Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Answer:
Energy produce in one year =20.49 x 10¹⁶ J/year
Explanation:
Given that
Plant produce 6.50 × 10⁸ J/s of energy.
It produce 6.50 × 10⁸ J in 1 s.
We know that
1 year = 365 days
1 days = 24 hr
1 hr = 3600 s
1 year = 365 x 24 x 3600 s
1 year = 31536000 s
So energy produce in 1 year = 31536000 x 6.50 × 10⁸ J/year
Energy produce in one year = 204984 x 10¹² J/year
Energy produce in one year =20.49 x 10¹⁶ J/year
Given:
Temperature of water, 
= 
=273 +(-6) =267 K
Temperature surrounding refrigerator, 
= 
=273 + 21 =294 K
Specific heat given for water, 
= 4.19 KJ/kg/K
Specific heat given for ice, 
= 2.1 KJ/kg/K
Latent heat of fusion, 
= 335KJ/kg
Solution:
Coefficient of Performance (COP) for refrigerator is given by:
Max 
= 
= 
= 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max 
= 
= 10.89
