I just solved for the second plugging in values I got w = -[ -93.6 * 10^3 + ((180^2)/2 - (120^2)/2) + ((60 * 9.81) - 0) - 300 * 10^3] which is equal to 384.0114 kJ/kg? Tell me where I've got it wrong if there are any errors. Thanks
This is the question "12kg of a fluid per minute goes through a reversible steady flow process. The properties of fluid of fluid at the inlet are p1 = 1.4bar, ρ1 = 25kg/m^3, C1 = 120m/s and u2 = 920 kJ/kg and the exit are p2 = 5.6bar, ρ2 = 5kg/m^3, C2 = 180m/s and u2 = 720kJ/kg. During the...
Apologies for my late reply. I've been working on the question and ran into S.I unit issues and I got -93.6kJ/kg for the first one.
The second one on the other hand, using the relation W = - [(u2 + P2V2) - (u1 + P1V1) + ((C2)2/2) - ((C1)2/2) + (g.Z2 - g.Z1 ) - Q]
Correct me if I'm wrong but...
OF COURSE, it is. I should take my time with formulas, thank you. Can I run the answer with you once I retry and i'd also try the second part again and see if I can figure that part out?
Homework Statement
12kg of a fluid per minute goes through a reversible steady flow process. The properties of fluid at the inlet are p1 = 1.4bar, ρ1 = 25kg/m3, C1= 120m/s and u1= 920kJ/kg and at the exit are p2= 5.6bar, ρ
2= 5 kg/m3, C2= 180m/s and u2
Homework Equations
u1 + P1V1 + (C1)2/2 +...
E = stress / strain
stress = F / A
Area = pi*(diameter)^2/4
strain= elongation/original length
making d subject of formula = sqrt(4*F*L/(E*pi*elongation))
plugging in values after conversion to S.I units gave 58617.25 * 10^-9m which is same as 0.586 * 10^-4m
Thanks, was wondering why it was there, I tried another go at the problem, by plugging in values and got (0.586 * 10^-4)m. Don't know how correct that is?
Homework Statement
A cylindrical rod of copper (E = 110GPa) having a yield strength of 240MPa is to be subjected to a load of 6660N. If the length of the rod is 380mm, what must be the diameter to allow an elongation of 0.5mm.
Homework Equations
E = stress/ strain
stress = Force/Area ; Area =...