The position of a particle is given by s = 0.27t
1 answer:
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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
Answer:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 × × coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 × × 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in
Answer:
(a) the rate of heat transfer to the coolant is Q = 139.71W
(b) the surface temperature of the shaft T = 40.97°C
(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW
Explanation:
See explanation in the attached files
