/*! 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 32 A single-pass counterflow exchan... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 32

A single-pass counterflow exchanger is required to \(\operatorname{cool} 7000 \mathrm{~kg} / \mathrm{h}\) of oil from 365 \(\mathrm{K}\) to \(330 \mathrm{~K}\). Cooling water is available at \(4000 \mathrm{~kg} / \mathrm{h}\) and \(290 \mathrm{~K}\). If the overall heat transfer coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), determine the surface area required. For the oil take \(c_{p}=2100 \mathrm{~J} / \mathrm{kg} \mathrm{K}\).

Short Answer

Expert verified
The required surface area for the heat exchanger is approximately 13.91 m虏.

Step by step solution

01

Calculate Heat Duty of Oil

First, we need to calculate the heat duty required to cool the oil from 365 K to 330 K. Use the formula: \[ Q = ext{mass flow rate of oil} \times c_p \times \Delta T \]where:- \( \text{mass flow rate of oil} = 7000 \text{ kg/h} = \frac{7000}{3600} \text{ kg/s} \)- \( c_p = 2100 \text{ J/kg K} \)- \( \Delta T = 365 \text{ K} - 330 \text{ K} = 35 \text{ K} \)Substitute these values into the equation:\[ Q = \frac{7000}{3600} \times 2100 \times 35 \approx 1.428 \times 10^5 \text{ W} \]
02

Determine Temperature Change for Water

Next, calculate the temperature change for the cooling water. Start with the energy balance: the heat lost by the oil must equal the heat gained by the water.The formula for the heat absorbed by the water is:\[ Q_w = \text{mass flow rate of water} \times c_{p(w)} \times \Delta T_w \]Since the energy transferred from the oil to the water is the same, we equate the heat duties:\[ \frac{7000}{3600} \times 2100 \times 35 = \frac{4000}{3600} \times c_{p(w)} \times \Delta T_w \]For simplifying, assume \( c_{p(w)} \) similar to water at around room temperature, which is \( 4186 \text{ J/kg K} \). Solve for \( \Delta T_w \):\[ \Delta T_w = \frac{1.428 \times 10^5}{(4000/3600) \times 4186} \approx 30.78 \text{ K} \]This implies the water heats to \( 290 + 30.78 = 320.78 \text{ K} \).
03

Calculate Log Mean Temperature Difference (LMTD)

Find the Log Mean Temperature Difference (LMTD) for the counterflow heat exchanger using:\[ \Delta T_{lm} = \frac{(T_{h,i} - T_{c,o}) - (T_{h,o} - T_{c,i})}{\ln\left(\frac{T_{h,i} - T_{c,o}}{T_{h,o} - T_{c,i}}\right)} \]where:- \( T_{h,i} = 365 \text{ K} \) (hot fluid initial temperature)- \( T_{h,o} = 330 \text{ K} \) (hot fluid outlet temperature)- \( T_{c,i} = 290 \text{ K} \) (cold fluid initial temperature)- \( T_{c,o} = 320.78 \text{ K} \) (cold fluid outlet temperature)Substitute the values:\[ \Delta T_{lm} = \frac{(365 - 320.78) - (330 - 290)}{\ln\left(\frac{365 - 320.78}{330 - 290}\right)} \approx 34.17 \text{ K} \]
04

Determine Required Surface Area

With the LMTD calculated, use the following formula to find the required surface area for the heat exchanger:\[ A = \frac{Q}{U \times \Delta T_{lm}} \]where:- \( Q = 1.428 \times 10^5 \text{ W}\)- \( U = 300 \text{ W/m}^2\text{ K} \)- \( \Delta T_{lm} \approx 34.17 \text{ K} \)Thus,\[ A = \frac{1.428 \times 10^5}{300 \times 34.17} \approx 13.91 \text{ m}^2 \]

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Key Concepts

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

Counterflow Exchanger
A counterflow heat exchanger is a type of heat exchanger where two fluids flow in opposite directions. This configuration allows for an efficient heat transfer process compared to other exchanger types like parallel flow. In a counterflow arrangement, the temperature gradient between the hot and cold fluids is maximized along the entire length of the exchanger.
Why is this configuration beneficial?
  • Maximal Temperature Difference: Because the flow directions are opposite, the hot fluid continuously transfers heat to the cold fluid, maintaining a large temperature difference.
  • Higher Efficiency: The higher temperature difference increases the overall heat transfer rate, making counterflow exchangers more efficient.
  • Effective Utilization: It can cool or heat fluids closer to their desired outlet temperatures than other configurations.
Heat Transfer Coefficient
The heat transfer coefficient, denoted by \( U \), is crucial in determining how effectively heat is transferred between two fluids. It represents the amount of heat (in watts) transferred per square meter of exchanger surface, per Kelvin temperature difference between the fluids.
Factors influencing \( U \):
  • Material Properties: Metals with higher thermal conductivity will generally allow for better heat transfer.

  • Surface Conditions: Clean surfaces enhance heat exchange, while fouling reduces it.

  • Flow Nature: Turbulent flow increases \( U \) compared to laminar flow, due to better mixing of fluids.
Understanding and optimizing the heat transfer coefficient helps in designing more efficient exchangers, reducing costs, and enhancing energy efficiency.
Log Mean Temperature Difference
The Log Mean Temperature Difference (LMTD) is a method to quantify the driving force in a heat exchanger. It accounts for the variation in temperature difference between fluids at each end of the exchanger. Calculating LMTD is essential for determining the required surface area for heat exchangers.
To calculate LMTD, use the formula:\[ \Delta T_{lm} = \frac{(T_{h,i} - T_{c,o}) - (T_{h,o} - T_{c,i})}{\ln\left(\frac{T_{h,i} - T_{c,o}}{T_{h,o} - T_{c,i}}\right)}\]
Key Components:
  • \( T_{h,i} \) and \( T_{h,o} \): Initial and outlet temperatures of the hot fluid.

  • \( T_{c,i} \) and \( T_{c,o} \): Initial and outlet temperatures of the cold fluid.
The LMTD provides a simplified way to evaluate the average temperature difference across the exchanger, crucial for heat exchanger design and analysis.
Heat Duty Calculation
Calculating the heat duty of a heat exchanger involves determining the energy required to bring about the desired temperature change in a fluid. The basic formula is:\[ Q = \dot{m} \times c_p \times \Delta T \]
Where:
  • \( \dot{m} \): The mass flow rate of the fluid.

  • \( c_p \): The specific heat capacity of the fluid, indicating how much heat is needed to change the temperature of a unit mass by one degree.

  • \( \Delta T \): The change in temperature.
Heat duty indicates the capacity of the heat exchanger and is critical in ensuring the exchanger is properly sized for its application.

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Most popular questions from this chapter

A geothermal power plant uses isobutane as the secondary working fluid. After expanding through the turbine, isobutane vapor condenses in a shell-and-tube condenser at \(325 \mathrm{~K}\). The condenser coolant is water at \(305 \mathrm{~K}\) supplied from a cooling tower at a rate of \(500 \mathrm{~kg} / \mathrm{s}\). The condenser shell contains 4000 tubes of 25 \(\mathrm{mm}\) O.D. and \(2 \mathrm{~mm}\) wall thickness; the overall heat transfer coefficient based on tube outside area is estimated to be \(450 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If it is desired to have a condenser effectiveness of \(80 \%\), determine (i) the outlet water temperature. (ii) the number of transfer units required. (iii) the length of the tube bundle. Also, if the thermal efficiency of the power cycle is \(30 \%\), determine the power output of the turbine.

An aircraft oil cooler is to be designed to reduce the oil temperature from \(390 \mathrm{~K}\) to \(365 \mathrm{~K}\). The oil (SAE 50 ) flow rate is \(1.5 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(140 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and the entering air temperature is \(310 \mathrm{~K}\), find the necessary transfer area for (i) counterflow. (ii) parallel flow. Assume balanced flow, \(C_{H}=C_{C}\).

A two-tube-pass, single-shell-pass heat exchanger transfers heat from the collector loop to the storage tank in a solar heating system for a building in Denver. The maximum heat transfer rate to be accommodated is \(24 \mathrm{~kW}\) (for a \(50 \mathrm{~m}^{2}\) collector at noon on a sunny day in fall or spring). Water at \(1 \mathrm{~kg} / \mathrm{s}\) from the storage tank enters the tubes at \(50^{\circ} \mathrm{C}\). The collector loop contains \(50 \%\) ethylene glycol aqueous solution flowing at \(0.66 \mathrm{~kg} / \mathrm{s}\) and enters the shell at \(67.5^{\circ} \mathrm{C}\). The tube bundle consists of 50 tubes of \(24 \mathrm{~mm}\) O.D. and \(2 \mathrm{~mm}\) wall thickness, for which an overall heat transfer coefficient based on outside area of \(835 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) is estimated. Determine the length of tubes required. For the glycol solution use \(c_{p}=3690 \mathrm{~J} / \mathrm{kg}\)

A \(6 \mathrm{~kg} / \mathrm{s}\) flow of sulfuric acid \(\left(c_{p}=1480 \mathrm{~J} / \mathrm{kg} \mathrm{K}\right)\) is to be cooled in a twostage counterflow h猫at exchanger. The hot acid at \(175^{\circ} \mathrm{C}\) is fed to a tank where it is stirred in contact with cooling coils; the continuous discharge from this tank at \(89^{\circ} \mathrm{C}\) flows into a second stirred tank and leaves the second tank at \(46^{\circ} \mathrm{C}\). Cooling water at \(20^{\circ} \mathrm{C}\) enters the cooling coil of the cold tank and leaves the cooling coil of the hot tank at \(80^{\circ} \mathrm{C}\). If the overall heat transfer coefficients are 52 and \(38 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) in the hot and cold tanks, respectively, determine the heat transfer surface areas required. Neglect heat losses to the surroundings, and assume that the acid temperature in each tank equals its outlet temperature.

An off-the-shelf heat exchanger with transfer surface area of \(50 \mathrm{~m}^{2}\) is required to cool \(4 \mathrm{~kg} / \mathrm{s}\) of oil at \(90^{\circ} \mathrm{C}\). A water supply of \(3 \mathrm{~kg} / \mathrm{s}\) at \(20^{\circ} \mathrm{C}\) is available. The overall heat transfer coefficient is expected to be approximately \(370 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). (i) Determine the oil outlet temperature for counterflow operation. (ii) If it is desired to have an outlet temperature \(3^{\circ} \mathrm{C}\) lower than the value obtained in part (i), determine the required change to the water flow rate. Take \(c_{p}=2030 \mathrm{~J} / \mathrm{kg} \mathrm{K}\) for the oil.
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