Answer: An electric power system is a network of electrical components deployed to supply, transfer, and use electric power.
Explanation:
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:
the correct answer is option B. W
Answer:
a) the log mean temperature difference (Approx. 64.5 deg C)
b) the rate of heat addition into the oil.
The above have been solved for in the below workings
Explanation:
Answer:
Cost = $2527.2 per month.
Explanation:
Given that
Discharge ,Q = 130 L/min
So

Cost = $0.45 per cubic meter
1 month = 30 days
1 days = 24 hr = 24 x 60 min
1 month = 30 x 24 x 60 min
1 month = 43,200 min
Lets x
x = 0.13 x 43,200

So the total cost = 5616 x 045 $
Cost = $2527.2 per month.