Momentum relation for control volume

[princejan7]

Homework Statement

http://postimg.org/image/i4p19540z/

Homework Equations

Resultant force on the control volume = Mass flow rate (Velocity outlets -Velocity inlets)

The Attempt at a Solution

I am just wondering if the 4cm is called depth, then what is the term for the "into the paper" dimension
Also, shouldn't there be two "v cos (theta/2)" terms in the x momentum expression since there are two streams leaving the control volume?thanks

 

[haruspex]

I am just wondering if the 4cm is called depth, then what is the term for the "into the paper" dimension

The question statement does not call the 4cm depth. But then, I'm not sure what the 4cm is supposed to represent. It seems to be the effective width of the flow, i.e. the region that's redirected by the wedge. The diagram make it look narrower than the wedge, which surprises me.

Also, shouldn't there be two "v cos (theta/2)" terms in the x momentum expression since there are two streams leaving the control volume?

That is taken into account by the 4cm width.

 

[princejan7]

The question statement does not call the 4cm depth. But then, I'm not sure what the 4cm is supposed to represent. It seems to be the effective width of the flow, i.e. the region that's redirected by the wedge. The diagram make it look narrower than the wedge, which surprises me.

could you also explain why mass flow rate per unit depth is calculated as density x velocity x 4cm

 

[haruspex]

could you also explain why mass flow rate per unit depth is calculated as density x velocity x 4cm

In time t, what volume goes by (width w, depth d, velocity v)?

 

[Chestermiller]

Let's suppose that the depth is d. Then the cross section of the channel perpendicular to the flow is 0.04d. The volumetric flow rate in the upstream region Q through this cross section is the fluid velocity times the area. So Q = 6(0.04)d. The mass flow rate is ρQ=6ρ(0.04)d.

In the region of the wedge, if the velocity is still 6 m/s, each channel surrounding the wedge has an opening (width) of 2 cm. This guarantees that the mass flow rate is conserved.

Chet

 

[princejan7]

In time t, what volume goes by (width w, depth d, velocity v)?

Let's suppose that the depth is d. Then the cross section of the channel perpendicular to the flow is 0.04d. The volumetric flow rate in the upstream region Q through this cross section is the fluid velocity times the area. So Q = 6(0.04)d. The mass flow rate is ρQ=6ρ(0.04)d.

In the region of the wedge, if the velocity is still 6 m/s, each channel surrounding the wedge has an opening (width) of 2 cm. This guarantees that the mass flow rate is conserved.

Chet

thanks, I finally understand that

I still don't understand this part

That is taken into account by the 4cm width.

How does the 4cm take into account the other stream; the second "mass flow rate* V *cos(theta/2)" term

 

[haruspex]

How does the 4cm take into account the other stream; the second "mass flow rate* V *cos(theta/2)" term

It's the combined (effective) width of the two streams, so it's included in the mass flow rate.

 

[Chestermiller]

How does the 4cm take into account the other stream; the second "mass flow rate* V *cos(theta/2)" term

That was from a quote by Haruspex that I don't understand either. But I can address the question you were asking. The rate of change of x momentum for the stream(s) impinging on the wedge is equal to the rate of x- momentum flow out of the control volume minus the rate of x momentum flow into the control volume. The rate of x-momentum flow out of the control volume is the mass flow rate times the x-component of velocity. Half the mass flow goes below the wedge, and half the mass flow goes above the wedge. Each of these streams has the same x component of velocity. So, if you want to treat the two streams separately, you can do that, but you have to take into account that half the total flow is present above and below.

Chet

 

[princejan7]

thanks everyone