Answer:
The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch
Explanation:
In this question, we are asked to calculate the expected settlement for the pole group given some information.
Please check attachment for complete solution and step by step explanation
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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:
hello your question lacks the required question attached below is the missing diagram
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
Explanation:
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
attached below is the detailed solution
Answer:
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