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A shell-and-tube heat exchanger with 1-shell pass and 14-tube passes is used to heat water in the tubes with geothermal steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) on the shell side. The tubes are thin-walled and have a diameter of \(2.4 \mathrm{~cm}\) and length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(3.9 \mathrm{~kg} / \mathrm{s}\). If the temperature difference between the two fluids at the exit is \(46^{\circ} \mathrm{C}\), determine (a) the rate of heat transfer, \((b)\) the rate of condensation of steam, and \((c)\) the overall heat transfer coefficient.

Short Answer

Expert verified
Question: Determine (a) the rate of heat transfer in the shell-and-tube heat exchanger, (b) the rate of condensation of steam, and (c) the overall heat transfer coefficient. Solution: (a) The rate of heat transfer, Q = 848.856 kW (b) The rate of condensation of steam, m_s = 0.385 kg/s (c) The overall heat transfer coefficient, U = 6.71 kW/(m²·K)

Step by step solution

01

Calculate the temperature change of water

To determine the rate of heat transfer, we need to know the temperature difference of water as it flows through the heat exchanger. Given the exit temperature difference of 46°C and the initial temperature of the water being 22°C, we can find the exit temperature of the water: 120°C - 46°C = 74°C. Now we can find the temperature change: ΔT = 74°C - 22°C = 52°C.
02

Calculate the rate of heat transfer (Q) through the water using the energy balance equation

To calculate the rate of heat transfer, we will use the energy balance equation: Q = mcΔT, where m is the mass flow rate of water, c is the specific heat capacity of water, and ΔT is the temperature change of water. We are given m = 3.9 kg/s, c = 4180 J/(kg·K), and we found ΔT = 52°C in the previous step. Thus, Q = 3.9 kg/s × 4180 J/(kg·K) × 52 K = 848,856 J/s or 848.856 kW.
03

Calculate the rate of heat transfer (Q) through the steam using the heat of vaporization (hfg)

The rate of heat transfer through the steam can be calculated by applying the energy balance, using the heat of vaporization (h_fg) as follows: Q = m_s×h_fg, where m_s is the mass flow rate of steam and h_fg = 2203 kJ/kg. Since the heat transferred from steam to water is the same, we have: m_s×h_fg = 848.856 kW.
04

Calculate the rate of condensation of steam (m_s)

Using the equation from Step 3, we can find the mass flow rate of steam, m_s: m_s = Q/h_fg = 848.856 kW / 2203 kJ/kg = 0.385 kg/s.
05

Calculate the overall heat transfer coefficient (U)

To find the overall heat transfer coefficient, we can use the heat transfer equation: Q = UAΔT_m, where U is the overall heat transfer coefficient, A is the heat transfer area, and ΔT_m is the logarithmic mean temperature difference. First, we must calculate the heat transfer area per tube: A_tube = πD×L×n_pass, where D is the diameter of the tube, L is the length per pass, and n_pass is the number of tube passes. We are given D = 0.024 m, L = 3.2 m, and n_pass = 14. Thus, A_tube = π×0.024 m×3.2 m×14 = 3.246 m². Next, we should calculate the logarithmic mean temperature difference, which could be done by assuming that the steam temperature does not change significantly and remain constant at 120°C and that the inlet and outlet temperatures of the water are 22°C and 74°C. We can calculate ΔT_m as follows: ΔT_m = [(120-74)-(120-22)] / ln((120-74)/(120-22)) = 39.09 °C. Finally, we can calculate the overall heat transfer coefficient, U, by rearranging the heat transfer equation, Q = UAΔT_m: U = Q / (A_tube × ΔT_m) = 848.856 kW / (3.246 m² × 39.09 K) = 6.71 kW/(m²·K). #Summary# In this exercise, our calculations yielded: (a) The rate of heat transfer, Q = 848.856 kW (b) The rate of condensation of steam, m_s = 0.385 kg/s (c) The overall heat transfer coefficient, U = 6.71 kW/(m²·K)

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Key Concepts

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

Shell-and-Tube Exchangers
Shell-and-tube heat exchangers are one of the most common types of heat exchangers you might come across in various industries. They consist of a large shell with a bundle of tubes inside it. These tubes are used to separate two different fluids that need to transfer heat between each other. In our scenario, the geothermal steam passes through the shell, while water flows through the tubes.

The shell-and-tube design is popular because it can handle high pressure and temperature differences well. You might find this type of exchanger being used for various applications from power plants to chemical processing facilities. It's versatile and can be designed to fit a wide variety of heat transfer needs by adjusting the number of passes, length, or materials of the tubes. Understanding how shell-and-tube heat exchangers work is essential to mastering thermal systems in engineering. They offer a reliable method to efficiently transfer heat energy.
Heat Transfer Rate Calculations
Calculating the heat transfer rate is one of the primary steps in analyzing the performance of a heat exchanger. This essentially tells us how much energy is being transferred from the hot fluid to the cool fluid. It’s often represented by the symbol Q, and is measured in watts (W) or kilowatts (kW).To find this value, we use the equation: \[Q = mc\Delta T\] where:
  • \(m\) is the mass flow rate of the fluid
  • \(c\) is the specific heat capacity
  • \(\Delta T\) is the change in temperature
For example, in this problem, we calculated the rate of heat transfer to be 848.856 kW. This indicates how much heat energy is moved from the steam to the water as it passes through the exchanger.Accurate heat transfer rate calculations are critical for designing efficient and effective heat exchangers. Missteps in these calculations might lead to undersized or oversized equipment which can be costly and inefficient.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, represented as \(U\), is a measure of a heat exchanger's ability to transfer heat between the two fluids within it. It takes into account all resistances to heat transfer including layer thicknesses, material properties, and convection processes inside the tubes and across the shell.To calculate \(U\), you will often use the equation: \[Q = U \cdot A \cdot \Delta T_m\] where:
  • \(A\) is the heat exchange area
  • \(\Delta T_m\) is the logarithmic mean temperature difference
In our example, we calculated \(U\) to be 6.71 kW/(m²·K). The value of \(U\) helps you understand how well your heat exchanger is performing. High values of \(U\) usually indicate efficient transfer of heat. Understanding the overall heat transfer coefficient is vital for heat exchanger design. It helps ensure the equipment is efficient and meets the performance criteria, avoiding energy losses and optimizing resource use.

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Most popular questions from this chapter

A 2-shell passes and 4-tube passes heat exchanger is used for heating a hydrocarbon stream \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) steadily from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). A water stream enters the shellside at \(80^{\circ} \mathrm{C}\) and leaves at \(40^{\circ} \mathrm{C}\). There are 160 thin-walled tubes, each with a diameter of \(2.0 \mathrm{~cm}\) and length of \(1.5 \mathrm{~m}\). The tube-side and shell-side heat transfer coefficients are \(1.6\) and \(2.5 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) Calculate the rate of heat transfer and the mass rates of water and hydrocarbon streams. (b) With usage, the outlet hydrocarbon-stream temperature was found to decrease by \(5^{\circ} \mathrm{C}\) due to the deposition of solids on the tube surface. Estimate the magnitude of fouling factor.

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the \(\varepsilon-\mathrm{NTU}\) method.

There are two heat exchangers that can meet the heat transfer requirements of a facility. Both have the same pumping power requirements, the same useful life, and the same price tag. But one is heavier and larger in size. Under what conditions would you choose the smaller one?

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).

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\). The rate of heat transfer between the two streams is (a) \(8.2 \mathrm{~kW}\) (b) \(23.7 \mathrm{~kW}\) (c) \(33.8 \mathrm{~kW}\) (d) \(44.8 \mathrm{~kW}\) (e) \(52.8 \mathrm{~kW}\)

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