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A heat exchanger contains 400 tubes with inner diameter of \(23 \mathrm{~mm}\) and outer diameter of \(25 \mathrm{~mm}\). The length of each tube is \(3.7 \mathrm{~m}\). The corrected log mean temperature difference is \(23^{\circ} \mathrm{C}\), while the inner surface convection heat transfer coefficient is \(3410 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the outer surface convection heat transfer coefficient is \(6820 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the thermal resistance of the tubes is negligible, determine the heat transfer rate.

Short Answer

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Question: Determine the heat transfer rate in a heat exchanger with given dimensions and heat transfer coefficients. Answer: To find the heat transfer rate, follow these steps: 1. Calculate the inner and outer surface area per tube using the formula \(A=2\pi rL\). 2. Convert the surface areas to m² from mm². 3. Calculate the total inner and outer surface areas by multiplying the surface area per tube with the number of tubes (400). 4. Calculate the overall heat transfer coefficient using the equation \(\frac{1}{U_{total}} = \frac{1}{h_i} + \frac{1}{h_o}\). 5. Calculate the heat transfer rate using the equation \(Q = U_{total} \cdot A_{total\_inner} \cdot \Delta T_{lm}\). The heat transfer rate (Q) is the final answer.

Step by step solution

01

Calculate the inner and outer surface area per tube

To calculate the surface area per tube, we need to find the inner and outer surface area for a single tube. The formula for the surface area of a cylinder is \(A=2\pi rL\), where \(r\) is the radius and \(L\) is the length. Let's find the surface area for both surfaces: Inner surface area per tube, \(A_i\): $$A_i = 2 \pi \cdot \frac{23 \,\mathrm{mm}}{2} \cdot 3.7 \,\mathrm{m}$$ Outer surface area per tube, \(A_o\): $$A_o = 2 \pi \cdot \frac{25 \,\mathrm{mm}}{2} \cdot 3.7 \,\mathrm{m}$$
02

Convert the surface areas to m²

As the diameters are provided in millimeters, we should convert the surface areas from \(\mathrm{mm}^2\) to \(\mathrm{m}^2\): $$A_i = \frac{A_i}{1000^2} \,\mathrm{m}^2$$ $$A_o = \frac{A_o}{1000^2} \,\mathrm{m}^2$$
03

Calculate total inner and outer surface areas

Next, we will find the total inner and outer surface area by multiplying the surface area per tube with the number of tubes (400): $$A_{total\_inner} = 400 \cdot A_i \,\mathrm{m}^{2}$$ $$A_{total\_outer} = 400 \cdot A_o \,\mathrm{m}^{2}$$
04

Calculate the overall heat transfer coefficient

As the thermal resistance of the tubes is negligible, we can use the equation for the overall heat transfer coefficient (\(U_{total}\)) based on the inner and outer heat transfer coefficients (\(h_i\) and \(h_o\)) and the inner and outer surface areas (\(A_i\) and \(A_o\)): $$\frac{1}{U_{total}} = \frac{1}{h_i} + \frac{1}{h_o}$$ Now, solving for \(U_{total}\): $$U_{total} = \frac{1}{\left(\frac{1}{h_i} + \frac{1}{h_o}\right)}$$ $$U_{total}=\frac{1}{\left(\frac{1}{3410\,\mathrm{W/m^2K}}+\frac{1}{6820\,\mathrm{W/m^2K}}\right)}$$
05

Calculate the heat transfer rate

Now we can determine the heat transfer rate (\(Q\)) using the overall heat transfer coefficient, total surface area of the heat exchanger, and the corrected log mean temperature difference (\(\Delta T_{lm}\)): $$Q = U_{total} \cdot A_{total\_inner} \cdot \Delta T_{lm}$$ $$Q = U_{total} \cdot 400 \cdot A_i \cdot 23^{\circ} \mathrm{C}$$ The heat transfer rate \(Q\) is the final answer.

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Key Concepts

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

Heat Exchanger Analysis
Understanding how heat exchangers work is crucial when dealing with systems designed to transfer heat from one fluid to another. Heat exchangers are used in a variety of applications, including refrigeration, air conditioning systems, and power plants. The analysis begins by determining the surface area available for heat transfer, taking into account the number of tubes and their dimensions. With the surface areas in hand, the next steps focus on the qualities of the surfaces that affect heat transfer, such as the material properties and surface convection heat transfer coefficients.
Through the use of a meticulous step by step method, which covers calculating the individual and total surface area of the tubes, converting these areas to appropriate units, and understanding the role of the heat transfer coefficients, we ensure the accuracy of our findings. This methodical approach is vital when instructing students how to analyze heat exchangers effectively.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient is a measure of the convective heat transfer between a solid surface and a fluid in contact with it. It's a critical aspect of evaluating how efficiently heat is transferred in a heat exchanger. For example, an inner surface convection heat transfer coefficient refers to the transfer on the fluid side that flows inside the tubes, while an outer surface coefficient refers to the transfer where the fluid flows on the outside.
To make studies easier to understand, it’s valuable to showcase real-life examples where these coefficients are used, such as in this exercise, where they directly influence the overall heat transfer coefficient calculation. Clarifying how these individual coefficients impact the overall efficiency drives home the significance of accurate measurement of the convection heat transfer coefficient.
Log Mean Temperature Difference
The log mean temperature difference (LMTD) is a pivotal concept that expresses the driving force behind the overall heat transfer in heat exchangers. It’s a theoretical temperature difference that averages the temperature gradients between the hot and cold sides over the length of the exchanger. Calculating the LMTD corrects for the variations in temperature difference that occur throughout the exchanger, providing a consistent value to use in heat transfer rate equations.
When explaining to students, use visuals or analogies, such as comparing the temperature difference to a car engine that drives the heat from the hot to the cold fluid. The greater the LMTD, the more 'horsepower' the exchanger has to transfer heat. Students should see the LMTD as a critical 'thermometer' that gauges the efficiency of their heat exchanger.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient is an encompassing measure that integrates the different modes and paths of heat transfer into one coefficient. It reflects the total resistance to heat flow through the various layers in a heat exchanger, including conduction through the tube material and convection on both the inside and outside of the tubes.
In simpler terms, it's like a summary of how good an insulator or conductor the entire system is. A higher overall coefficient indicates a more effective heat exchanger. When guiding students, it's essential to emphasize the relationship between the overall coefficient and the individual convection heat transfer coefficients. They must understand that diminishing these individual resistances enhances overall heat transfer, analogous to widening a bottleneck to increase water flow. Breaking down complex equations into real-world applications can significantly improve student comprehension.

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Most popular questions from this chapter

Saturated liquid benzene flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(75^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) by using a source of cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(3.5 \mathrm{~kg} / \mathrm{s}\) and \(15^{\circ} \mathrm{C}\) through a \(20-\mathrm{mm}-\) diameter tube of negligible wall thickness. The overall heat transfer coefficient of the heat exchanger is estimated to be \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the specific heat of the liquid benzene is \(1839 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and assuming that the capacity ratio and effectiveness remain the same, determine the heat exchanger surface area for the following four heat exchangers: \((a)\) parallel flow, \((b)\) counter flow, \((c)\) shelland-tube heat exchanger with 2 -shell passes and 40-tube passes, and \((d)\) cross-flow heat exchanger with one fluid mixed (liquid benzene) and other fluid unmixed (water).

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.

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) \(1393 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} /\right.\) \(\mathrm{kg} \cdot \mathrm{K})\) to \(\mathrm{cool}\) 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, what is the number of transfer units (NTU) of this radiator? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5

Hot exhaust gases of a stationary diesel engine are to be used to generate steam in an evaporator. Exhaust gases \(\left(c_{p}=1051 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enter the heat exchanger at \(550^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) while water enters as saturated liquid and evaporates at \(200^{\circ} \mathrm{C}\left(h_{f g}=1941 \mathrm{~kJ} / \mathrm{kg}\right)\). The heat transfer surface area of the heat exchanger based on water side is \(0.5 \mathrm{~m}^{2}\) and overall heat transfer coefficient is \(1780 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer, the exit temperature of exhaust gases, and the rate of evaporation of water.

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