Momentum relation for control volume

In summary, the question discusses the calculation of the resultant force on a control volume, which is equal to the mass flow rate multiplied by the difference between the velocities at the outlets and inlets. The question also mentions the dimensions of the flow, specifically the depth and "into the paper" dimension, and asks about the inclusion of two "v cos (theta/2)" terms in the x momentum expression. The summary also includes a discussion about the effective width of the flow and its role in calculating the mass flow rate, as well as an explanation of the rate of change of x momentum for the streams impinging on the wedge.
  • #1
princejan7
93
0

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
 
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  • #2
princejan7 said:
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.
princejan7 said:
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.
 
  • #3
haruspex said:
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
 
  • #4
princejan7 said:
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)?
 
  • #5
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
 
  • #6
haruspex said:
In time t, what volume goes by (width w, depth d, velocity v)?
Chestermiller said:
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
 
  • #7
princejan7 said:
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.
 
  • #8
princejan7 said:
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
 
  • #9
thanks everyone
 

What is momentum relation for control volume?

Momentum relation for control volume is a fundamental concept in fluid mechanics that describes the relationship between the forces acting on a fluid and the resulting change in momentum within a specific volume of the fluid. It is based on the law of conservation of momentum, which states that the total momentum of a closed system remains constant unless acted upon by an external force.

How is momentum relation for control volume derived?

The momentum relation for control volume is derived from the Navier-Stokes equations, which are the governing equations for fluid flow. These equations are based on the principles of conservation of mass, momentum, and energy, and can be applied to any fluid flow situation. By applying these equations to a control volume, we can determine the change in momentum within that volume due to external forces.

What is the significance of the momentum relation for control volume?

The momentum relation for control volume is important because it allows us to analyze and predict the behavior of fluids in various engineering applications. By understanding the forces and changes in momentum within a control volume, we can design systems and structures that can withstand or control the forces of fluid flow.

What are the limitations of the momentum relation for control volume?

The momentum relation for control volume is based on certain assumptions, such as the fluid being incompressible and the flow being steady and laminar. These assumptions may not hold true in all fluid flow situations, and thus the results obtained from the momentum relation may not be entirely accurate. Additionally, the momentum relation does not take into account the effects of turbulence, which can significantly impact the behavior of fluids.

How is the momentum relation for control volume applied in real-world situations?

The momentum relation for control volume is widely used in various engineering fields, such as aerospace, mechanical, and civil engineering. It is applied in the design of pumps, turbines, and other fluid machinery, as well as in the analysis of fluid flow in pipes, channels, and other structures. It is also used in the design of vehicles, such as airplanes and ships, to ensure their stability and performance in various fluid flow conditions.

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