A 1 -shell and 2-tube type heat exchanger has an overall heat transfer coefficient of \(300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). The shell side fluid has a heat capacity rate of \(20,000 \mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\), while the tube side fluid has a heat capacity rate of \(40,000 \mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\). The inlet temperatures on the shell side and tube side are \(200^{\circ} \mathrm{F}\) and \(90^{\circ} \mathrm{F}\), respectively. If the total heat transfer area is \(100 \mathrm{ft}^{2}\), determine \((a)\) the heat transfer effectiveness and \((b)\) the actual heat transfer rate in the heat exchanger.

Short Answer

Expert verified
Answer: (a) The heat transfer effectiveness is approximately 0.662. (b) The actual heat transfer rate is approximately 1,459,280 Btu/h.

Step by step solution

01

Calculate the minimum heat capacity rate and heat capacity rate ratio

We are given the heat capacity rates for the shell side fluid (\(C_{s} = 20,000 \mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\)) and the tube side fluid (\(C_{t} = 40,000 \mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\)). The minimum heat capacity rate (\(C_{min}\)) is the smallest of the two heat capacity rates: \(C_{min} = \min(C_{s}, C_{t}) = 20,000 \mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\) The heat capacity rate ratio (\(C_r\)) is the ratio of the minimum heat capacity rate to the maximum heat capacity rate: \(C_r = \frac{C_{min}}{\max(C_{s}, C_{t})} = \frac{20,000}{40,000} = 0.5\)
02

Calculate the number of transfer units (NTU)

The number of transfer units (NTU) is given by the formula: \(NTU = \frac{U \cdot A}{C_{min}}\) where \(U\) is the overall heat transfer coefficient, \(A\) is the total heat transfer area, and \(C_{min}\) is the minimum heat capacity rate. We are given that \(U = 300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) and \(A = 100 \mathrm{ft}^{2}\). \(NTU = \frac{300 \cdot 100}{20,000} = 1.5\)
03

Determine the heat transfer effectiveness

For a 1-shell and 2-tube type heat exchanger, the heat transfer effectiveness (\(\epsilon\)) can be calculated using the following formula: \(\epsilon = \frac{1 - e^{-NTU(1+C_r)}(1+C_r)}{1 + C_r}\) Substituting the values of \(NTU\) and \(C_r\): \(\epsilon = \frac{1 - e^{-1.5(1+0.5)}(1+0.5)}{1 + 0.5}\) \(\epsilon \approx 0.662\) So, the heat transfer effectiveness is approximately 0.662.
04

Calculate the actual heat transfer rate

The actual heat transfer rate (\(q\)) can be determined using the heat transfer effectiveness (\(\epsilon\)), the minimum heat capacity rate (\(C_{min}\)), and the temperature difference between the inlet temperatures of the shell side and tube side fluids (\(T_{s_in} - T_{t_in}\)): \(q = \epsilon \cdot C_{min} \cdot (T_{s_in} - T_{t_in})\) We have already calculated the heat transfer effectiveness (\(\epsilon \approx 0.662\)) and the minimum heat capacity rate (\(C_{min} = 20,000 \mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\)). The inlet temperatures of the shell side and tube side fluids are \(T_{s_in} = 200^{\circ} \mathrm{F}\) and \(T_{t_in} = 90^{\circ} \mathrm{F}\), respectively. \(q \approx 0.662 \cdot 20,000 \cdot (200 - 90)\) \(q \approx 1,459,280 \mathrm{Btu/h}\) So, the actual heat transfer rate in the heat exchanger is approximately \(1,459,280 \mathrm{Btu/h}\). In conclusion, (a) the heat transfer effectiveness of the heat exchanger is approximately 0.662 and (b) the actual heat transfer rate is approximately \(1,459,280 \mathrm{Btu/h}\).

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Key Concepts

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

Overall Heat Transfer Coefficient
The overall heat transfer coefficient, denoted by the symbol U, is a measure of how well a heat exchanger can transfer heat between two fluids. It's essentially a combination of several factors: the material conductance, the convection heat transfer at the fluid-solid interfaces, and any fouling resistances. This coefficient plays a crucial role in quantifying the rate at which heat is exchanged.

In our exercise, the overall heat transfer coefficient is given as 300 Btu/h•ft²•°F. To utilize this value, we employ the formula for calculating the number of transfer units (NTU), where U is multiplied by the total heat transfer area (A) and divided by the minimum heat capacity rate (Cmin). Understanding the concept of U allows students to diagnose how changes in the heat exchanger design or operation can impact the heat transfer rate.
Minimum Heat Capacity Rate
When analyzing a heat exchanger, the minimum heat capacity rate, Cmin, is a critical value that influences the overall behavior of the heat exchange process. It is the lower value between the heat capacity rates of the two fluids involved. The heat capacity rate is a product of a fluid's mass flow rate and its specific heat and represents the fluid's ability to absorb or release heat.

In our worked example, the minimum heat capacity rate is 20,000 Btu/h•°F, which is the smaller value between the shell side and tube side fluids. This figure is significant because it sets a limit on the maximum amount of heat the heat exchanger can transfer. Beyond comprehending how to calculate Cmin, students should appreciate its importance in determining heat exchanger performance and the engineering considerations that might arise from this limitation.
Number of Transfer Units (NTU)
The concept of the Number of Transfer Units (NTU) provides a dimensionless measure of the size of a heat exchanger relative to the flow rate and thermal properties of the fluids involved. The NTU approach is a valuable tool for determining the heat exchanger's effectiveness, which is a measure of its thermal performance compared to the maximum possible heat transfer.

As seen in the exercise, the NTU is calculated by the formula NTU = U • A / Cmin. A higher NTU value indicates a larger capacity of the heat exchanger to allow heat transfer between the fluids. In the example, we computed an NTU of 1.5 and used it to find the heat exchanger's effectiveness. Students should understand that the NTU helps predict how changing variables like size or flow rates might affect the efficacy of the heat transfer process.

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Most popular questions from this chapter

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) to cool the engine-coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(30^{\circ} \mathrm{C}\) air through this radiator at a rate of \(10 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, the effectiveness of the radiator is \(0.4\). Determine \((a)\) the outlet temperature of the air and (b) the rate of heat transfer between the two fluids.

A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

In a one-shell and eight-tube pass heat exchanger, the temperature of water flowing at rate of \(50,000 \mathrm{lbm} / \mathrm{h}\) is raised from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\). Hot air \(\left(c_{p}=0.25 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) that flows on the tube side enters the heat exchanger at \(600^{\circ} \mathrm{F}\) and exits at \(300^{\circ} \mathrm{F}\). If the convection heat transfer coefficient on the outer surface of the tubes is \(30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\), determine the surface area of the heat exchanger using both LMTD and \(\varepsilon-\mathrm{NTU}\) methods. Account for the possible fouling resistance of \(0.0015\) and \(0.001 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on water and air side, respectively.

Describe the cardiovascular counter-current mechanism in the human body.

In a chemical plant, a certain chemical is heated by hot water supplied by a natural gas furnace. The hot water \(\left(c_{p}=\right.\) \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) is then discharged at \(60^{\circ} \mathrm{C}\) at a rate of \(8 \mathrm{~kg} / \mathrm{min}\). The plant operates \(8 \mathrm{~h}\) a day, 5 days a week, 52 weeks a year. The furnace has an efficiency of 78 percent, and the cost of the natural gas is \(\$ 1.00\) per therm ( 1 therm \(=105,500 \mathrm{~kJ})\). The average temperature of the cold water entering the furnace throughout the year is \(14^{\circ} \mathrm{C}\). In order to save energy, it is proposed to install a water-to-water heat exchanger to preheat the incoming cold water by the drained hot water. Assuming that the heat exchanger will recover 72 percent of the available heat in the hot water, determine the heat transfer rating of the heat exchanger that needs to be purchased and suggest a suitable type. Also, determine the amount of money this heat exchanger will save the company per year from natural gas savings.

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