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Fluid mechanics: Bernoulli's equation and conservation of mass

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Homework Statement


A mechanical servo-mechanism comprising of a movable piston-cylinder within a vertical cylinder operates based on a venturi contraction in a horizontal 350mm diameter pipe that delivers a fluid of relative density 0.95. The upper end of the 100mm diameter vertical cylinder is connected by a pipe to the throat of the venturi, while the lower end of the cylinder is connected by a pipe to the pipe inlet. The piston in the cylinder is to be lifted vertically when the flow through the venturi exceeds 0.15m3 /s, such that it will activate a controller to throttle back the flow rate. Note that the piston rod is known to have a diameter of 20mm and that it passes through both ends of the vertical cylinder. Neglecting friction, calculate the diameter required of the venturi throat if the gross effective load on the piston rod is 180N.
Ans: 0.162m
upload_2018-8-24_16-8-57.png

Homework Equations


Benoulli equation & conservation of mass

The Attempt at a Solution


I tried solving using Bernoulli's equation and conservation of mass.

$$0.15=0.175^2\pi* v_p=r_t^2*\pi*v_t$$
where the subscript p represents location at pipe inlet while t represents location at throat

Area of cylinder is,$$(0.05^2-0.01^2)\pi = 0.00024\pi$$
Pressure difference in area of cylinder is,
$$P_p -P_t=180/(0.0024\pi)=23873$$
$$P_p + 0.5\rho v_p^2 = P_t + 0.5\rho v_t^2 +\rho *g(0.175-r_t)$$
$$\rho=950, \ \ g=9.81, \ \ v_p=1.56$$
$$r_t=\sqrt{\frac{0.15}{\pi*v_t}}$$
After subbing in, $$25028.96=475*v_t^2+9319.5(0.175-\sqrt{\frac{0.15}{\pi*v_t}})$$

I can only solved after I can get velocity at throat. But I am stuck at the above part.
 

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Answers and Replies

  • #2
stockzahn
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Your post would be easier to read if you'd use symbols without plugging in preliminary numbers.

What you've achieved so far:

1) Mass conservation:

$$\dot{V}=r_p^2 \pi v_p = r_t^2 \pi v_t $$

2) The necessary pressure difference to generate the required force ##F_{Piston}##:

$$p_p-p_t=\left( r_{out}^2-r_{in}^2\right)\pi F_{Piston}=A_{Piston} F_{Piston} $$

The Attempt at a Solution


I tried solving using Bernoulli's equation and conservation of mass.

$$P_p + 0.5\rho v_p^2 = P_t + 0.5\rho v_t^2 +\rho *g(0.175-r_t)$$
3) Bernoulli:

$$p_p+\frac{v_p^2\rho}{2}=p_t+\frac{v_t^2\rho}{2}\underbrace{+\rho g \left(r_p-r_t\right)}_{?}$$

Why did you add this last term to the RHS of the Bernoulli equation?
 
  • #3
254
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Your post would be easier to read if you'd use symbols without plugging in preliminary numbers.

What you've achieved so far:

1) Mass conservation:

$$\dot{V}=r_p^2 \pi v_p = r_t^2 \pi v_t $$

2) The necessary pressure difference to generate the required force ##F_{Piston}##:

$$p_p-p_t=\left( r_{out}^2-r_{in}^2\right)\pi F_{Piston}=A_{Piston} F_{Piston} $$



3) Bernoulli:

$$p_p+\frac{v_p^2\rho}{2}=p_t+\frac{v_t^2\rho}{2}\underbrace{+\rho g \left(r_p-r_t\right)}_{?}$$

Why did you add this last term to the RHS of the Bernoulli equation?
I think because the height difference for the tube at the throat compared to the tube at the pipe inlet will have gravitational energy which will affect the flow in the system. The rp is supposed to be on the LHS and I bring it to the RHS to include the gravitational effects due to height difference.
 
  • #4
stockzahn
Homework Helper
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I think because the height difference for the tube at the throat compared to the tube at the pipe inlet will have gravitational energy which will affect the flow in the system. The rp is supposed to be on the LHS and I bring it to the RHS to include the gravitational effects due to height difference.
Where is (the line of) the center of mass of the flow at the entrance (position ##p##) and where is it at the most narrow cross section (position ##t##)?

Edit: In the drawing it is marked very well...
 
  • #5
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Where is (the line of) the center of mass of the flow at the entrance (position ##p##) and where is it at the most narrow cross section (position ##t##)?

Edit: In the drawing it is marked very well...
Thanks! However, in reality, it is supposed to include the gravitational effects right? This is because the tubes used to measure the pressure are not placed in the center!
 
  • #6
stockzahn
Homework Helper
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Thanks! However, in reality, it is supposed to include the gravitational effects right? This is because the tubes used to measure the pressure are not placed in the center!
If you want to take into account the pressure distribution over the cross section - yes. But since you treat the problem as 1-dimensional to solve the Bernoulli equation you don't take it into account. I suppose you could use the difference of the fluid height to correct the pressures at the small pipes connected with the piston:

$$p_p + \rho g 2 r_p - \left(p_t+\rho g 2 r_t\right)= p_p-p_t+\rho g 2 \left(r_p-r_t\right)=\left( r_{out}^2-r_{in}^2\right)\pi F_{Piston}=A_{Piston} F_{Piston} $$

This would lead to the same problem you had (a equation you have to solve numerically). However neglecting the hydrostatic head with the given geometry results in a mistake of about 10 % (a little less).

Edit: If there is no flow the difference in hydrostatic head exists as well, therefore it cancels out before and during the measurement. So maybe the problem statement should say: a force of 180 N compared to the fluid at rest.
 
  • #7
254
5
If you want to take into account the pressure distribution over the cross section - yes. But since you treat the problem as 1-dimensional to solve the Bernoulli equation you don't take it into account. I suppose you could use the difference of the fluid height to correct the pressures at the small pipes connected with the piston:

$$p_p + \rho g 2 r_p - \left(p_t+\rho g 2 r_t\right)= p_p-p_t+\rho g 2 \left(r_p-r_t\right)=\left( r_{out}^2-r_{in}^2\right)\pi F_{Piston}=A_{Piston} F_{Piston} $$

This would lead to the same problem you had (a equation you have to solve numerically). However neglecting the hydrostatic head with the given geometry results in a mistake of about 10 % (a little less).

Edit: If there is no flow the difference in hydrostatic head exists as well, therefore it cancels out before and during the measurement. So maybe the problem statement should say: a force of 180 N compared to the fluid at rest.
Thanks a lot!
 
  • #8
19,918
4,094
If you want to take into account the pressure distribution over the cross section - yes. But since you treat the problem as 1-dimensional to solve the Bernoulli equation you don't take it into account. I suppose you could use the difference of the fluid height to correct the pressures at the small pipes connected with the piston:

$$p_p + \rho g 2 r_p - \left(p_t+\rho g 2 r_t\right)= p_p-p_t+\rho g 2 \left(r_p-r_t\right)=\left( r_{out}^2-r_{in}^2\right)\pi F_{Piston}=A_{Piston} F_{Piston} $$

This would lead to the same problem you had (a equation you have to solve numerically). However neglecting the hydrostatic head with the given geometry results in a mistake of about 10 % (a little less).

Edit: If there is no flow the difference in hydrostatic head exists as well, therefore it cancels out before and during the measurement. So maybe the problem statement should say: a force of 180 N compared to the fluid at rest.
The system is hydroststic in the vertical direction, so the vertical pressure variations cancel out.
 
  • #9
stockzahn
Homework Helper
Gold Member
498
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The system is hydroststic in the vertical direction, so the vertical pressure variations cancel out.
I think that's what I tried to say with my last sentence:

Edit: If there is no flow the difference in hydrostatic head exists as well, therefore it cancels out before and during the measurement. So maybe the problem statement should say: a force of 180 N compared to the fluid at rest.
Or do you mean something else and I did get it wrong?
 
  • #10
19,918
4,094
I didn't quite follow, so I thought I would clarify a little more.
 

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