In this video we are going to work through an example in which we are going to calculate
the air pressure at 40,000 feet above sea level using three different assumptions.
One, that the fluid is incompressible.
Two, that the fluid is compressible and isothermal.
Three, the more realistic case in which we have a compressible gas, air, and a non-isothermal
gradient and a temperature gradient for our elevation.
So a non-isothermal case.
The derivations for the equations we are going to use have been done in a previous screencast.
We are just going to work through this example to show you the differences in values that
we are going to get using these three assumptions.
In the incompressible case we are saying that our air is an incompressible fluid and that
we are looking for some pressure, P1, related to a pressure at another height.
Here we have our height of our air.
We are using the specific weight of air in this case.
That is something you would look up and our value is 12.014 N/m^3.
We need to convert our elevation 40,000 feet to meters.
Which is about 12,190 m.
We also need to know one of the pressures to solve this.
We know that our atmospheric pressure at sea level is 101.33 kPa.
Our relationship for the incompressible case is that P1 is going to equal P2, the pressure
at the ground minus the height and specific weight of our fluid above it.
We can plug in some numbers and makes sure that our units are the same.
On the right side we have N/m^2 which is the same as a pascal and here we have kPa so we
will have to divide these by 1000 Pa to make sure that our units are the same.
We get a pressure of -45.13 kPa which is an unrealistic answer.
In this case assuming that the gas is incompressible is a bad assumption.
Let's take a look at the case where we assume that the gas is compressible but also that
there is no change in temperature from sea level to 40,000 feet.
In this case we are given the variables as the same.
The only real difference is the equation we are using.
Our relationship for pressure and height for a compressible gas is going to be P1 is equal
to P2 times the exponential of negative g for the gravitational constant over (R times
T_o) times our change in elevation.
Here we are using 9.807 m/s^2 as our gravitational constant.
Also the gas constant I am using 286.9 J/(kg*K) and the temperature at sea level of 288.15
K. You will find that everything inside of the exponential ends up being unitless.
We get a pressure P1 that is equal to 23.9 kPa.
A much more realistic pressure and one that is lower that at sea level which is something
that we would expect.
Now the real question is how much does it differ if we take into consideration that
there may be a temperature gradient from sea level to 40,000 ft.
For a non-isothermal case we have a relationship between the pressure at P1 and P2 using the
following relationship derived in another video.
Here T_o is our temperature at sea level.
Again we are using g, our gravitational constant, R our gas constant.
We have this new term beta which is known as the lapse rate.
Basically the rate of change in temperature with elevation.
That is something you will have to look up for our system.
For our first later here, the first 40,000 ft above sea level we found that beta can
be written as 0.00650 K/m.
This is a small change in temperature each meter of elevation but this can be significant
when looking at a large change in elevation.
I will plug in numbers for our equation.
When I solve this I get a pressure at 40,000 feet as 18.67 kPa.
Now you can compare our two cases, the non-isothermal and isothermal case and see there is a pretty
significant difference.
Almost 20% difference between the two.
So which one is more accurate in this case?
If we look at a table of the altitude vs. absolute atmospheric pressure you will see
the value reported for 40,000 feet is 18.7 which is basically what we calculated for
our non-isothermal case.
Hopefully this gives you an idea of the vast difference that can and how important it is
to make the appropriate assumptions when calculating your static fluid pressure.