Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Answer:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
Answer:
89375 N
Explanation:
Rearrange the formula for normal stress for F:


Convert given values to base units:
275 MPa =
Pa
325
= 0.000325 
Substituting in given values:
F =
N
The first one and the second i’m glad to help