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Chapter 11: Problem 121

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in a 1 -shell and 2-tube heat exchanger with a surface area of \(0.5 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(0.5 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(15^{\circ} \mathrm{C}\), determine \((a)\) the heat transfer effectiveness, \((b)\) the outlet temperature of the cold water, and \((c)\) the heat transfer rate for the heat exchanger.

Short Answer

Expert verified
Using the given values for the mass flow rate, specific heat, and inlet temperature of the cold water, along with the calculated value for the maximum heat transfer rate (q_max), we can determine the heat transfer effectiveness (ε) from the expressions mentioned in Step 3 and 4: q_actual = m_cc_pc(T_co - T_ci) ε = (q_actual) / (q_max) To calculate ε, we need to find the value of q_actual, which requires the value of the outlet temperature of cold water (T_co). Unfortunately, without knowing the value of T_co or ε, we cannot find the heat transfer effectiveness directly. Additional information would be required to solve for ε. b) What is the outlet temperature of the cold water? As mentioned in part (a), we don't have enough information to calculate T_co or ε directly. We need more information to solve this part. c) What is the heat transfer rate for the heat exchanger? Once again, without having the value of T_co or ε, it isn't possible to directly calculate the actual heat transfer rate (q_actual).

Step by step solution

01

Calculate the maximum possible heat transfer

To calculate the maximum possible heat transfer, we need to find the temperature difference between the hot fluid (water vapor) and the cold fluid (water) at the inlet. The hot fluid temperature is \(100^{\circ}\mathrm{C}\), and the cold fluid temperature is \(15^{\circ}\mathrm{C}\). The maximum possible temperature difference is: \(\Delta T_{max} = T_{hot} - T_{cold} = 100^{\circ}\mathrm{C} - 15^{\circ}\mathrm{C} = 85^{\circ}\mathrm{C}\).
02

Determine the heat transfer effectiveness

The heat transfer effectiveness is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate. We can calculate the heat transfer effectiveness as follows: \(\varepsilon = \frac{q_{actual}}{q_{max}}\) To find \(q_{max}\), we multiply the heat transfer surface area \((A)\), overall heat transfer coefficient \((U)\), and the maximum temperature difference \((\Delta T_{max})\): \(q_{max} = UA \Delta T_{max}\) We are given \(A = 0.5\mathrm{~m}^2\) and \(U = 2000\mathrm{~W/m^2\cdot K}\). Now let's calculate \(q_{max}\): \(q_{max} = (2000\mathrm{~W/m^2\cdot K})(0.5\mathrm{~m}^{2})(85^{\circ}\mathrm{C}) = 85000\mathrm{~W}\)
03

Calculate the actual heat transfer rate

We know the mass flow rate of the cold water \((m_{c})\) is \(0.5\mathrm{~kg/s}\) and the specific heat of the cold water \((c_{pc})\) is \(4179\mathrm{~J/kg\cdot K}\). We can derive the actual heat transfer rate, \(q_{actual}\), from the following equation: \(q_{actual} = m_{c}c_{pc}(T_{co}-T_{ci})\), where \(T_{co}\) is the outlet temperature of the cold water and \(T_{ci}\) is the inlet temperature of the cold water \((15^{\circ}\mathrm{C})\). Let's rearrange the equation to determine \(T_{co}\) later: \(T_{co} = \frac{q_{actual}}{m_{c}c_{pc}} + T_{ci}\) First, we need to find the heat transfer effectiveness \((\varepsilon)\). From the previous step, we calculated \(q_{max}\). After rearranging the heat transfer effectiveness formula, we can find \(q_{actual}\): \(q_{actual} = \varepsilon q_{max}\)
04

Find the outlet temperature of the cold water

Finally, we substitute the values of \(q_{actual}\) and the given values for \(m_{c}\), \(c_{pc}\), and \(T_{ci}\) into the rearranged equation for \(T_{co}\): \(T_{co} = \frac{\varepsilon q_{max}}{m_{c}c_{pc}} + T_{ci}\) Now you can calculate the outlet temperature of the cold water, which will be part \((b)\) of the exercise. With \(T_{co}\), you can now return to step 3 and calculate the actual heat transfer rate, which is part \((c)\) of the exercise. Finally, the heat transfer effectiveness \((\varepsilon)\) itself is part \((a)\) of the exercise.

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Key Concepts

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

Heat Transfer Rate
Understanding the concept of heat transfer rate is fundamental to evaluating the performance of a heat exchanger. The heat transfer rate, denoted as \(q\), represents the amount of heat energy transferred per unit time and is measured in watts (W). In the context of a heat exchanger, the actual heat transfer rate \(q_{actual}\) involves the mass flow rate of the fluid, its specific heat \(c_p\), and the temperature difference between the inlet and the outlet.
To calculate \(q_{actual}\), the formula \(q_{actual} = m_c c_{pc}(T_{co} - T_{ci})\) is used, where \(m_c\) is the mass flow rate of the fluid, \(c_{pc}\) is the specific heat capacity of the fluid, \(T_{co}\) is the outlet temperature, and \(T_{ci}\) is the inlet temperature. By substituting into this formula, you can determine how much energy is being transferred to the fluid as it travels through the heat exchanger.
Outlet Temperature Calculation
Calculating the outlet temperature of the cold water in a heat exchanger, denoted as \(T_{co}\), is vital for understanding how effectively the heat exchanger is increasing the fluid's temperature. To do this, you need to rearrange the formula for the actual heat transfer rate to solve for \(T_{co}\).
By isolating \(T_{co}\) from the heat transfer rate equation, we get \(T_{co} = \frac{q_{actual}}{m_c c_{pc}} + T_{ci}\), where \(q_{actual}\) is derived from the effectiveness of the heat exchanger and the calculated maximum heat transfer rate \(q_{max}\). With these values, you can accurately determine the temperature of the fluid after it has passed through the heat exchanger, giving you insight into the exchanger's ability to heat the fluid.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, symbolized as \(U\), is a crucial factor in assessing a heat exchanger's efficiency. This coefficient quantifies how well heat is transferred through the exchanger materials from one fluid to another and depends on the nature of the heat transfer process, the physical properties of the fluids, and the characteristics of the heat exchanger itself.
To evaluate the heat exchanger performance, we utilize \(U\) along with the heat exchange surface area \(A\) and the driving temperature difference \(\triangle T_{max}\) to calculate the maximum possible heat transfer rate, \(q_{max} = U A \triangle T_{max}\). The overall heat transfer coefficient serves as an indicator of the quality of the heat exchanger. A higher \(U\) value indicates a more efficient heat transfer, often allowing the exchanger to be smaller for the same heat transfer rate — a crucial aspect in heat exchanger design and operation.

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Most popular questions from this chapter

How do heavy clothing and extreme environmental conditions affect the cardiovascular counter-current exchanger?

Water is boiled at \(150^{\circ} \mathrm{C}\) in a boiler by hot exhaust gases \(\left(c_{p}=1.05 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) that enter the boiler at \(400^{\circ} \mathrm{C}\) at a rate of \(0.4 \mathrm{~kg} / \mathrm{s}\) and leaves at \(200^{\circ} \mathrm{C}\). The surface area of the heat exchanger is \(0.64 \mathrm{~m}^{2}\). The overall heat transfer coefficient of this heat exchanger is (a) \(940 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(1056 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1145 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(1230 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(1393 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

In a textile manufacturing plant, the waste dyeing water \(\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to preheat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(15^{\circ} \mathrm{C}\) at the same flow rate in a double-pipe counter-flow 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.

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Oil is being cooled from \(180^{\circ} \mathrm{F}\) to \(120^{\circ} \mathrm{F}\) in a 1 -shell and 2-tube heat exchanger with an overall heat transfer coefficient of \(40 \mathrm{Btu} / \mathrm{h} \mathrm{ft} 2{ }^{\circ} \mathrm{F}\). Water \(\left(c_{p c}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) enters at \(80^{\circ} \mathrm{F}\) and exits at \(100^{\circ} \mathrm{F}\) with a mass flow rate of \(20,000 \mathrm{lbm} / \mathrm{h}\), determine (a) the log mean temperature difference and \((b)\) the surface area of the heat exchanger.
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