Thermofluids U tube manometer wind tunnel question

In summary, the conversation involves calculating the air speed in a wind tunnel based on the deflection of a water manometer, using Bernoulli's equation and the ideal gas law. The pressure difference between inside and outside the tunnel is 100 pa, leading to a calculated air speed of 13 m/s. The conversation also explores various theories and equations to understand the relationship between speed and pressure in a moving fluid.
  • #1
P111ltl
10
0
1. Homework Statement
Air at 20 degrees celsius and 100kpa is drawn into a wind tunnel. If the deflection h of a water manometer that is connected to a hole in the wall of the test section to measure the gauge pressure in the flow is 98mm. Calculate air speed in test section in m/s ignore fluid friction .
Take R of air to be 287 j/kgK and density of water to be 1000 kg/m3 and g= 9.81

part b) if the air speed is 45 m/s calculate the deflection h in the manometer.


2. Homework Equations
I have been having trouble with this for ages now and am really stuck have been looking at Pressure= density xg x height giving 961.38 Pa then using pv=nrt v=87.469 using this method unsure if I am barking up the wrong tree?

for part b) again using p=nRT/v = 1868.68 Pa then using P=densityxgxdelta h giving 190mm or 0.19m however this doesn't sound right to me as the 45 m/s air speed is causing a greater deflection that the previously worked out air speed so i am sure something is not quite right but what? I am converting 20 degrees to 293 K and using that any suggestion?


If anyone needs a drawing i will attempt one in paint.

Thanks very much in advance
 
Physics news on Phys.org
  • #2
P111ltl said:
1. Homework Statement
Air at 20 degrees celsius and 100kpa is drawn into a wind tunnel. If the deflection h of a water manometer that is connected to a hole in the wall of the test section to measure the gauge pressure in the flow is 98mm. Calculate air speed in test section in m/s ignore fluid friction .
Take R of air to be 287 j/kgK and density of water to be 1000 kg/m3 and g= 9.81

part b) if the air speed is 45 m/s calculate the deflection h in the manometer.


2. Homework Equations
I have been having trouble with this for ages now and am really stuck have been looking at Pressure= density xg x height giving 961.38 Pa then using pv=nrt v=87.469 using this method unsure if I am barking up the wrong tree?

for part b) again using p=nRT/v = 1868.68 Pa then using P=densityxgxdelta h giving 190mm or 0.19m however this doesn't sound right to me as the 45 m/s air speed is causing a greater deflection that the previously worked out air speed so i am sure something is not quite right but what? I am converting 20 degrees to 293 K and using that any suggestion?
This appears to be a Bernoulli principle problem. What is the relationship between change in speed and change in pressure of a moving fluid?

AM
 
  • #3
static pressure+ dynamic pressure = total pressure
Ps+ (densityx V2/2)= total pressure so therefore as the speed increases the total pressure increases, How does this apply to a deflection in a manometer?
 
  • #4
P111ltl said:
static pressure+ dynamic pressure = total pressure
Ps+ (densityx V2/2)= total pressure so therefore as the speed increases the total pressure increases, How does this apply to a deflection in a manometer?
The manometer reading measures the pressure difference between the two ends. One end is in the tunnel and the other is open to the atmosphere outside the tunnel. That pressure difference is enough to support a 98mm column of water.

AM
 
  • #5
I understand now about the manometer gives an indication as to the pressure in the test section.
Are any of my initial theries correct or do i need to rethink them? I take it there's a pressure drop and the speed increasees in the test area creating the deflection h. How do i go about working out the velocity from the values given in the Q stated initially? .
Thanks
 
  • #6
P111ltl said:
I understand now about the manometer gives an indication as to the pressure in the test section.
Are any of my initial theries correct or do i need to rethink them? I take it there's a pressure drop and the speed increasees in the test area creating the deflection h. How do i go about working out the velocity from the values given in the Q stated initially? .
Bernoulli's equation gives is the relationship between speed of a fluid and its pressure.

AM
 
  • #7
How do i work out the pressure knowing that 98 mm is the deflection? I take it i then use this value of pressure in bernoullis equation to work out the velocity .Then for b rearrange this equation for the finding of the deflection with the value of 40 for the velocity.?
thanks
 
  • #8
For this I got:

1/2 u^2+P/ρ+gh=constant

the constant being wind tunnel pressure of 100Kpa.

Ended up with u = 3.963 m/s
 
  • #9
a13x said:
For this I got:

1/2 u^2+P/ρ+gh=constant

the constant being wind tunnel pressure of 100Kpa.

Ended up with u = 3.963 m/s
I get a different figure.

The wind tunnel pressure not 100Kpa. That is the pressure of the air before it is brought into the tunnel, ie it is the pressure outside the tunnel. From Bernouilli's equation:

[tex]\frac{1}{2}\rho v_{inside}^2 + P_{inside} + \rho gh_{inside} = \frac{1}{2}\rho v_{outside}^2 + P_{outside} + \rho gh _{outside}[/itex]

But because the air outside is stationary and there is no change in vertical height:

[tex]\frac{1}{2}\rho v_{inside}^2 = P_{outside} - P_{inside} = \Delta P[/itex]

The pressure difference is 100 pa (98 mm water).

Using the ideal gas law: [itex]\rho = P/RT[/itex] so [itex]\rho = 10^5/(287*293) = 1.19 \tex{Kg/m}^3[/itex]

[tex]v = \sqrt{2\Delta P/\rho} = \sqrt{2*100/1.19} = [/tex] 13 m/sec.

AM
 
  • #10
Can definitely see where I went wrong with this. Didn't assume the pressure was the outside pressure.

Thanks for the help. made things seem a lot simpler.
 

Related to Thermofluids U tube manometer wind tunnel question

1. What is a U tube manometer and how is it used in a wind tunnel?

A U tube manometer is a device used to measure the difference in pressure between two points in a fluid. In a wind tunnel, it is typically used to measure the difference in pressure between the stagnation point (where the air flow is at its highest speed) and the static point (where the air flow is at rest). This measurement can then be used to calculate the airspeed in the wind tunnel.

2. How does a U tube manometer work?

A U tube manometer works by using the principle of hydrostatics, which states that the pressure at any point in a fluid at rest is equal in all directions. The U-shaped tube is filled with a liquid, typically water or mercury, and connected to the points where pressure is being measured. The difference in the levels of the liquid in the two arms of the tube shows the difference in pressure between the two points.

3. What is the significance of the height difference between the two arms of a U tube manometer?

The height difference between the two arms of a U tube manometer is directly proportional to the pressure difference between the two points being measured. This allows for a more accurate measurement of the pressure difference, as well as the airspeed in the wind tunnel.

4. How is the airspeed calculated using a U tube manometer in a wind tunnel?

The airspeed can be calculated using the Bernoulli's principle, which states that the total energy of a fluid remains constant along a streamline. Using the pressure difference measured by the U tube manometer and the known density of the fluid, the airspeed in the wind tunnel can be calculated using the equation v = √(2Δp/ρ), where v is the airspeed, Δp is the pressure difference, and ρ is the density of the fluid.

5. Are there any limitations to using a U tube manometer in a wind tunnel?

While a U tube manometer is a simple and effective way to measure pressure differences in a wind tunnel, it does have some limitations. The liquid used in the manometer must have a high enough density to provide an accurate measurement, and the height difference between the two arms of the tube must not be too large, as it can cause instability in the measurement. Additionally, the manometer must be calibrated and maintained regularly for accurate readings.

Similar threads

  • Advanced Physics Homework Help
Replies
7
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Mechanical Engineering
Replies
4
Views
3K
  • Mechanical Engineering
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
4K
  • Advanced Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
20K
Back
Top