Which of the following is not a relationship set between elements in a sketch
1 answer:
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Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg
Answer:
(a) 65.04%
(b) 16.91%
Solution:
As per the question:
At inlet:
Pressure of the compressor, P = 140 kPa
Temperature, T = = 263 K
Isentropic work, W = 700 kPa
At outlet:
Pressure, P' = 700 kPa
Temperature, T' = = 333 K
Now, from the steam table;
At the inlet , at a P = 700 kPa, T =:
h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K
At outlet, at P = 140 kPa, T =:
h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K
Also in isentropic process, s = and
at 700kPa
(a) Isentropic efficiency of the compressor,
(b) The temperature of the environment, = 273 + 27 = 300 K
Availability at state 1,
Similarly for state 2,
Now, the efficiency of the compressor as per the second law;
= 67.57%