### 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.