# Understanding Convection Heat Transfer in a Convection Heat Exchanger

• Joshua Pham
In summary: Do you remember the open system (control volume) version of the 1st law of thermodynamics (that you must have learned in your thermodynamics course)?Yes, I remember the open system version of the 1st law of thermodynamics. It states that the total internal energy (aka the total energy of the system) is conserved.
Joshua Pham
Homework posted in wrong forum, so no template
Hello guys, in the question attached, my understanding is that there is a heat transfer
that heats the fluid from an initial
at the input, to
at the output. This heat transfer is via convection from walls of temperature
.
Firstly,
.

Because the walls are
and the fluid is of a lower temperature, this heat transfer is via convection from the walls. The correct answer says that
. Why is this? Why not
? or
. The fluid next to the wall is heated and changes until it reaches
at the outlet, so how do we know which temperature to use in the convection equation where heat is being transferred from the walls to the fluid?

They expect you to assume that the overall heat transfer coefficient is h.

Joshua Pham
Hello Chestermiller; really apologise I didn't fill out the homework template. What does the over heat transfer coefficient ##h_c## have anything to do with the temperature difference being ##T_h - T_o##? I would argue that the temperature difference used in ##q = h_c A \Delta T## should be ##T_h - T_{average\ inside}##?

Chestermiller said:
The heat flux is equal to the overall heat transfer coefficient times ##(T_h-T_0)## (i.e., the overall temperature difference).

I'm still not quite convinced. Why would ##T_h - T_0## be the overall temperature difference?? Would the "overall" temperature difference be the average temperature difference? Which is $$T_h - \frac{T_0 - T_i}{2}$$ ?

Oh. I see what you are getting at. It goes like this:
$$\rho V c\frac{dT_0}{dt}=\dot{m}c(T_i-T_o)+hA(T_h-T_0)$$
This assumes that the chamber is well-mixed so that the average temperature in the chamber ##T_0## is always equal to the outlet temperature. So this is a transient heat transfer problem, in which you are solving for the average/outlet temperature as a function of time. So, at any given time, ##T_0## is assumed to be the temperature thorughout the chamber.

BvU and Joshua Pham
Ah! Thanks so much!

Hey Chestermiller, could you provide more insight into how you derived that equation? Reading online I've only seen the expression $$Ah(T-T_{\infty}) = -\rho Vc \frac{dT}{dt}$$ after googling transient heat convection transfers.

Joshua Pham said:
Hey Chestermiller, could you provide more insight into how you derived that equation? Reading online I've only seen the expression $$Ah(T-T_{\infty}) = -\rho Vc \frac{dT}{dt}$$ after googling transient heat convection transfers.
This is the same as the equation I wrote, except for the term involving flow into and out of the chamber (which was not a feature of the problem in your reference).

Chestermiller said:
This is the same as the equation I wrote, except for the term involving flow into and out of the chamber (which was not a feature of the problem in your reference).

Thanks Chestermiller.
The way I derived your expression is,
$$\dot{Q_{in}} - \dot{Q_{out}} = \frac{d(mc(T_o - T_i))}{dt}$$
but there is no ##\dot{Q_{out}}## if you consider the inner chamber as the system and there is no heat loss to the surroundings. So,
$$\dot{Q_{in}} = \frac{d(mc(T_o - T_i))}{dt}$$
Next, this heat transfer rate can only come from convection of heat from the walls interfacing the inner and outer chambers which is given by,
$$q = h_cA(T_h - T_o)$$
These must equal and thus,
$$\frac{d(mc(T_o - T_i))}{dt} = h_c A(T_h - T_o)$$
Now to obtain your expression, you need to subject the left term to the product rule. But why?? Aren't ##T_i## and ##T_o## constants? When I did a thermodynamics course, that left term would just turn into, $$\dot{m}c(T_o - T_i)$$

Joshua Pham said:
Thanks Chestermiller.
The way I derived your expression is,
$$\dot{Q_{in}} - \dot{Q_{out}} = \frac{d(mc(T_o - T_i))}{dt}$$
but there is no ##\dot{Q_{out}}## if you consider the inner chamber as the system and there is no heat loss to the surroundings. So,
$$\dot{Q_{in}} = \frac{d(mc(T_o - T_i))}{dt}$$
Next, this heat transfer rate can only come from convection of heat from the walls interfacing the inner and outer chambers which is given by,
$$q = h_cA(T_h - T_o)$$
These must equal and thus,
$$\frac{d(mc(T_o - T_i))}{dt} = h_c A(T_h - T_o)$$
Now to obtain your expression, you need to subject the left term to the product rule. But why?? Aren't ##T_i## and ##T_o## constants? When I did a thermodynamics course, that left term would just turn into, $$\dot{m}c(T_o - T_i)$$
You must have forgotten a lot from your thermo course. For the material in the tank, m and c are constants, as is ##T_i## (but not ##T_0##, of course, which is a function of time). You left out the terms for enthalpy flowing into and out of the tank.

Do you remember the open system (control volume) version of the 1st law of thermodynamics (that you must have learned in your thermodynamics course)? If so, please write it down for me.

Chet

Chestermiller said:
You must have forgotten a lot from your thermo course. For the material in the tank, m and c are constants, as is ##T_i## (but not ##T_0##, of course, which is a function of time). You left out the terms for enthalpy flowing into and out of the tank.

Do you remember the open system (control volume) version of the 1st law of thermodynamics (that you must have learned in your thermodynamics course)? If so, please write it down for me.

Chet

$$\dot{Q} - \dot{W} = \sum_{out} \dot{m} (h+ke+pe) -\sum_{in} \dot{m} (h+ke+pe)$$
Hence,
$$\dot{Q} = \sum_{out} \dot{m} h -\sum_{in} \dot{m} h$$

So for convection,
$$hA_c(T_h - T_o) = \dot{m}(h_2-h_1) = \dot{m}c_p(T_o-T_i)$$?

Why would ##T_i## be a constant but not ##T_o##??

Joshua Pham said:
$$\dot{Q} - \dot{W} = \sum_{out} \dot{m} (h+ke+pe) -\sum_{in} \dot{m} (h+ke+pe)$$
Hence,
$$\dot{Q} = \sum_{out} \dot{m} h -\sum_{in} \dot{m} h$$

So for convection,
$$hA_c(T_h - T_o) = \dot{m}(h_2-h_1) = \dot{m}c_p(T_o-T_i)$$?
This is the steady state version. You are dealing with a transient problem. So you have to include the rate of change of internal energy within the control volume (chamber).

Chestermiller said:
This is the steady state version. You are dealing with a transient problem. So you have to include the rate of change of internal energy within the control volume (chamber).

Ah! Sorry

Here goes,
$$\dot{Q} + \dot{m}(h_1 - h_2) = \dot{m}(u_2 - u_1)$$

Heat is from convection so,
$$hA_c(T_h-T_o) = \dot{m}(c_p + c_v)(T_o-T_i)$$

Is that correct?
Why would ##T_i## be a constant but not ##T_o##?

Joshua Pham said:
Ah! Sorry

Here goes,
$$\dot{Q} + \dot{m}(h_1 - h_2) = \dot{m}(u_2 - u_1)$$

Heat is from convection so,
$$hA_c(T_h-T_o) = \dot{m}(c_p + c_v)(T_o-T_i)$$

Is that correct?
No. The rate of change in internal energy for the chamber is incorrect. It is $$\rho V\frac{du}{dt}=\rho Vc\frac{dT_0}{dt}$$
Also note that, for an incompressible liquid, the heat capacities at constant volume and pressure are the same, and are equal to the parameter c in your problem statement.
Why would ##T_i## be a constant but not ##T_o##?
##T_i## is the initial temperature, and also the temperature of the inlet stream to the chamber which, according to the problem statement is constant. ##T_0## is the temperature of the liquid inside of the chamber, and also the temperature of the liquid coming out of the chamber. The liquid is being heated, so its temperature is getting higher.

Joshua Pham
Chestermiller said:
No. The rate of change in internal energy for the chamber is incorrect. It is $$\rho V\frac{du}{dt}=\rho Vc\frac{dT_0}{dt}$$
Also note that, for an incompressible liquid, the heat capacities at constant volume and pressure are the same, and are equal to the parameter c in your problem statement.

##T_i## is the initial temperature, and also the temperature of the inlet stream to the chamber which, according to the problem statement is constant. ##T_0## is the temperature of the liquid inside of the chamber, and also the temperature of the liquid coming out of the chamber. The liquid is being heated, so its temperature is getting higher.
Thank you for that! I understand it much better now! In my thermo course, I think we just used steam tables for pure substances and the ideal gas equation for air. So I'd never dealt with a situation where the temperature was a function of time. I'm doing a second course later this year which will cover that. This is from a linear systems course.

## 1. What is a convection heat exchanger?

A convection heat exchanger is a device that transfers heat from one fluid to another through the process of convection. It consists of two separate fluid streams, the hot fluid and the cold fluid, that are brought into contact with each other in order to transfer heat.

## 2. How does a convection heat exchanger work?

A convection heat exchanger works by bringing the two fluids in close contact with each other through the use of a series of tubes or plates. The hot fluid, being at a higher temperature, transfers its heat to the colder fluid through convection, resulting in a temperature change in both fluids.

## 3. What are the different types of convection heat exchangers?

There are two main types of convection heat exchangers: parallel-flow and counter-flow. In a parallel-flow heat exchanger, the two fluids flow in the same direction, while in a counter-flow heat exchanger, the two fluids flow in opposite directions. There are also cross-flow heat exchangers, where the two fluids flow perpendicular to each other.

## 4. What are the advantages of using a convection heat exchanger?

Convection heat exchangers offer several advantages, including efficient heat transfer, compact design, and the ability to handle high temperatures and pressures. They are also versatile and can be used in a variety of industries, such as HVAC, chemical processing, and power generation.

## 5. How do you maintain a convection heat exchanger?

To maintain a convection heat exchanger, regular cleaning and inspection are necessary. This includes checking for any leaks or corrosion, as well as removing any buildup of debris or scale on the heat transfer surfaces. It is also important to monitor the flow rates and temperatures of the two fluids to ensure optimal heat transfer efficiency.

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