Answer:
d = 2.69 mm
Explanation:
Assuming the cable is rated with a factor of safety of 1.
The stress on the cable is:
σ = P/A
Where
σ = normal stress
P: load
A: cross section
The section area of a circle is:
A = π/4 * d^2
Then:
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
Replacing:
0.106 inches = 2.69 mm
Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer:
A) W' = 15680 KW
B) W' = 17113.87 KW
Explanation:
We are given;
Temperature at state 1; T1 = 290 K
Temperature at state 3; T3 = 1100 K
Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw
Pressure of air into compressor; P_c = 95 kPa
Pressure of air into turbine; P_t = 760 kPa
A) The power assuming constant specific heats at room temperature is gotten from;
W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in
Now, we don't have T4 and T2 but they can be gotten from;
T4 = [T3 × (r_p)^((1 - k)/k)]
T2 = [T1 × (r_p)^((k - 1)/k)]
r_p = P_t/P_c
r_p = 760/95
r_p = 8
Also,k which is specific heat capacity of air has a constant value of 1.4
Thus;
Plugging in the relevant values, we have;
T4 = [(1100 × (8^((1 - 1.4)/1.4)]
T4 = 607.25 K
T2 = [290 × (8^((1.4 - 1)/1.4)]
T2 = 525.32 K
Thus;
W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000
W' = 0.448 × 35000
W' = 15680 KW
B) The power accounting for the variation of specific heats with temperature is given by;
W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in
From the table attached, we have the following;
At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K
At T3 = 1100 K, h3 = 1161.07 KJ/K
At T1 = 290 K, h1 = 290.16 KJ/K
At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K
Thus;
W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000
W' = 17113.87 KW
