1. Aug 21, 2007

### P111ltl

1. The problem statement, all variables and given/known data
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. Relevant 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 im 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 doesnt 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.

2. Aug 23, 2007

### Andrew Mason

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. Aug 23, 2007

### P111ltl

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. Aug 23, 2007

### Andrew Mason

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. Aug 23, 2007

### P111ltl

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 theres 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. Aug 24, 2007

### Andrew Mason

Bernoulli's equation gives is the relationship between speed of a fluid and its pressure.

AM

7. Aug 24, 2007

### P111ltl

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. Aug 30, 2007

### a13x

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. Aug 30, 2007

### Andrew Mason

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:

$$\frac{1}{2}\rho v_{inside}^2 + P_{inside} + \rho gh_{inside} = \frac{1}{2}\rho v_{outside}^2 + P_{outside} + \rho gh _{outside}[/itex] But because the air outside is stationary and there is no change in vertical height: [tex]\frac{1}{2}\rho v_{inside}^2 = P_{outside} - P_{inside} = \Delta P[/itex] The pressure difference is 100 pa (98 mm water). Using the ideal gas law: $\rho = P/RT$ so $\rho = 10^5/(287*293) = 1.19 \tex{Kg/m}^3$ [tex]v = \sqrt{2\Delta P/\rho} = \sqrt{2*100/1.19} =$$ 13 m/sec.

AM

10. Aug 30, 2007

### a13x

Can definately see where I went wrong with this. Didn't assume the pressure was the outside pressure.

Thanks for the help. made things seem alot simpler.