Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
A) 
B)
Explanation:
Given data:
P-1 = 100 lbf/in^2
degree f


effeciency = 80%
from steady flow enerfy equation

where h1 and h2 are inlet and exit enthalpy
for P1 = 100 lbf/in^2 and T1 = 500 degree F


for P1 = 40 lbf/in^2


exit enthalapy h_2


from above equation
[1 Btu/lbm = 25037 ft^2/s^2]

b) amount of entropy


at ![h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2](https://tex.z-dn.net/?f=h_2%20%3D%201197.77%20Btu%2Flbm%20%5B%5Ctex%5D%20%20and%20%5Btex%5DP_2%20%3D%2040%20lbf%2Fin%5E2)


Answer:
The inlet match number is 3.42.
Explanation:
Answer:
oof
Explanation:
I don't know but please don't report me
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