Recent content by Perodamh

Aah, that's awesome all the same, thanks a lot.

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

Thanks, i'd work towards an answer for the second question, but what do you think of the first, I got -93.6kJ/kg?

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 +...

Neat, thanks so much

I skipped the square root part so uhm 0.765 * 10^-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 =...

Okay, thanks

o my bad, the length is 2.5, just asked the lecturer again, for the area does that mean the length * width or still width * thickness

The Area is the width * thickness? I've crosschecked and that's how the question is.