Problema 11-44 Cengel. Método LMTD.

Marco Antonio Chávez Rojo
25 May 202003:29

Summary

TLDRThis script demonstrates the application of the logarithmic mean temperature difference method in heat transfer calculations. It details a carbon dioxide cooling process in a double-pipe counterflow heat exchanger, where 720 kg/h of CO2 is cooled from 150°C to 40°C. The cold fluid, water, enters at 10°C and exits at a calculated temperature. The script guides through the steps of determining the heat flow, using the specific heat capacity (cp), mass flow rate, and temperature differences. It concludes with the calculation of the overall heat transfer coefficient, resulting in a value of 2311, providing a clear example of thermal engineering principles.

Takeaways

  • 🔍 The script discusses a simple problem to illustrate the application of the logarithmic mean temperature difference method in heat transfer calculations.
  • 🌡️ The problem involves a carbon dioxide flow being cooled from 150°C to 40°C at a rate of 720 kg/h in a double-pipe heat exchanger with counter-flow.
  • 💧 Water enters the heat exchanger at 10°C at a rate of 540 kg/h and is heated by the carbon dioxide.
  • 📏 The outer diameter of the inner tube is 2.5 cm, and its length is 6 meters, which are key dimensions for calculating the heat transfer area.
  • ⏱️ The heat transfer rate is calculated using the mass flow rate, specific heat capacity (cp), and the temperature difference between the inlet and outlet of the hot fluid.
  • 🔄 The counter-flow arrangement is important as it affects the temperature gradients and the calculation of the outlet temperature of the cold fluid.
  • ⚖️ The logarithmic mean temperature difference is used to account for the varying temperature differences along the length of the heat exchanger.
  • 📐 The area of the heat transfer surface is calculated by multiplying the tube's diameter by its length, which is essential for determining the overall heat transfer coefficient.
  • 🔢 The overall heat transfer coefficient is determined by rearranging the heat transfer equation and solving for the coefficient, resulting in a value of 2311 in this case.
  • 🔧 The problem emphasizes the importance of congruence in the calculations to ensure the accuracy of the heat transfer coefficient.
  • 📝 The script provides a step-by-step approach to solving heat transfer problems using the logarithmic mean temperature difference method, which is crucial for designing and analyzing heat exchangers.

Q & A

  • What problem is being discussed in the script?

    -The script discusses a problem involving the application of the logarithmic mean temperature difference method in a heat exchanger scenario to calculate the total heat transfer coefficient.

  • What is the hot fluid in this problem and what are its properties?

    -The hot fluid is carbon dioxide (CO2) with a specific heat capacity (cp) of 2200 J/kg·K, an inlet temperature of 150°C, and an outlet temperature of 40°C. It flows at a rate of 720 kg/h.

  • What is the flow rate of the hot fluid in kg/s?

    -The flow rate of the hot fluid is 720 kg/h, which is equivalent to 0.2 kg/s (720 kg/3600 s).

  • What is the purpose of the heat exchanger in this scenario?

    -The purpose of the heat exchanger is to cool down the hot fluid (carbon dioxide) by transferring heat to the cold fluid (water) through a double-pipe counterflow heat exchanger.

  • What is the cold fluid in this problem and its flow rate?

    -The cold fluid is water with an inlet temperature of 10°C. It flows at a rate of 540 kg/h.

  • What are the dimensions of the inner tube in the heat exchanger?

    -The inner tube has an outer diameter of 2.5 centimeters and a length of 6 meters.

  • How is the total heat transfer calculated?

    -The total heat transfer is calculated by multiplying the mass flow rate of the hot fluid (720 kg/h) by its specific heat capacity (2200 J/kg·K) and the temperature difference (150°C - 40°C).

  • What is the concept of the logarithmic mean temperature difference method?

    -The logarithmic mean temperature difference (LMTD) method is a way to calculate the average temperature difference between the hot and cold fluids in a heat exchanger, which is used to determine the heat transfer rate.

  • How is the outlet temperature of the cold fluid determined?

    -The outlet temperature of the cold fluid is determined by dividing the total heat transfer by the mass flow rate of the cold fluid (540 kg/h), its specific heat capacity, and then adding it to the inlet temperature of the cold fluid.

  • What is the calculated global heat transfer coefficient?

    -The calculated global heat transfer coefficient is 2311 W/m²·K, which represents the overall efficiency of heat transfer in the heat exchanger.

  • Why is it important to calculate the surface area of the tube?

    -Calculating the surface area of the tube is important because it is used in conjunction with the LMTD to determine the overall heat transfer coefficient, which is a key parameter in the design and analysis of heat exchangers.

Outlines

00:00

🔍 Introduction to the Log Mean Temperature Difference Method

This paragraph introduces a problem-solving scenario using the logarithmic mean temperature difference method. The problem involves a carbon dioxide flow cooling from 150°C to 40°C at a rate of 720 kilograms per hour as it passes through a heat exchanger. The heat exchanger is a counter-current double-pipe system where water enters at 10°C and exits at an unknown temperature. The goal is to calculate the total heat transfer coefficient, given the properties of the hot and cold fluids, their flow rates, and the dimensions of the heat exchanger.

Mindmap

Keywords

💡Logarithmic Mean Temperature Difference (LMTD)

The Logarithmic Mean Temperature Difference is a method used in heat transfer calculations to determine the average temperature difference between two fluids in a heat exchanger. In the video, it is applied to calculate the heat transfer between the hot carbon dioxide stream and the cold water stream in a counter-current heat exchanger. The LMTD is essential for determining the heat transfer rate and is calculated using the temperatures at the inlet and outlet of both fluids.

💡Heat Exchanger

A heat exchanger is a device designed to transfer heat between two or more fluids at different temperatures. The video discusses a double-pipe counter-current heat exchanger, where one fluid is cooled while the other is heated. The heat exchanger is central to the problem presented in the video, as it is the apparatus through which the heat transfer is calculated.

💡Carbon Dioxide (CO2)

In the context of the video, carbon dioxide is the fluid being cooled in the heat exchanger. It is mentioned with a flow rate of 720 kilograms per hour and an initial temperature of 150 degrees Celsius. The heat transfer process involves cooling this carbon dioxide, which is a key part of the video's demonstration of the heat transfer calculation.

💡Specific Heat Capacity (cp)

Specific heat capacity, denoted as 'cp' in the script, is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius. It is a crucial parameter in heat transfer calculations. In the video, the specific heat capacity of the hot fluid (carbon dioxide) and the cold fluid (water) are given as 2200 and 4200 J/(kg·K), respectively, and are used to calculate the heat flow.

💡Flow Rate

The flow rate refers to the volume or mass of a substance that moves through a given area per unit of time. In the script, the flow rates of both the carbon dioxide and the water are provided, which are 720 kg/h and 540 kg/h, respectively. These values are necessary for calculating the heat transfer in the heat exchanger.

💡Temperature Gradient

A temperature gradient is the rate of change of temperature with respect to a change in position. In the video, the temperature gradients of the hot and cold fluids are essential for calculating the LMTD and, subsequently, the heat transfer. The script mentions the temperatures at the inlet and outlet of the fluids, which create the temperature gradients.

💡Counter-Current Flow

Counter-current flow is a configuration in which two fluids flow in opposite directions in a heat exchanger. This setup is used in the video to maximize the temperature difference between the fluids, thereby enhancing heat transfer. The script describes the heat exchanger as having a counter-current flow, which affects the calculation of the temperature differences.

💡Thermal Coefficient

The thermal coefficient, or the overall heat transfer coefficient, is a measure of the efficiency of heat transfer between the fluids in a heat exchanger. The script aims to calculate this coefficient, which is influenced by the materials of the heat exchanger, the flow rates, and the temperature differences.

💡Tube Diameter

The tube diameter is a critical dimension in heat exchanger design, affecting the surface area available for heat transfer. In the video, the inner tube's outer diameter is given as 2.5 centimeters, which, along with the tube's length, is used to calculate the surface area for heat transfer.

💡Surface Area

Surface area is the total area of the surface of an object. In the context of heat exchangers, the surface area is important for calculating the rate of heat transfer. The script calculates the surface area of the inner tube using its diameter and length, which is necessary for determining the overall heat transfer coefficient.

💡Global Heat Transfer Coefficient

The global heat transfer coefficient is the overall coefficient that accounts for all the resistances to heat transfer in a system. In the video, the final calculated value of 2311 represents this coefficient, indicating the overall efficiency of the heat exchange process.

Highlights

Introduction to the logarithmic mean temperature difference method for carbon flow.

Carbon cooling from 150 to 40 degrees Celsius at a rate of 720 kg/h.

Heat exchanger with counter-current flow and water entering at 10 degrees Celsius.

Calculation of heat transfer using mass flow rate, specific heat, and temperature difference.

Determination of the outlet temperature of the cold fluid using the heat transfer equation.

Use of the logarithmic mean temperature difference for heat exchange calculations.

Identification of the hot and cold fluid temperatures for the heat exchange process.

Calculation of the area of the heat exchanger tube based on diameter and length.

Determination of the global heat transfer coefficient using the area and temperature difference.

The importance of congruence in the logarithmic mean temperature difference method.

Practical application of the method in a double-pipe heat exchanger.

Conversion of mass flow rate to volumetric flow rate for heat transfer calculations.

Explanation of the counter-current flow effect on the temperature profiles of the fluids.

Calculation of the temperature difference at both ends of the heat exchanger.

Use of the cold fluid's inlet temperature as the hot fluid's outlet temperature in counter-current flow.

Final calculation of the global heat transfer coefficient to be 2311.

Completion of the heat exchanger analysis using the logarithmic mean temperature difference method.

Transcripts

play00:01

vamos a hacer este problema muy simple

play00:04

pero solamente para ilustrar cómo se

play00:06

aplica el método de la media logarítmica

play00:10

de la diferencia de temperaturas me dice

play00:13

un flujo del carbono carbono

play00:19

y aquí está el cp se enfría a razón de

play00:21

720 kilogramos por hora desde 150 hasta

play00:25

40 grados celsius al pasar por el tubo

play00:27

interior de la mente el cambiador de

play00:29

calor del tubo doble a contraflujo entra

play00:32

agua aquí está el cp al intercambiador a

play00:36

10 grados celsius a razón de 540

play00:39

kilogramos por hora el diámetro exterior

play00:42

del tubo interior es de 2.5 centímetros

play00:44

y su longitud es de 6 metros calculé el

play00:48

coeficiente total de transferencia de

play00:50

calor queremos la 1 ya están máquinas

play00:55

datos que nos dé el problema vean el

play00:59

fluido caliente primero

play01:01

que esté el cp 2200

play01:05

temperatura de entrada son 150 la de

play01:07

salida 40 el flujo mágico son 720

play01:12

kilogramos en 3.600 segundos

play01:19

es un 0.2 bien

play01:23

tengo las temperaturas de entre de

play01:25

sanidade

play01:26

caliente tengo todo para calcular el cp

play01:32

perdón en el flujo de calor

play01:36

es el flujo mágico por el cp por la

play01:38

diferencia de temperaturas tenemos esto

play01:40

ese mismo calor obviamente es el que se

play01:43

transfiere al fluido frío y con eso

play01:47

puedo determinar la temperatura de

play01:48

salida del giro fresh simplemente

play01:51

dividiendo el calor entre m cp y suman

play01:55

aso a la temperatura de entrada

play01:58

listo ya que tengo las 4 temperaturas

play02:02

entonces ya puedo calcular las

play02:04

diferencias de temperatura en un extremo

play02:06

y en otro miren aquí tengo de 4 - b 13

play02:12

y temperatura de salida porque como está

play02:15

a contraflujo la temperatura de entrada

play02:17

de uno de ellos es la de salida del otro

play02:20

aquí escribimos escribió al revés puse

play02:23

la fría menos la caliente se van a dar

play02:25

cuenta que no hay diferencia siempre con

play02:28

esa congruente también el del tractor es

play02:30

del tate 2 lo hago

play02:33

así en el frío menos el caliente

play02:36

entonces cuando haga

play02:39

la diferencia el promedio logarítmica de

play02:42

todos modos me va a quedar una cantidad

play02:44

positiva entonces no hay problema

play02:47

como se haga siempre y cuando se sea

play02:49

congruente ya que tengo del tate

play02:54

hay que calcular distinto

play02:56

hay que calcular el área superficial

play02:58

tenemos un tubo de seis metros de 2.5

play03:01

centímetros de diámetro y por el

play03:04

diámetro por la longitud es el área

play03:07

y finalmente despejando de la relación

play03:11

con igual a ahora por del trate

play03:16

tengo el coeficiente global de

play03:18

transferencia que es 2311 y listo

play03:22

terminando

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相关标签
Heat ExchangeLogarithmic MeanTemperature DifferenceCarbon FluxThermal CalculationEngineering ProcessFluid DynamicsThermal EngineeringHeat TransferEngineering Math
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