Recent content by Brinkley23

No, what worries me with that is, my tutors have not provided any material for me to learn adiabatic reversible expansion, so I can't see them wanting me to use it in my answer (I have emailed them). Im sure you are correct, but when I have an issue, they normally just reply "refer to your...

Ok, so I am back to square 1. Is the Ideal Gas Law what I should be using then still?

ok, so I've have completely started again, for part 1 I have now used Gay-Lussac's law. p1/t1 = p2/t2 Therefore t2 = t1p2/p1 t2= (323)(28000000) / 32000000 = 282.6k for part 2 I have rearranged the ideal gas law to solve for n. n=pV / RT P=28000000pa V=0.45m^3 R=8.314 T=282.6K dry air =...

units for R=8.314 JK^-1mol^-1

I believe I have used the correct units, my notes say that: p = pa v = m^3 n= moles T= K R=universal gas constant 8.314 original given pressure value given was 280bar, converted to 28000000pa original volume was 450L, converted to 0.45m^3 using these figures to work out temperature the value...

For part 1. I need to find the temperature, I rearranged the equation so that: T = pV/nR (28000000)(0.45) / (28.96)(8.314) This gives me a value of 52331.4 I am unsure where the units I have used in the equation are correct. I have used pa for pressure, or should I be using kpa? this would...

Thanks for replying Lnewqban, After spending some more time familiarising myself with the equations I seemed to understand it much better and I too came up with those final values, which confirms I must have done something right and corrected my initial mistakes. Thanks again!

Hi, I have made an attempt to answer the 4 following questions: Car mass 400kg accelerates from 20-200km/h over a distance of 120m. A resistive force of 1000N is to be assumed during acceleration. determine: a) the average acceleration from 20km/h to 200km/h b) time taken to accelerate from...

Hi guys see below thread (duplicated by mistake). Many Thanks

That makes sense...but I think I made an error with original workings out. My full revised workings are below: Tensile stress (direct force/area) 6928 / 2.01 x 10∧-4 = 34467661.7 Pa (34.47 MPa) Shear stress (shear force/area) 4000 / 2.01 x 10∧-4 = 19900498 Pa (19.90 MPa) TASK - "Determine...

Thanks for replying, Would I be calculating that by taking the Ultimate shear stress (300N) and the shear stress (199mPa)? Assuming yes, FOS 300/199 = 1.51 FOS tensile - 14.51 FOS shear - 1.51 It asks me to determine the factor of safety in operation?

Hi, I need to work out the FOS for a 16mm diameter bolt with a force of 8KN exerted on it. I have already worked out: direct stress = 34.47mPa shear stress = 199mPa Information given: Ultimate Tensile stress = 500NM/m2 Ultimate Shear stress = 300MN/m2 The FOS calculation I have been given to...