How can this be solved when a piston diameters are D1=10cm and D2=4cm. When the pressure chamber 2 is 2000kPa and the pressure chamber 3is 700kPa, what is the pressure in chamber 1, in kPa ? Figure P1-55E, http://web.cecs.pdx.edu/~derekt/Classes/ME%20321%20Spring%202011/Hmwk1.pdf
1 Pa = 1 N/m^2 First you want to change your units into m and Pa so, D1 = 0.1 m; D2 = 0.04 m P2 = 2,000,000 Pa; P3 = 700,000 Pa Now you want to find the area of the cross sections, so A1 = pi*(0.1/2)^2 A2 = pi*(0.04/2)^2 Now to find P1, we want to do force balance, so we need to find the forces, F1 = P1*A1 (up) F2 = P2*A2 (down) F3 = P3*(A1-A2) (down) F1 = F2 + F3 --> P1*A1 = P2*A2 + P3*(A1-A2) Now solve for P1, and your answer is in Pa so transform back into kPa.
yh, A= pi *D^2/4 ?
Area = pi*r^2; radius = diameter/2
P1 (down is the atmospheric pressure), P2 (up) and P3 (middle) ? How is F1=F2+F3 ? http://web.cecs.pdx.edu/~derekt/Classes/ME%20321%20Spring%202011/Hmwk1.pdf a picture of the question is in Fig P1-55, p.42, q1-57
Up and down are completely relative, but as long as you are consistent that should be ok. Here I'm defining up as the the top of the page. For your question about F1 = F2 + F3, this is derived from the fact that all forces acting on the piston must sum to zero. Since F1 (caused by P1) is opposing F2 and F3 (caused by P2 and P3), you get 0 = F1 - F2 - F3 --> F1 = F2 + F3.
P1 is below the piston so it wants to push it up, while P2 and P3 are on top of the piston and want to push it down.
how is F3 = P3*(A1-A2) ?
P3 only effects the area the piston not covered by the smaller shaft. Thats why we have to subtract the area P3 doesn't interact with (A1 - A2). You can think about it as a larger circle and we subtract the area of the smaller circle to get the area of the rim.
yh, P1*A1=7135.787N, therefore, P1=908kPa
awesome, thanx
np
wh@ uni do you attend ?
Texas A&M
kwl
I'm from England
Cool, are you in college?
yh, Coventry University, studying mecha
awesome, nice to meet you!
u2
Correct answe is 900KPa
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