When bending metal, the material on the outside of the curve stretches while the material on the inside of the curve compresses.
1 answer:
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Answer: (a) 36.18mm
(b) 23.52
Explanation: see attachment
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ
Answer:
.
Explanation:
Given that
L= 50 m
Pressure drop = 130 KPa
copper tube is 3/4 standard type K drawn tube.
From standard chart ,the dimension of 3/4 standard type K copper tube given as
Outside diameter=22.22 mm
Inside diameter=18.92 mm
Dynamic viscosity for kerosene
We know that
Where Q is volume flow rate
L is length of tube
is inner diameter of tube
ΔP is pressure drop
μ is dynamic viscosity
Now by putting the values
So flow rate is .