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'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 =...
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...
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...