Ah right. Saw my mistake, the cross terms should be a preceded with a minus sign. Thanks for the assistance once again nvn. Now its on to failure criterion!
Homework Statement
A long cylindrical boiler shell is 1.75 m in diameter and has a wall thickness of 12 mm. Treating the boiler as a thing shell, predict what internal pressures will produce yield in the shell according to the: (a) Rankine, (b) Max. Principal Strain, (c) Tresca and (d) Von...
That can be eliminated when doing the manometry equations by adding a ρgx term to both equations which then cancel out when you equate the pressures.
For instance:
Pleft = P300 mm + ρgx + ρgh, where h = 0.300 m.
Pright = P150 mm + ρgx + SGmercuryρgh
Equating the two eliminates the ρgx...
Okay. So then if I understand correctly, the pressure in the pipe is (denoted here with the subscript 5):
P5 = Pgauge + 0.600*ρg
P5 = 389886 Pa
Wikipedia, not something I rely on too heavily. Got burn't back in my 1st year of varsity when citing wikipedia as a reference.
@nvn: The pressure after the pump is given as 384 kPa = 384x103 Pa. Unless I have missed the boat and assumed too much.
Its a tutorial question. I used a specific gravity of 13.6 for mercury.
I'll just scan in my written work, its going to be much easier than using clunky BBCode and typing it out.
Okay.
The subscript 1 pertains to the 300 mm pipe after the pump and the subscript 2 pertains to the 150 mm pipe in the venturi.
v1A1 = v2A2
=> v1 = v2(D2/D1)2
Bernoulli:
H1 = H2
=> z1 + P1/γ + v12/2g = z2 + P2/γ + v22
=> z1 = z2 therefore they cancel.
=> 384 000/9810 +...
Q was computed by using Q = vA.
My method of approach was:
Find the pressure inside the smaller part of the venturi using manometry and the given information. I then used manometry to get the pressure at the left hand manometer. As the right hand tube is exposed to the atmosphere it eliminates...
Very well.
The flow rate I got was 0.158 m^3/s.
From this, the differential head I got was 42.16m.
The equation for pump power I used was:
Powerpump = ρgΔHQ
You're getting there. ;-)
1. Axial Force, Shear Force and Moment. (Torsion is a moment about the axis of a member)
2. Shear Stress yes, but normal not so much. The word is Direct Stresses.
3. Yep, but typically its represented as N/mm^2. Yes you do divide the axial force by the...
Oh I know how to deal with it, but you need to figure that one out. ;-)
Here are a couple of questions that you should ask yourself:
Q: What types of forces do you get? (Hint: There are 3, but 2 have the word "force" attached to them where the other doesn't.)
Q: What type of stresses do these...
That is purely your choice, just remember which one you made positive and which one you made negative. However, by convention tensile stresses are usually taken as +ve.
Just remember stress is taken as a vector (though this is not 100% correct) and therefore it has a magnitude and direction...
@nvn: That would be the standard equation, for example:
Q300 mm = Q150 mm
∴ v300 mmA300 mm = v150 mmA150 mm
∴ v300 mm = v150 mm(D150 mm/D300 mm)2
Edit:
Bah, right figured it out. Thank you.
I get a value of roughly 65.3 kW.