How can I calculate momentum efflux from boundary data of a 2D nozzle?

gauthamnarayan · October 11, 2015, 14:49

I am trying to calculate the momentum efflux(N/m) from a 2d axis-symmetric case of a CD nozzle. I have the U,V momentums and pressure data. I have tried to use trapezoidal rule to line integrate the momentum at the exit. This is resulting in extremely low values. There might be an error in the variables I am integrating.

Can someone please guide me as to how I could calculate the momentum efflux from the boundary data of a 2D case.

FMDenaro · October 11, 2015, 16:17

Quote:

Originally Posted by gauthamnarayan

I am trying to calculate the momentum efflux(N/m) from a 2d axis-symmetric case of a CD nozzle. I have the U,V momentums and pressure data. I have tried to use trapezoidal rule to line integrate the momentum at the exit. This is resulting in extremely low values. There might be an error in the variables I am integrating.

Can someone please guide me as to how I could calculate the momentum efflux from the boundary data of a 2D case.

Given the total momentum flux (rho vv + p I + 2 mu Grad v ), you just project it along the normal unit vector to the exit boundary and you get a vector flux having x and y components. Now you can integrate along the exit line using the trapezoidal rule. Of course, the solution must be accurate

gauthamnarayan · October 12, 2015, 00:01

Quote:

Originally Posted by FMDenaro

Given the total momentum flux (rho vv + p I + 2 mu Grad v ), you just project it along the normal unit vector to the exit boundary and you get a vector flux having x and y components. Now you can integrate along the exit line using the trapezoidal rule. Of course, the solution must be accurate

Dear sir,

I have the UMomentum from the solver. I have tried to integrate $\int$ Pdx using the trapezoidal rule. If this is correct, then I have a fundamental doubt. Please bear with me as it may seem obvious.

The units of momentum efflux is N/m. which I have understood as Force/unit length of boundary. But,

Force = Momentum/Time

Please can you explain how integral of momentum with length, gives a result in N/m.

Thank you very much.

FMDenaro · October 12, 2015, 03:16

Quote:

Originally Posted by gauthamnarayan

Dear sir,

I have the UMomentum from the solver. I have tried to integrate $\int$ Pdx using the trapezoidal rule. If this is correct, then I have a fundamental doubt. Please bear with me as it may seem obvious.

The units of momentum efflux is N/m. which I have understood as Force/unit length of boundary. But,

Force = Momentum/Time

Please can you explain how integral of momentum with length, gives a result in N/m.

Thank you very much.

As you see, the momentum flux has the dimension of the pressure (N/m^2) therefore just a line integration gives dimension (N/m)

LuckyTran · October 18, 2015, 11:13

Are you working in cartesian x,y,z or cylindrical coordinates? And are you off by a factor of 2pi? If it's a 2D axisymmetric then the integral should be over $\int$ ()dA or $\int$ ()rdrdtheta, or $\int$ ()rdr and multiply by 2pi. But you still need to weight the integrand by r.

gauthamnarayan · October 26, 2015, 07:15

Quote:

Originally Posted by LuckyTran

Are you working in cartesian x,y,z or cylindrical coordinates? And are you off by a factor of 2pi? If it's a 2D axisymmetric then the integral should be over $\int$ ()dA or $\int$ ()rdrdtheta, or $\int$ ()rdr and multiply by 2pi. But you still need to weight the integrand by r.

Sorry to reply late.

I am using a cartesian coordinate system. I was actually of by a huge factor. Anyways I was able to solve it. Turns out i was supposed to also multiply it with velocity.

This particular solver outputs momentum as rho*u and not rho*u*u.
So actually i was ending up with mass flow integral and not that of momentum efflux.

Thanks for your reply.

LuckyTran · October 26, 2015, 20:32

That's great. I guess as a final note for others that might read this thread in the future I'd like to point out:

If you are talking about a particular fluid particle at a given location r(x,y,z) then rho*u is the momentum of the fluid particle at that location. So it is correct to say that rho*u is the momentum of a fluid particle.

The confusion is when you integrate fluxes through an area element dA, there is a dot products of the velocity vector with the area vector dA. i.e. mass flux is the integral of simply rho, the rate of efflux of mass density is mass flux which is then the product of rho with the dot product of u and dA. It's important to differentiate what is the property being advected (rho for mass, rho*u for momentum, rho*h for enthalpy) and the rate at which it is being advected (u dotted with dA).

October 11, 2015, 14:49	How can I calculate momentum efflux from boundary data of a 2D nozzle?	#1
gauthamnarayan New Member gautham narayan Join Date: Apr 2015 Posts: 16 Rep Power: 11	I am trying to calculate the momentum efflux(N/m) from a 2d axis-symmetric case of a CD nozzle. I have the U,V momentums and pressure data. I have tried to use trapezoidal rule to line integrate the momentum at the exit. This is resulting in extremely low values. There might be an error in the variables I am integrating. Can someone please guide me as to how I could calculate the momentum efflux from the boundary data of a 2D case.

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October 18, 2015, 11:13		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	Are you working in cartesian x,y,z or cylindrical coordinates? And are you off by a factor of 2pi? If it's a 2D axisymmetric then the integral should be over $\int$ ()dA or $\int$ ()rdrdtheta, or $\int$ ()rdr and multiply by 2pi. But you still need to weight the integrand by r.

October 26, 2015, 20:32		#7
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	That's great. I guess as a final note for others that might read this thread in the future I'd like to point out: If you are talking about a particular fluid particle at a given location r(x,y,z) then rhou is the momentum of the fluid particle at that location. So it is correct to say that rhou is the momentum of a fluid particle. The confusion is when you integrate fluxes through an area element dA, there is a dot products of the velocity vector with the area vector dA. i.e. mass flux is the integral of simply rho, the rate of efflux of mass density is mass flux which is then the product of rho with the dot product of u and dA. It's important to differentiate what is the property being advected (rho for mass, rhou for momentum, rhoh for enthalpy) and the rate at which it is being advected (u dotted with dA).