Video: Calculating the Flow Rate of a Fluid in a Heat Exchanger

In the heat exchanger shown in the accompanying diagram, a fluid flowing into the exchanger’s outer pipe has a specific heat capacity of 4 kJ/kg ⋅ °C. Fluid enters the pipe at the point 2 with a flow rate of 3.00 kg/s and at a temperature of 10°C. The fluid exits the pipe at the point 4 at a temperature of 15°C. Another fluid with a specific heat capacity of 2 kJ/kg ⋅ °C enters the exchanger’s inner pipe at the point 1. The second fluid enters this pipe at a temperature of 90°C and leaves the pipe at the point 3, at a temperature of 30°C. What is the flow rate of the fluid that moves through the inner pipe?

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Video Transcript

In the heat exchanger shown in the accompanying diagram, a fluid flowing into the exchanger’s outer of pipe has a specific heat capacity of four kilojoules per kilogram degrees Celsius. Fluid enters the pipe at the point two with a flow rate of 3.00 kilograms per second at a temperature of 10 degrees Celsius. The fluid exits the pipe at the point four at a temperature of 15 degrees Celsius. Another fluid with a specific heat capacity of two kilojoules per kilogram degrees Celsius enters the exchanger’s inner pipe at the point one. The second fluid enters this pipe at a temperature of 90 degrees Celsius and leaves the pipe at the point three at a temperature of 30 degrees Celsius. What is the flow rate of the fluid that moves through the inner pipe?

Okay, so first of all, this is the heat exchanger that’s mentioned in the question. And we’re told that it consists of two pipes: the outer pipe, which consists of all of this bit and all of this bit, and the inner pipe, which consists of this bit. Now we’re told that a fluid flowing into the exchanges out of pipe has a specific heat capacity of four kilojoules per kilogram degrees Celsius. And so we can say that for the fluid in the outer pipe or the outer fluid, the specific heat capacity 𝑐 sub o standing for heat capacity sub outer is equal to four kilojoules per kilogram degrees Celsius. We’ve also been told that this fluid enters the pipe at the point two and exits the pipe at the point four.

So essentially it enters this way and then leaves this way. We’ve also been told that it flows at a flow rate of 3.00 kilograms per second. So we’ll say that the flow rate of the outer pipe 𝐹 sub o is equal to 3.00 kilograms per second. Finally, we’ve been told that the initial temperature of the fluid in the outer pipe is 10 degrees Celsius and it leaves the outer pipe with a temperature of 15 degrees Celsius. So that’s all the information that we have about the fluid in the outer pipe. Let’s then look at the fluid flowing in the inner pipe. We’re told that that fluid has a specific heat capacity of two kilojoules per kilogram degrees Celsius. So we can say that 𝑐 sub i specific heat capacity of the fluid in the inner pipe is two kilojoules per kilogram degrees Celsius. We’re also told that it enters the pipe at the point one and leaves at the point three. So essentially it’s flowing this way through the inner pipe.

We’re told that the second fluid enters the pipe at a temperature of 90 degrees Celsius and leaves at a temperature of 30 degrees Celsius. So we can say that the temperature of the fluid in the inner pipe at the start is 90 degrees Celsius and at the finish is 30 degrees Celsius. We’ve been asked to find the flow rate of the fluid that moves through the inner pipe. In other words, we’ve been asked to find 𝐹 sub i, the flow rate of the inner fluid. Now the way that we’re gonna do this is to realize that in a heat exchanger, the whole point is that heat is exchanged from one fluid to the other. And therefore, in a perfect scenario, which is what we’re assuming this to be, any heat lost by one fluid is gained by the second fluid.

Now in this case, we can see that the temperature of the outer fluid is increasing from start to finish. It goes from 10 degrees Celsius to 15 degrees Celsius. Therefore, this fluid is the one gaining heat. And the inner fluid, which has a temperature that goes down from start to finish, is the one that’s losing heat. So how are we going to go about working out the heat exchange by either of these fluids? Well, we could recall the definition of specific heat capacity. Specific heat capacity is equal to the heat exchanged per unit mass to increase the temperature of a substance by one degree Celsius. And so we can use this equation and rearrange it to give us the heat exchanged by one of the fluids, which is equal to the mass of the fluid multiplied by the specific heat capacity of the fluid multiplied by the temperature change of the fluid.

Now in this case for the outer fluid, the heat exchange which we’ll call Δ𝑄 sub o is equal to the mass of the fluid, which we we’ll come back to in a second but we’ll also call this 𝑚 sub o, multiplied by 𝑐 sub o, the specific heat capacity of the fluid, multiplied by the temperature change of the fluid, which is given by subtracting the starting temperature from the final temperature. So we’ve got this quantity, the specific heat capacity. We’ve got the final temperature. And we’ve got the initial temperature as well. What we don’t quite have though is the mass of the fluid. But we do have the flow rate. We know that every second, 3.00 kilograms of fluid are flowing through the outer pipe. So what we could do is to modify our equation over here very slightly so that on the right-hand side of the equation instead of having the mass of the fluid we have the mass transferred per second, which is basically the flow rate.

But then if we’re going to divide the right-hand side by the unit of the second then we have to do the same thing on the left. In other words then, the heat transferred per second is equal to the mass transferred of that fluid per second multiplied by its specific heat capacity multiplied by the difference in temperature. And now at this point, we’ve got everything we need to work out the right-hand side of this equation. We plug all the values in: the mass flow rate, which by the way is exactly the same as 𝐹 sub o, and we’re plug in the specific heat capacity and the final and initial temperatures. At this point, we can see that this parenthesis is going to have a unit of degrees Celsius. So that unit of degrees Celsius cancels with this unit of degrees Celsius in the denominator. As well as this, this kilogram unit cancels with this kilogram unit. And we’re gonna be left with kilojoules per second. In other words, this is the heat transferred per second to the fluid in the outer pipe, which is exactly what we expected to find on the left-hand side.

So evaluating the stuff on the right-hand side of the equation, we find that 60 kilojoules of heat are transferred to the outer fluid every second. Now why is this useful? Well this is useful because we said earlier that any heat transferred to this outer fluid must have come from the inner fluid. In other words, every second, the inner fluid will be transferring 60 kilojoules of heat to the outer fluid. So the heat gained by the outer fluid every second is the same as the heat lost by the inner fluid every second. In other words, we can say that the change in heat of the inner fluid per second is equal to negative 60 kilojoules per second, because this time the inner fluid is losing heat, therefore the negative sign. But then we can say for the inner fluid this time that the heat transferred per second is equal to the mass of the fluid transferred per second multiplied by the specific heat capacity of the inner fluid multiplied by the change in temperature of the inner fluid, so same equation as before but this time we’ve got the subscripts i rather than o.

And then we can remember that the mass transferred per second of the fluid is the same as the flow rate 𝐹 sub i, which is what we’re trying to find. So we rearrange the equation to find this. So we can say that the heat exchange per second by the inner fluid divided by the specific heat capacity of the inner fluid multiplied by the temperature change of the inner fluid is equal to the mass of the inner fluid transferred every second, or in other words the fluid flow rate. At this point, we simply sub in the values on the left-hand side of the equation, which leaves us with negative 60 kilojoules per second — that’s the heat transferred by the inner fluid because heat is lost remember — divided by two kilojoules per kilogram degrees Celsius. That’s the specific heat capacity of the inner fluid. And also in the denominator is the temperature change: 30 degrees Celsius minus 90 degrees Celsius.

Now once again this parenthesis is gonna have a unit of degrees Celsius. So that cancels with this per degrees Celsius. And this unit of kilojoules canceled with this unit of kilojoules. So what we have is a unit of kilograms in the denominator of this fraction, but that fraction itself is in the denominator. So this unit is going to move up. And then we have to divide that by this unit of seconds. So our final unit is going to be kilograms per second, just as we expect for flow rate. And of course, this parenthesis is going to be negative 30 degrees Celsius minus 90 degrees Celsius is negative 60 degrees Celsius. So the negative sign in that parenthesis cancels out with this negative sign. So overall our answer is going to be positive. And when we do evaluate it, we find our answer to be 0.5 kilograms per second. And at this point, we found our final answer. We figured out that 0.5 kilograms of fluid flows to the inner pipe every second. In other words, the flow rate of the fluid is 0.5 kilograms per second.

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