A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and Blie in t
1 answer:
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Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
T₁ = 300 K
= 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + (T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
= 1400×
= 1007.6 K
T₅ = T₄ + ( - T₄)/
= 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + (T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
= 1400×
= 1007.6 K
T₇ = T₆ + ( - T₆)/
= 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b.
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Answer:
0.16 micron per day
Explanation:
Given:
The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m
Initial tensile stress, σ₁ = 120 MPa
Final stress = 30 MPa
now from Griffith's equation, we have
where,
Gc and E are the material constants
now,
for the initial stage
........{1}
and for the final case
............{2}
on dividing 1 by 2, we get
or
a₂ = 4² × 0.1 × 10⁻⁶ m
or
a₂ = 1.6 micron
Now,
the change from 0.1 micron to 1.6 micron took place in 10 days
therefore, the rate at which the crack is growing =
or
average rate of change of crack = 0.16 micron per day