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
Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
Answer:
<em><u>THE ANSWER IS: B</u></em>
Explanation:
I took the Unit test and the answer is B
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
.