Problema 11-44 Cengel. Método LMTD.
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
🔍 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)
💡Heat Exchanger
💡Carbon Dioxide (CO2)
💡Specific Heat Capacity (cp)
💡Flow Rate
💡Temperature Gradient
💡Counter-Current Flow
💡Thermal Coefficient
💡Tube Diameter
💡Surface Area
💡Global Heat Transfer Coefficient
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
vamos a hacer este problema muy simple
pero solamente para ilustrar cómo se
aplica el método de la media logarítmica
de la diferencia de temperaturas me dice
un flujo del carbono carbono
y aquí está el cp se enfría a razón de
720 kilogramos por hora desde 150 hasta
40 grados celsius al pasar por el tubo
interior de la mente el cambiador de
calor del tubo doble a contraflujo entra
agua aquí está el cp al intercambiador a
10 grados celsius a razón de 540
kilogramos por hora el diámetro exterior
del tubo interior es de 2.5 centímetros
y su longitud es de 6 metros calculé el
coeficiente total de transferencia de
calor queremos la 1 ya están máquinas
datos que nos dé el problema vean el
fluido caliente primero
que esté el cp 2200
temperatura de entrada son 150 la de
salida 40 el flujo mágico son 720
kilogramos en 3.600 segundos
es un 0.2 bien
tengo las temperaturas de entre de
sanidade
caliente tengo todo para calcular el cp
perdón en el flujo de calor
es el flujo mágico por el cp por la
diferencia de temperaturas tenemos esto
ese mismo calor obviamente es el que se
transfiere al fluido frío y con eso
puedo determinar la temperatura de
salida del giro fresh simplemente
dividiendo el calor entre m cp y suman
aso a la temperatura de entrada
listo ya que tengo las 4 temperaturas
entonces ya puedo calcular las
diferencias de temperatura en un extremo
y en otro miren aquí tengo de 4 - b 13
y temperatura de salida porque como está
a contraflujo la temperatura de entrada
de uno de ellos es la de salida del otro
aquí escribimos escribió al revés puse
la fría menos la caliente se van a dar
cuenta que no hay diferencia siempre con
esa congruente también el del tractor es
del tate 2 lo hago
así en el frío menos el caliente
entonces cuando haga
la diferencia el promedio logarítmica de
todos modos me va a quedar una cantidad
positiva entonces no hay problema
como se haga siempre y cuando se sea
congruente ya que tengo del tate
hay que calcular distinto
hay que calcular el área superficial
tenemos un tubo de seis metros de 2.5
centímetros de diámetro y por el
diámetro por la longitud es el área
y finalmente despejando de la relación
con igual a ahora por del trate
tengo el coeficiente global de
transferencia que es 2311 y listo
terminando
Voir Plus de Vidéos Connexes
Heat Transfer (27) - Heat transfer in internal flows in tubes
Thermal Conductivity, Stefan Boltzmann Law, Heat Transfer, Conduction, Convecton, Radiation, Physics
Specific Heat of a Metal Lab
GCSE Physics - Internal Energy and Specific Heat Capacity #28
Heat Engines and Thermal Efficiency| Grade 9 Science Quarter 4 Week 7
L22 | Heat Loss & Gain
5.0 / 5 (0 votes)