9. Technician A says to examine each bearing carefully for chips, pits, and scratches during servicing. Technician B says that d
2 answers:
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Answer:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = .................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q =
Q =
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W
Answer:
a)
b)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
