Let’s talk about the first law of
thermodynamics. The goal is to take all the forms of energy, like work,
heat, and total energy and relate them together. The first law of thermodynamics states that
Energy cannot be neither created nor destroyed during a process, it can only change forms.
For example, from kinetic to heat, potential to kinetic, etc. This means that all the energy can
be accounted for during any process. It’s really just a different way of saying the
conservation of energy.
If we have a system where energy enters the system and energy leaves the system,
the net change in total energy of that system is equal to the difference between the energy that
enters and the energy that leaves. In simple terms, the change in total energy of a system
is equal the total energy entering a system minus the energy leaving the system.
We can represent that like this and it’s called energy balance.
If we wanted to figure out the energy change of a system, all we need to do is
subtract the initial energy from the final energy. We can represent that like this.
Keep in mind energy exists in numerous forms, like internal energy, kinetic, potential, and more.
We can represent the total energy of a system excluding electric and magnetic energy,
using the letter E and we can show the change in total energy of a system like this.
Here U represents internal energy, KE represents kinetic energy and PE
represents potential energy. Note that in cases where the system is stationary, in other words,
there is no change in velocity or elevation, then the change in kinetic and potential energy is 0.
So you just end up with this. Lastly, we need to talk about
mechanisms of energy transfer. There are 3 ways in which energy transfers to or from a system.
Heat (Q), work (W) and mass flow (m). These topics were discussed in detail in the previous videos,
so if you need a refresh, please check the description. Let’s do a quick summary.
When there is a heat gain in the system, the energy of the molecules goes up
meaning that internal energy of the system increases, and when there is a heat loss,
the energy of the molecules goes down so the internal energy of the system decreases.
Work transfer to a system increases the energy of a system and work transfer
from a system decreases the energy of a system. So things like turbines produce
work while things like compressors consume work. Lastly, we have mass flow. When mass enters a
system, the energy of the system increases, and when mass leaves the system, the energy of the
system decreases. For example, when oil gets too hot, you can take some out and replace the amount
taken with cold oil to reduce the temperature quickly. I got that tip from Gordan Ramsay.
We can summarize this by writing out our equation like this. Note that here, we are
saying the quantity that leaves the system and the quantity that enters the system is the same.
Of course, if there is no temperature difference, then this is zero,
if there is no work transfer, then is this zero, and if there is no mass flow, then this is zero.
If we were to write this in a compact form, then we can write it like this.
The Units would be kilojoules. And if we wanted to write this in rate form, then it would look
like this. So we’re saying everything per unit time. Think of it as energy transfer that happens
per second, or minute, etc. and that’s equal to the rate of change in the internal system,
so kinetic energy change per second, or potential energy change per second, and
so forth. The units for this would be kilowatts. Lastly, if we needed to express this on a unit
mass basis, we can write it like this. The units would be kJ per kilogram.
Now let’s move on to some examples, starting with some easy ones to some hard ones
to see how we can apply what we discussed. Let’s look at this problem where we have a house
and it’s found that during the winter months, it will lose heat at a rate of 60000 btu per
hour. When we turn on the lights and appliances, the heat generated would be 6000 btu per hour.
We need to figure out the power required to maintain the house at a constant temperature
if we use heaters in kilowatts. So the entire house will be our system.
The energy going out would be 60000 btu/h, and the energy coming in would be the 6000 btu/h plus the
amount our heater needs to create. Remember that the total energy has to be constant in the house
which means the difference between energy in and energy out must be equal to 0.
In other words, energy in must equal energy out. That means our heater needs to create 54000 btu/h.
Now we need this answer in Kw so 1 kw is equal to 3412 btu/h. We divide our value
and we get our answer. So the power rating required for our heater would 15.83 kw.
In this problem, we have a room that’s at a temperature of 20 degrees Celsius. Then we turn on
a few electric devices. We need to figure out the rate of increase of the energy content in the room
if there is no heat transfer through the walls.. The entire room will be our system. Since there is
no heat transfer through the walls, our energy out will be zero. Also note that the temperature given
is unnecessary for this question. The energy in would be the addition of all the electrical
devices. Now the rate of increase of energy would be equal to the energy in minus the energy out.
So the rate of increase in energy is 1650 W. Let’s take a look at this problem where we have
a 60W fan that circulates air through ducts. To keep the flow going, the fan needs to raise the
pressure of the air by 50 pascals. We need to find the highest possible average flow velocity
in the duct. Let’s write down what we know. The power of the fan is 60 W, that’s energy going
in, and we can write it in j/s. Remember, 1 watt is 1 j/s. We also have the pressure and since this
fan needs to raise the pressure by 50 pascals, that means the pressure at the start of the area
vs the end of the area must have a difference of 50 pascals, so that’s delta P. Lastly, the
flow section has a diameter of 30 cm, or 0.3 m. The steps we will take to solve this problem
is first, we need to figure out the volume flow rate. Once we have that, we can figure out the
velocity by dividing the volume flow rate by the cross-sectional area. So how do we figure out the
volume flow rate? For that, we can write an energy equation. Our energy in must equal the energy out.
So what energy is going in? We have the power that goes in and we have the flow work, in other words,
flow energy, that goes in. And on the other side, we have the flow work that goes out.
Don’t forget flow work is Pressure times volume and volume is equal to mass times specific volume.
So rate flow work is equal to mass flow rate times pressure times the specific volume.
If this is unfamiliar to you, please watch the video on energy forms and mass flow.
Now we can bring the work flow to the other side. The specific volume and the mass flow is the same
throughout the area we are looking at, so we can pull them out and write it like this. Notice when
we do this, we get P2 minus P1, so that’s the difference in pressure between the two points,
which is delta P. If you think back to the previous video, you might remember that mass
flow is equal to volume flow divided by specific volume. So if we isolate it for volume flow,
we get this. Now you can see that what we have here is volume flow, so let’s replace it. Now
all we need to do is isolate for volume flow and start plugging in our values. We do need to
use a unity conversion ratio since the units for volume flow needs to be cubic meter per second.
Remember that 1 joule is equal to 1 pascal cubic meter. Let’s solve and get we our volume flow.
To figure out the average flow velocity, we need to think back to our mass flow equation.
We can get rid of the density, so we are left with just volume flow. Now we isolate it for
the average velocity. The cross-sectional area can be found using this equation.
Let’s plug in the diameter and solve. Now we can use this value in our equation.
Solving gives us the highest possible average flow velocity in the duct.
Let’s take a look at one last example. We have a gasoline pump that uses 3.8 kw of power, and it
creates a pressure difference of 7 kpa between the inlet and outlet of the pump. We need to figure
out the maximum possible volume flow rate. Let’s write down what we know. The power
usage is 3.8 kW, or 3.8 kj/s. The pressure difference is 7 kpa, so that’s delta p.
Now, we need to write our energy equation. This question is very similar to the previous question,
so we can go through this fairly quickly. So what energy goes in and what energy comes out?
We have the power that goes into the motor, flow energy that goes in and flow energy that goes out.
We can isolate the equation to write it like this. Now we can replace mass flow multiplied
by specific volume, which is equal to volume flow. Let’s isolate for the volume flow. Now
we can plug in the values we know. Don’t forget to use a unity conversion ratio since we need our
answer to be in cubic meters per second. Solving gives us the maximum possible volume flow rate.
That should cover the types of problems you will face when it comes to the first law of
thermodynamics. I hope this video helped, thanks for watching and best of luck with your studies!