/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Water at a rate of \(45,500 \mat... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 10

Water at a rate of \(45,500 \mathrm{~kg} / \mathrm{h}\) is heated from 80 to \(150^{\circ} \mathrm{C}\) in a heat exchanger having two shell passes and eight tube passes with a total surface area of \(925 \mathrm{~m}^{2}\). Hot exhaust gases having approximately the same thermophysical properties as air enter at \(350^{\circ} \mathrm{C}\) and exit at \(175^{\circ} \mathrm{C}\). Determine the overall heat transfer coefficient.

Short Answer

Expert verified
The overall heat transfer coefficient for the heat exchanger is approximately 0.042 kJ/(s路m虏路K).

Step by step solution

01

Determine the rate of heat transfer

To do this, we'll need to calculate the amount of heat the water absorbs using the mass flow rate of the water and the change in temperature. The formula for calculating the heat transfer rate (Q) is: \(Q = m \cdot c_p \cdot \Delta T\) where: - \(m\) is the mass flow rate of water (\(45,500 \frac{\mathrm{kg}}{\mathrm{h}}\)) - \(c_p\) is the specific heat capacity of water (\(4.18 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\)) - \(\Delta T\) is the change in temperature of the water (\(150^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C}\))
02

Calculate the heat transfer rate

Now let's calculate the heat transfer rate: \(Q = 45,500 \frac{\mathrm{kg}}{\mathrm{h}} \cdot 4.18 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \cdot (150^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C})\) \(Q = 45,500 \frac{\mathrm{kg}}{\mathrm{h}} \cdot 4.18 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \cdot 70 \mathrm{K}\) First, let's convert the mass flow rate from kg/h to kg/s: \(45,500 \frac{\mathrm{kg}}{\mathrm{h}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}} = 12.639 \frac{\mathrm{kg}}{\mathrm{s}} \) Now, plug in the value and perform the calculation: \(Q = 12.639 \frac{\mathrm{kg}}{\mathrm{s}} \cdot 4.18 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} \cdot 70 \mathrm{K}\) \(Q 鈮� 3,709.1 \frac{\mathrm{kJ}}{\mathrm{s}}\)
03

Determine the Log Mean Temperature Difference (LMTD)

Now, we need to calculate the Log Mean Temperature Difference (LMTD) to determine the overall heat transfer coefficient later. The formula for LMTD is: LMTD = \(\frac{\Delta T_1 - \Delta T_2}{\ln{(\frac{\Delta T_1}{\Delta T_2})}}\) where: - \(\Delta T_1\) is the temperature difference between the hot gases and cold water at the inlet - \(\Delta T_2\) is the temperature difference between the hot gases and cold water at the outlet \(\Delta T_1 = T_{g,inlet} - T_{w,inlet} = 350^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C} = 270 \mathrm{K}\) \(\Delta T_2 = T_{g,outlet} - T_{w,outlet} = 175^{\circ} \mathrm{C} - 150^{\circ} \mathrm{C} = 25 \mathrm{K}\) Now we can calculate the LMTD: LMTD = \(\frac{270 \mathrm{K} - 25 \mathrm{K}}{\ln{(\frac{270 \mathrm{K}}{25 \mathrm{K}})}} 鈮� 95.83 \mathrm{K}\)
04

Calculate the overall heat transfer coefficient

Now, we use the heat transfer rate, LMTD, and surface area to determine the overall heat transfer coefficient (U). The formula is: \(U = \frac{Q}{A \cdot \text{LMTD}}\) where: - U is the overall heat transfer coefficient - Q is the heat transfer rate, calculated in Step 2 - A is the total surface area of the heat exchanger (925 m虏) - LMTD is the Log Mean Temperature Difference, calculated in Step 3 \(U = \frac{3,709.1 \frac{\mathrm{kJ}}{\mathrm{s}}}{925 \mathrm{m}^2 \cdot 95.83\, \mathrm{K}} 鈮� 0.042 \frac{\mathrm{kJ}}{\mathrm{s} \cdot \mathrm{m}^2 \cdot \mathrm{K}}\) The overall heat transfer coefficient is approximately 0.042 kJ/(s路m虏路K).

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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, often denoted by **U**, represents how effectively heat is transferred across different materials and fluids in a heat exchanger. It takes into account all resistances to heat transfer, including conduction through the material of the heat exchanger and convection on both the hot and cold sides. The unit for this coefficient is typically watts per square meter Kelvin (W/m虏路K), but it can also appear as kJ/s路m虏路K in some contexts.
To calculate the overall heat transfer coefficient, you use the formula:
  • \(U = \frac{Q}{A \cdot \text{LMTD}}\)
where:
  • \(Q\) is the heat transfer rate
  • \(A\) is the surface area of the heat exchanger
  • LMTD is the log mean temperature difference, a crucial component calculated based on the specific setup.
This coefficient is essential to evaluate the efficiency of heat exchangers in both industrial and domestic applications, ensuring energy savings by maximizing thermal energy transfer.
Heat Exchangers
Heat exchangers are devices designed to efficiently transfer heat between two or more fluids or between a solid surface and a liquid or gaseous fluid, or between different solid surfaces. They play a crucial role in many industries, supporting processes like heating, cooling, and thermal energy recovery.
Heat exchangers come in various forms, each suited to specific applications:
  • Shell and Tube Heat Exchangers: Utilize tubes enclosed within a cylindrical shell; one fluid flows through the tubes, while the other flows outside them in the shell.
  • Plate Heat Exchangers: Consist of multiple thin plates packed together, creating channels for fluids to flow alternately between them.
  • Air-cooled Heat Exchangers: Transfer heat from a fluid to the air via finned tubes.
Understanding the type and function of different heat exchangers helps in determining their efficiency and suitability for particular industrial or residential procedures.
Log Mean Temperature Difference
The Log Mean Temperature Difference (LMTD) is a critical concept in heat exchanger design, reflecting the driving force for heat transfer. LMTD handles varying temperature differences along the heat exchanger's length, providing a single average temperature difference for calculation purposes.
The calculation is:
  • \( \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)} \)
Where:
  • \(\Delta T_1\) is the temperature difference between the hot and cold fluids at one end.
  • \(\Delta T_2\) is the temperature difference at the opposite end.
The LMTD accounts for these differences and is a more accurate measure than simply taking an arithmetic mean. Calculating LMTD is crucial in determining the efficiency and size of heat exchangers, ensuring they meet desired thermal performance standards.
Heat Transfer Rate
Heat transfer rate, often denoted as **Q**, is the quantity of heat energy exchanged between the hot and cold fluids within a heat exchanger over a specific time period. It is measured in watts (W) or kilojoules per second (kJ/s), reflecting how fast energy moves from one fluid or process to another.
You compute this rate using the formula:
  • \( Q = m \cdot c_p \cdot \Delta T \)
where:
  • \(m\) is the mass flow rate of the fluid
  • \(c_p\) is the specific heat capacity of the fluid
  • \(\Delta T\) is the temperature change experienced by the fluid
Understanding the heat transfer rate isn't just about calculating numbers; it offers insights into the thermal dynamics of a system, guiding optimizations and adjustments for improved energy efficiency.

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Most popular questions from this chapter

A steel tube \((k=50 \mathrm{~W} / \mathrm{m}\) - \(\mathrm{K})\) of inner and outer diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{a}=26 \mathrm{~mm}\), respectively, is used to transfer heat from hot gases flowing over the tube \(\left(h_{\mathrm{h}}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) to cold water flowing through the tube \(\left(h_{c}=8000 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\right)\). What is the cold-side overall heat transfer coefficient \(U_{c}\) ? To enhance heat transfer, 16 straight fins of rectangular profile are installed longitudinally along the outer surface of the tube. The fins are equally spaced around the circumference of the tube, each having a thickness of \(2 \mathrm{~mm}\) and a length of \(15 \mathrm{~mm}\). What is the corresponding overall heat transfer coefficient \(U_{c}\) ?

A novel design for a condenser consists of a tube of thermal conductivity \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) with longitudinal fins snugly fitted into a larger tube. Condensing refrigerant at \(45^{\circ} \mathrm{C}\) flows axially through the inner tube, while water at a flow rate of \(0.012 \mathrm{~kg} / \mathrm{s}\) passes through the six channels around the inner tube. The pertinent diameters are \(D_{1}=10 \mathrm{~mm}, D_{2}=14 \mathrm{~mm}\), and \(D_{3}=50 \mathrm{~mm}\), while the fin thickness is \(t=2 \mathrm{~mm}\). Assume that the convection coefficient associated with the condensing refrigerant is extremely large. Determine the heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\).

In open heart surgery under hypothermic conditions, the patient's blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube, counterflow heat exchanger of length \(0.5 \mathrm{~m}\) be used for this purpose, with the thin-walled inner tube having a diameter of \(55 \mathrm{~mm}\). The specific heat of the blood is \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If water at \(T_{h j}=60^{\circ} \mathrm{C}\) and \(\dot{m}_{h}=0.10 \mathrm{~kg} / \mathrm{s}\) is used to heat blood entering the exchanger at \(T_{c A}=18^{\circ} \mathrm{C}\) and \(\dot{m}_{c}=0.05 \mathrm{~kg} / \mathrm{s}\), what is the temperature of the blood leaving the exchanger? The overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) The surgeon may wish to control the heat rate \(q\) and the outlet temperature \(T_{c, 0}\) of the blood by altering the flow rate and/or inlet temperature of the water during the rewarming process. To assist in the development of an appropriate controller for the prescribed values of \(\hat{m}_{c}\) and \(T_{c \jmath}\), compute and plot \(q\) and \(T_{c, \rho}\) as a function of \(\dot{m}_{h}\) for \(0.05 \leq \dot{m}_{\mathrm{h}} \leq 0.20 \mathrm{~kg} / \mathrm{s}\) and values of \(T_{h, l}=50,60\), and \(70^{\circ} \mathrm{C}\). Since the dominant influence on the overall heat transfer coefficient is associated with the blood flow conditions, the value of \(U\) may be assumed to remain at \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Should certain operating conditions be excluded?

Saturated process steam at 1 atm is condensed in a shell-and-tube heat exchanger (one shell, two tube passes). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) with an average velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). The tubes are thin walled and made of copper with a diameter of \(14 \mathrm{~mm}\) and length of \(0.5 \mathrm{~m}\). The convective heat transfer coefficient for condensation on the outer surface of the tubes is \(21,800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the number of tubes/pass required to condense \(2.3 \mathrm{~kg} / \mathrm{s}\) of steam. (b) Find the outlet water temperature. (c) Find the maximum possible condensation rate that could be achieved with this heat exchanger using the same water flow rate and inlet temperature. (d) Using the heat transfer surface area found in part (a), plot the water outlet temperature and steam condensation rate for water mean velocities in the range from 1 to \(5 \mathrm{~m} / \mathrm{s}\). Assume that the shell-side convection coefficient remains unchanged.

The compartment heater of an automobile exchanges heat between warm radiator fluid and cooler outside air. The flow rate of water is large compared to the air, and the effectiveness, \(\varepsilon\), of the heater is known to depend on the flow rate of air according to the relation, \(\varepsilon \sim \dot{m}_{\text {air }}^{-0.2}\). (a) If the fan is switched to high and \(\dot{m}_{\text {air }}\) is doubled, determine the percentage increase in the heat added to the car, if fluid inlet temperatures remain the same. (b) For the low-speed fan condition, the heater warms outdoor air from 0 to \(30^{\circ} \mathrm{C}\). When the fan is turned to medium, the airflow rate increases \(50 \%\) and the heat transfer increases \(20 \%\). Find the new outlet temperature.
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