A coil having a resistance of 10 ohms and an inductance of 4 H is switched across a 20W dc source. Calculate (a) time required b
1 answer:
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Answer:
a) 4160 V
b) 12 kW and 81 kVAR
c) 54 kW and 477 kVAR
Explanation:
1) The phase voltage is given as:
The complex power S is given as:
The line current I is given as:
The phase voltage at the sending end is:
The magnitude of the line voltage at the source end of the line (
b) The Total real and reactive power loss in the line is:
The real power loss is 12000 W = 12 kW
The reactive power loss is 81000 kVAR = 81 kVAR
c) The sending power is:
The Real power delivered by the supply = 54000 W = 54 kW
The Reactive power delivered by the supply = 477000 VAR = 477 kVAR
Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
