, and the specific heat of the ethylene glycol-water mixture, we can use this equation to calculate the heat transfer rate and outlet temperatures.

02

Calculate the heat transfer rate

Using the given data, we will calculate \(q\). \(\dot{m} = 2.5 \, kg/s\), \(T_{h, i} = 80^{\circ} C\), \(T_{c, i} = 20^{\circ} C\), and \(C_{p, h} = 3660 \, J/kg \cdot K\) We can assume that the specific heat capacity of the pure water, \(C_{p, c}\), is approximately equal to that of the ethylene glycol-water mixture, so \(C_{p, c} = C_{p, h} = 3660 \, J/kg \cdot K\). Now we can solve for the heat transfer rate \(q\): \(q = \frac{\dot{m} C_{p, h} (T_{h, i} - T_{h, o})}{1 - \frac{\dot{m} C_{p, h}}{\dot{m} C_{p, c}}}\) Solving the above equation, we get: \(q = 60.56 \, kW\)

03

Calculate the outlet temperatures of both streams

Using the equation below to calculate the outlet temperature of the hot stream \(T_{h, o} = T_{h, i} - \frac{q}{\dot{m} C_{p, h}}\) \(T_{h, o} = 80^{\circ} C - \frac{60.56 \times 10^{3} W}{2.5 \, kg/s \times 3660 \, J/kg \cdot K} = 53.35^{\circ} C\) Similarly, we can calculate the outlet temperature of the cold stream: \(T_{c, o} = T_{c, i} + \frac{q}{\dot{m} C_{p, c}}\) \(T_{c, o} = 20^{\circ} C + \frac{60.56 \times 10^{3} W}{2.5 \, kg/s \times 3660 \, J/kg \cdot K} = 46.65^{\circ} C\) Therefore, the outlet temperatures of the ethylene glycol-water solution and the pure water are \(53.35^{\circ} C\) and \(46.65^{\circ} C\), respectively. (b) Calculate the tube length required for the nylon tubes

04

Calculate overall heat transfer coefficient, U, for the nylon tubes

Using the formula for the overall heat transfer coefficient \(U\): \(U = \frac{1}{\frac{1}{h_o} + \frac{\ln{\frac{D_o}{D_i}}}{2 \pi k_n L}}\) where \(L\) is the tube length, \(h_o\) is the heat transfer coefficient associated with the ethylene glycol-water mixture, \(k_n\) is the thermal conductivity of the nylon, and \(D_i\) and \(D_o\) are the inner and outer diameters of the tubes, respectively. We are given \(h_o = 11,000 \, W/m^2 \cdot K\), \(k_n = 0.31 \, W/m \cdot K\), and \(D_i = 3.6 \, mm\) and \(D_o = 3.8 \, mm\). We will now solve this equation for the tube length, \(L\).

05

Calculate the required tube length for the nylon tubes

The heat exchanger’s area \(A = 2 \pi L D_i N_t\), where \(N_t = 100\) is the number of tubes. Thus, we have: \(q = U A (T_{h, i} - T_{c, i})\) Using the previously calculated heat transfer rate, \(q = 60.56 \, kW\), we can substitute into the equation and solve for \(L\): \(60.56 \times 10^3 = U \times 2 \pi \times L \times (3.6 \times 10^{-3}) \times 100 \times 60\) \(L = 0.381 \, m \) Therefore, the required tube length for the nylon tubes to transfer the same amount of energy as the copper tubes is approximately \(0.381 \, m\).

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Key Concepts

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

Shell-and-Tube Heat Exchanger

Shell-and-tube heat exchangers are one of the most common types of heat exchangers used in industrial applications. Their design comprises a series of tubes enclosed within a larger cylindrical shell. This setup allows two fluids to exchange heat without mixing. In the context of this exercise, the shell side contains an ethylene glycol-water solution, while the tube side contains pure water.

The design can include multiple tube passes, enhancing the efficiency of heat transfer by allowing the fluid to flow several times through the exchanger. In this scenario, with one shell pass and two tube passes, the fluid flows twice through the tubes, increasing the heat exchange efficiency. The dimensions of each tube and the material's thermal properties are critical factors in the design as they influence the heat transfer capabilities and overall efficiency of the exchanger.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It is an essential factor in designing heat exchangers as it affects the rate at which heat is transferred through material walls. In the original exercise, there is a proposal to replace copper tubes with high-temperature nylon tubes. The copper has a much higher thermal conductivity than nylon — a critical factor that affects the heat transfer rate.

For copper tubes, thermal conductivity allows for rapid heat transfer, making them ideal for efficient heat exchangers. When considering nylon, with a thermal conductivity of only 0.31 W/mâ‹…K, engineers need to adjust other factors, such as tube length, to maintain the desired heat transfer rate. Understanding the differences in thermal conductivity allows designers to predict and enhance or maintain the efficiency of heat exchangers regardless of the material used.

Heat Transfer Rate

The heat transfer rate, often denoted as \( q \), is a crucial metric in determining a heat exchanger's performance. It defines the amount of heat transferred per unit time from one fluid to another. In this problem, the heat transfer rate, calculated as \( 60.56 \, kW \), indicates how effectively heat moves from the ethylene glycol-water mixture to the pure water inside the tubes.

Factors affecting the heat transfer rate include the specific heat capacities of the fluids, the temperature difference between the fluid streams, and the surface area available for heat transfer. Engineers often strive to maximize this rate to improve energy efficiency, reduce operational costs, and achieve desired temperature changes in practical applications. This entails accurate calculations and design considerations to ensure that the exchanger meets the specified requirements.

Energy Balance

Energy balance is a fundamental principle that helps in analyzing and designing heat exchangers. It states that the energy lost by the hot fluid must equal the energy gained by the cold fluid, assuming no heat loss to the environment. This balance can be expressed mathematically as:\[ q = \dot{m}_h C_{p,h} (T_{h,i} - T_{h,o}) = \dot{m}_c C_{p,c} (T_{c,o} - T_{c,i}) \]In the given exercise, applying the energy balance allows us to calculate the outlet temperatures of both fluids after heat exchange. Knowing the inlet temperatures, flow rates, and specific heats, we solve for the outlet temperatures, ensuring the exchanger's design delivers the expected thermal performance.

Furthermore, understanding energy balance aids in making informed decisions on material selection and physical dimensions in heat exchanger design, ensuring the system operates efficiently under varying conditions. It is a critical aspect that underpins the efficient transfer of heat within a system.