Answer:
A. smallest wire is No. 12
54 volts
Ohms law. E= I x R
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:
Rubber block is not shown. I have attached an image of it.
Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
Answer:
Yes
Explanation:
Given Data
Temprature of source=750°c=1023k
Temprature of sink =0°c=273k
Work produced=3.3KW
Heat Rejected=4.4KW
Efficiency of heat engine(η)=
and
Heat Supplied 

η=
η=42.85%
Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink
η=1-
η=1-
η=73.31%
Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.