Explain why the following acts lead to hazardous safety conditions when working with electrical equipmenta. Wearing metal ring o
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Answer:
See explaination
Explanation:
please see attachment for the detailed diagram used in solving the given problem.
It is attached as an attachment.
Answer:
I)E= 40.95 GPa
II)E=5.29 GPa
Explanation:
I)
Given that
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ = 72.5 GPa ,V₂=0.55
Longitudinal moduli given as ;
E= E₁V₁+E₂V₂
E= 2.41 x 0.45 + 72.5 x 0.55 GPa
E= 40.95 GPa
II)
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ =230 GPa ,V₂=0.55
Transverse moduli given as:
E=5.29 GPa
Answer:
POWER INPUT = 82.989 KW
Explanation:
For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg
For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg
Heat loss Q = 0.05w
Inlet diameter = 3 cm
exit diamter = 1.5 cm
volume of tank will be v = area * velocity
velocity at inlet
velocity at outlet
steady flow energy equation
solving wc = 1830.64 kJ/kg
wc in KWH
we know that