Answer:
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Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using variable spec
ific heats, determine (a) the isentropic efficiency, (b) the exit velocity, and (c) the entropy generation.
1 answer:
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Answer:
F₁ = 1500 N
F₂ = 750 N
= 500 N
Explanation:
Given :
Power transmission, P = 7.5 kW
= 7.5 x 1000 W
= 7500 W
Belt velocity, V = 10 m/s
F₁ = 2 F₂
Now we know from power transmission equation
P = ( F₁ - F₂ ) x V
7500 = ( F₁ - F₂ ) x 10
750 = F₁ - F₂
750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )
∴F₂ = 750 N
Now F₁ = 2 F₂
F₁ = 2 x F₂
F₁ = 2 x 750
F₁ = 1500 N , this is the maximum force.
Therefore we know,
= 3 x
where is centrifugal force
=
/ 3
= 1500 / 3
= 500 N
Answer:
It will be equivalent to 338.95 N-m
Explanation:
We have to convert 250 lb-ft to N-m
We know that 1 lb = 4.45 N
So foe converting from lb to N we have to multiply with 4.45
So 250 lb = 250×4.45 =125 N
And we know that 1 feet = 0.3048 meter
Now we have to convert 250 lb-ft to N-m
So
So 250 lb-ft = 338.95 N-m