Recent content by fayan77

Thank you

the only ones I could combine are the ones at connection B. However, that 14 Kip force only act on member AB correct? If so then from member BCD I use ##\Sigma##MD = 0: 10(5)-By(10)-Bx(12) = 0 50 = 10By+12Bx from member BCD 168 = 12Bx-9By from member AB solving for Bx and By...

Ax Ay Bx By Dx Dy 6 unknowns

But then I'm still left with 3 equations 4 unknowns. member AB x positive to the right y positive up moment positive counter clockwise. ##\Sigma##Fx=0: Ax+14-Bx = 0 ##\Sigma##Fy=0: Ay-By = 0 ##\Sigma##MA=0: -14(12) + Bx(12) - By(9) = 0 same with other member

Homework Statement Homework Equations Using dimensional plane in the vertical y and horizontal x. ##\Sigma##Fx = 0 ##\Sigma##Fy = 0 ##\Sigma##Mz = 0 The Attempt at a Solution I have detached the 2 members at B and analyze them both separately. The 14 kip force is throwing me off though...

OOOOOHHHH I see, problem states energy head for section 1 as a whole is 80 ft I was fixated that 80ft was y1... got it Thank you. On another note whenever I am solving a problem with a two reservoir system we assume velocity heads are zero because of their size the dh/dt is basically 0...

y2 = 12 P2/##\gamma## = 18*144/62.4 =41.5 velocity head @ section 2 = (Q/A)2/64.4 = .14 solving for hL = 80 - 12 - 41.5 - .14 = 26.36 ft why is P1 = 0? and why is velocity head @ section 1 = 0?

So my assumptions are correct? velocity heads are the same and that pressure at the top = 0? And while I have you here, does Darcy's equation and Heinz-Williams yield the same loss? can I use them in combination to solve for a particular problem?

potential energy of a static column of water (height) + velocity head + pressure head... the equation I have written above

Im looking for energy loss in terms of feet not power loss. I have a lot of information Pressure heights flow rate length diameter no roughness what is throwing me off is that if I use general energy equation are my assumptions correct that P1=0 and that velocity heads are eqaual? If i use darcy...

Homework Statement Homework Equations y1 + P1/##\gamma##+v1/2g=y2 + P2/##\gamma##+v2/2g+hL The Attempt at a Solution I used the above equation and assumed that P1=0 and both velocity heads are equal. Is that valid? However I did not need to use the geometry of the pipe so I think I am doing...

Divide Bernoulli's equation by ##\gamma## and include head losses. Everything is in terms of head (length) ##\gamma## = ##\rho##g alpha is coefficient correction

ok, So I start off with Bernoulli's equation solving for P1. Because v1 = v2 the kinetic energy term will cancel out leaving us with P1 + ##\rho##gh1 = P2 + ##\rho##gh2 P1 +(1.94)(32.2)(80) = 2592 + (1.94)(32.3)(12) P1 = -1655.8 psf This does not make sense to me ( having a negative pressure at...

...That is the problem exercise. Do you want me to type it instead? I would like some direction so I can try it and post my work. Am I correct to use Bernoulli's equation to find P1? and will velocities be the same in both sections?

Homework Statement Homework Equations y1+P1/##\gamma##+##\alpha##v12/2g = y2+P2/##\gamma##+##\alpha##v22/2g + hL The Attempt at a Solution so it gives me pressure at p2 but not p1. I tried using Bernoulli's equation to find p1 i did plugged into energy equation above and got an answer but...