Answer:
W = - 184.8 kW
Explanation:
Given data:
We know that work is done as
for
density of air is 1.22 kg/m^3 and
for
W = -[14 + 1.708[400-300]]
W = - 184.8 kW
Answer:
Area required = 9.5 ft²
Explanation:
Step by step explanation is given in the attached document.
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑ = mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
Explanation:
Given that:
From process 1 → 2
Process 2 → 3
The volume is constant i.e
Process 3 → 1
P = constant i.e the compression from state 1
Now, to start with 1 → 2
The work-done for the process 1 → 2 through adiabatic expansion is:
We know that 1 bar =
∴
For process 2 → 3
Since V is constant
Thus:
W = PΔV = 0
For process 3 → 1
W = PΔV
The net work-done now for the entire system is :
The sketch of the processes on p -V coordinates can be found in the image attached below.
Answer:
Tech A is correct
Explanation:
Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.
Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.