Surface Integral Problem (Answer Already Given)

[McAfee]

Homework Statement

The problem is to calculate the flux emanating from the exhaust of a jet engine.

The air gas velocity from a jet engine varies linearly from a maximum of 300 m/s at the center of the circular exhaust opening to zero at the edges. If the exhaust diameter is 1.6 m, find the exhaust flow.



The answer is 201 meters cubed over second

Homework Equations

y=mx +b

The Attempt at a Solution

b=300 m/s at center
x =radius in meters = r
m=slope=0-300/(0.8-0)=-375 cycles/s, 0.8 is radius of circular opening
y=B=-(375*x)+300
m/s

2: Compute Flux, Psi

flux(non-uniform flux denisity) = Double integral B dot dS
dS = surface area of circle=pi*r^2 in this case.
limits are 0-0.8 of both integrands
variable of interest is r. similar to integrating acceleration twice to get position

I know I need to ulimately set up a double integral

 

[LCKurtz]

Homework Statement

Since it varies linearly I can derive an equation that gives me the flow rate as a function of position.

The answer is 201 meters cubed over second

Homework Equations

y=mx +b

The Attempt at a Solution

b=300 m/s at center
x =radius in meters = r
m=slope=0-300/(0.8-0)=-375 cycles/s, 0.8 is radius of circular opening
y=B=-(375*x)+300
m/s

2: Compute Flux, Psi

flux(non-uniform flux denisity) = Double integral B dot dS
dS = surface area of circle=pi*r^2 in this case.
limits are 0-0.8 of both integrands
variable of interest is r. similar to integrating acceleration twice to get position

I know I need to ulimately set up a double integral

It would help if you would give a complete statement of the problem. Is this fluid in a circular pipe? What do you know about the flow rate aside from being 300 m/s at the center? Where did the 375 come from? Where did that linear equation come from? Don't make us guess what the problem is.

 

[McAfee]

It would help if you would give a complete statement of the problem. Is this fluid in a circular pipe? What do you know about the flow rate aside from being 300 m/s at the center? Where did the 375 come from? Where did that linear equation come from? Don't make us guess what the problem is.

Sorry about that sir. I just edited the original post.

 

[LCKurtz]

Homework Statement

The problem is to calculate the flux emanating from the exhaust of a jet engine.

The air gas velocity from a jet engine varies linearly from a maximum of 300 m/s at the center of the circular exhaust opening to zero at the edges. If the exhaust diameter is 1.6 m, find the exhaust flow.



The answer is 201 meters cubed over second

Homework Equations

y=mx +b

The Attempt at a Solution

b=300 m/s at center
x =radius in meters = r
m=slope=0-300/(0.8-0)=-375 cycles/s, 0.8 is radius of circular opening
y=B=-(375*x)+300
m/s

Let's keep the variable $r$. I don't know why you would label the units cycles/sec. So you correctly have $B=-375r+300$

2: Compute Flux, Psi

flux(non-uniform flux denisity) = Double integral B dot dS
dS = surface area of circle=pi*r^2 in this case.
limits are 0-0.8 of both integrands
variable of interest is r. similar to integrating acceleration twice to get position

I know I need to ulimately set up a double integral

$dS$ is not the surface area of the circle. $dS$ represents the differential element of surface area. That would be $dydx$ in rectangular coordinates. But since this is a circular area you need the $dS$ element in polar coordinates.

What is $dS$ in polar coordinates? (Answer that).

Then set up the integral using your formula. Think about the correct limits for the variables. Come back and tell us what you get.

 

[member 428835]

201.062 you mean?

 

[member 428835]

LCKurtz is correct, keep it up, youll get it

 

[McAfee]

Thanks guys. I will work on it later today and I will get back to you guys. Once again thanks

 

[McAfee]

Let's keep the variable $r$. I don't know why you would label the units cycles/sec. So you correctly have $B=-375r+300$



$dS$ is not the surface area of the circle. $dS$ represents the differential element of surface area. That would be $dydx$ in rectangular coordinates. But since this is a circular area you need the $dS$ element in polar coordinates.

What is $dS$ in polar coordinates? (Answer that).

Then set up the integral using your formula. Think about the correct limits for the variables. Come back and tell us what you get.

Ok I figured it out guys. Thanks to LCKurtz for explaining it to me.

The function I had is correct -375r+300. Since it was a circle it was easier for me to put into circular coordinates. So I had to change from dxdy to r dr dΘ

Whole set-up:
∫∫-375r+300 r dr dΘ from 0 to 0.8 then from 0 to 2π.

Next we can separate the integrals so:
∫1 dΘ from 0 to 2∏ and ∫-375r^2+300r dr from 0 to 0.8

Then rest is easy to solve.

And the final answer is 201.062