• Support PF! Buy your school textbooks, materials and every day products Here!

Turbine Problem

  • Thread starter pyroknife
  • Start date
  • #1
613
3
I attached the file.

I just wanted to clarify something.

All i need to find is the density @ inlet&outlet and the diameter at the outlet.

The heat loss they have shown has no impact on the values that were asked for right?

density can be found by pressure/(RT) where R=gas constant and T=temperature
And the diameter can be found by setting the mass flow at the inlet = mass flow at the outlet.
 

Attachments

Answers and Replies

  • #2
1,197
0
The problem states the inlet pressure is 45 psi. But the diagram shows 160 psi at inlet. The 45 psi is shown at the outlet. Something is amiss.

But to answer your question, if you know P and T, you can determine density. Knowing that you know the mass flow. The heat loss has no impact on the values asked. Conservation of mass will determine exit area.
 
  • #3
613
3
Yeah I noticed that typo too. The prof said to assume it is 160 psi. Okay I thought so.

Basically my calculations for the density was pressure/(RT)
Where R=gas constant of air and T=temperature in rankine


I just said set (density at inlet)*(Velocity at inlet)*(Area of inlet)=(density at outlet)*(Velocity at outlet)*(Area of outlet) and solved for the exit area.

But jst out of curiosity, all turbines have circular cross sections right?
 
  • #4
1,197
0
A round shape is the easiest to machine and therefore costs less. Exhaust pipes are subject to thermal stresses due to piping wanting to expand radially and axially at startup. If it were square or had any corners, you would have built in 'stress risers' that would make it more prone to failure. Round is best for any number of reasons.
 
  • #5
613
3
A round shape is the easiest to machine and therefore costs less. Exhaust pipes are subject to thermal stresses due to piping wanting to expand radially and axially at startup. If it were square or had any corners, you would have built in 'stress risers' that would make it more prone to failure. Round is best for any number of reasons.
Oh okay, I just solved the problem, but something weird happened.

The way the picture is drawn, the outlet area should be bigger, but the area I got issmaller than the inlet.
 
Last edited:
  • #6
1,197
0
Yes, I also end up with a larger exit area. But rho*A*V = constant. Working fluid is air. If you know pressure and temperature, you then know density. The trapezoid symbols used for compressors and turbines in thermo texts are meant to infer the working fluid is compressed or expanded. The numbers in this problem are strange.
 
  • #7
613
3
Yes, I also end up with a larger exit area. But rho*A*V = constant. Working fluid is air. If you know pressure and temperature, you then know density. The trapezoid symbols used for compressors and turbines in thermo texts are meant to infer the working fluid is compressed or expanded. The numbers in this problem are strange.
Thanks. did you get diameter = 0.453 ft?

I talked to a friend. He says turbines are supposed to have a smaller outlet area than inlet. And that the figure is drawn like a diffuser instead of a tirbine.
 
  • #8
1,197
0
I do not calculate 0.453 ft^2. I get a little over 23 in^2 which is 0.16 ft^2.
 
  • #9
613
3
First I calculated the two densities
At 1: density=P1/(RT1)
P1=(160 lb/in^2 * 12in^2 / 1ft^2)
R=(1716 ft*lbf/(slug*R)
T1=320+460 R
P1= .0172 slug/ft^3

P2=45 lb/in^2 * 12in^2/1ft^2
R=1716 ftlb/(slugR)
T2=50+460 R
Density 2=.0074 slug/ft^3

COnservation of mass:
(.0172)(30/144 ft^2)(100ft/s)=.0072 slug/ft^3 * 300 ft/s * A2
A2=.1614 ft^2
Diameter=.453 ft

Crap! I just wrote this all out for no reason. I meant I got diameter=.453 ft. I got Area=.16 ft^2 as well.
 

Related Threads for: Turbine Problem

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
992
  • Last Post
Replies
0
Views
2K
Replies
1
Views
783
  • Last Post
Replies
1
Views
1K
Replies
13
Views
14K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
531
Top