The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×
m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is , the containers volume is V, waters mass is m.
= 150× /1
= 0.15 / Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 / Kg and 0.13 / Kg.
= 350+(400-350) 
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
= = 0.15 / Kg
In saturated water labels for = 40°C.
= 0.001008 / Kg
= 19.515 / Kg
The final state is two phase region < < .
In saturated water labels for = 40°C.
= 
= 7.3851 kPa
= 7.3851 kPa
Answer:
The elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2
Answer:
No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.
