TFS practice problem of the week...

[Slay the P.E.]

Happy Thursday, all. Here is your problem for the week:

A venturi meter is installed in a 2-in diameter, schedule 40 steel pipe (ID = 2.067 in). The fluid in the pipe is ambient temperature No. 4 fuel oil with a specific gravity SG=0.85. The meter is provided with a U-tube manometer. From past experience with this meter, it is known that for a manometer fluid height h of 8 inches, the oil flow rate is 400 gallons per minute. The average fluid velocity (ft/min) in the pipe just upstream of the meter when the height h is 4 inches is most nearly:

(A) 27

(B) 283

(C) 1,225

(D) 1,623

 

[Vel2018]

Happy Thursday, all. Here is your problem for the week:

A venturi meter is installed in a 2-in diameter, schedule 40 steel pipe (ID = 2.067 in). The fluid in the pipe is ambient temperature No. 4 fuel oil with a specific gravity SG=0.85. The meter is provided with a U-tube manometer. From past experience with this meter, it is known that for a manometer fluid height h of 8 inches, the oil flow rate is 400 gallons per minute. The average fluid velocity (ft/min) in the pipe just upstream of the meter when the height h is 4 inches is most nearly:

(A) 27

(B) 283

(C) 1,225

(D) 1,623

View attachment 10957

I got D.

 

[Slay the P.E.]

I got D.

Correct.

 

[mongolianbbq]

This one is confusing me. You aren't given the density of the manometer fluid or the dimensions of the narrow section... are you supposed to use some sort of similarity ratio? Even if you do and find the flow rate for h=4 that way, you dont have A2. Am I going the right direction?

 

[Vel2018]

This one is confusing me. You aren't given the density of the manometer fluid or the dimensions of the narrow section... are you supposed to use some sort of similarity ratio? Even if you do and find the flow rate for h=4 that way, you dont have A2. Am I going the right direction?

As the question said, it used to measure 8in of h at 400GPM. That should tell you something, try again youll get it.

 

[mckenz007]

This one is confusing me. You aren't given the density of the manometer fluid or the dimensions of the narrow section...

Doesn’t say but if you assume the manometer fluid is mercury it works out!

 

[Vel2018]

Doesn’t say but if you assume the manometer fluid is mercury it works out!

Probably, but if you don't know the answer yet, its gonna hunt you for a while jajaja

Solution is very short xD

 

[MikeGlass1969]

See equation 18.41 in MERM....  

 

[Slay the P.E.]

See equation 18.41 in MERM....  

@mckenz007 It will work out independently of the density of the manometer fluid.

Look at the equation for flow through a meter. You can see that it simply states that flow rate is proportional to the square root of h. Right? In other words

Q = K  (h)^(1/2)​

where K is some constant (Side note and interesting potential exam-type question: What are the units of this constant K?)

We know what Q is when h=8... what is it when h=4? Use the above form of the equation to eliminate K and solve for Q when h=4 

 

[Vel2018]

What I did was use the basic equation. 

using v^2/2g=h(rhofluid-rhoOil)/rhoOil there could be a coefficient of flow based on the Q equation of venturi with manometer, but that doesn't matter since everything will turn to a constant.

So there you get the K if you move h=8in on the left side of the equation. Then use that K with an h of 4in to get v2. 

K does not have a unit. 

 

[Vel2018]

@mckenz007 It will work out independently of the density of the manometer fluid.

Look at the equation for flow through a meter. You can see that it simply states that flow rate is proportional to the square root of h. Right? In other words

Q = K  (h)^(1/2)​

where K is some constant (Side note and interesting potential exam-type question: What are the units of this constant K?)

We know what Q is when h=8... what is it when h=4? Use the above form of the equation to eliminate K and solve for Q when h=4 

Would that be really a potential question? Since this problem can be solved in many ways. 

If I use the standard Q=CfAo(2gh(rho2-rho1)/rho1)^1/2 I get the unit of K to be ft^5/s^2

But if I do it with velocity I get K without a unit. It's basically the same, one using velocity and one using Q. 

 

[Slay the P.E.]

K does not have a unit. 

The units of K has to be something like gpm/(in^(1/2)) for the dimensional homogeneity of the expression Q = K * sqrt(h)

In this case in fact K = [400 gpm / sqrt(8 in)] therefore K = 141.4 gpm/(in^(1/2))

 

[Slay the P.E.]

Would that be really a potential question? Since this problem can be solved in many ways. 

If I use the standard Q=CfAo(2gh(rho2-rho1)/rho1)^1/2 I get the unit of K to be ft^5/s^2

But if I do it with velocity I get K without a unit. It's basically the same, one using velocity and one using Q. 

Well, but the question I posed was about the constant of proportionality between flow rate and square root of manometer height. By the way you might have a typo because the units would be ft^(5/2)/s

The constant of proportionality between upstream velocity and square root of manometer height has different units, but it is also not dimensionless.

 

[Slay the P.E.]

This discussion about units reminded me of this problem we wrote and briefly considered for including it in our practice exam:

~~~~~~~~~~~~~~~~~~~~~~



In fluid mechanics, the Weber number for a fluid is a dimensionless parameter, defined as:

We =  ( ρ V^2 L)/σ​

where ρ is the fluid’s density, V is a velocity, L is a characteristic length, and σ is the surface tension. Since the Weber number is dimensionless, the units for surface tension in the SI system must be:

(A) kg/(m·s) 

(B) N/m

(C) m/N

(D) s^2/kg

~~~~~~~~~~~~~~~~~~
 

 

[MikeGlass1969]

(B)

 

[mongolianbbq]

@mckenz007 It will work out independently of the density of the manometer fluid.

Look at the equation for flow through a meter. You can see that it simply states that flow rate is proportional to the square root of h. Right? In other words

Q = K  (h)^(1/2)​

where K is some constant (Side note and interesting potential exam-type question: What are the units of this constant K?)

We know what Q is when h=8... what is it when h=4? Use the above form of the equation to eliminate K and solve for Q when h=4 

This is what I did and I got Q, but I got stuck trying to figure out the area of the throat in order to solve for velocity. Unless I read the question wrong and "just upstream of the meter" means the area before the diameter reduction? I guess the meter starts at the larger diameter since thats where the manometer is introduced? If I do that I get 1673 ft/m.

Units of K would be gpm / in^2  (which could be reduced further to in^(5/2)/minute)

 

[mongolianbbq]

I agree with Mike for the weber problem. Even though I looked up surface tension to be force/length in about 30 seconds, I also broke up the problem into units replacing a kg*m/s^2 with N along the way and came up with N/m.

 

[Slay the P.E.]

I agree with Mike for the weber problem. Even though I looked up surface tension to be force/length in about 30 seconds, I also broke up the problem into units replacing a kg*m/s^2 with N along the way and came up with N/m.

That’s the flaw in this problem, I think. One can just look up surface tension and notice it is force per unit length. I think for future editions I would say “and σ is a material property” instead of “and σ is the surface tension”

 

[Slay the P.E.]

Happy Thursday, TFS peeps.

Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 squared meters. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity.

The mass flow rate of the air (kg/min) is nearest:
(A)  79
(B)  245
(C)  2230
(D)  4728

 

[mckenz007]

D?