Answer:
(b)False
Explanation:
We know that specific Internal energy of gas u=
T
and specific enthalpy of gas h=
T
If we take the case of air we know that
=0.707 KJ/Kg=K ,
=1.005 KJ/Kg=K
If we take A fixed temperature T=300 K
so u=212.1 KJ/ kg ,h=301.5 KJ/kg
So we can say that specific enthaply of gas is always greater than its specific internal energy.
Explanation:
Strengthening by grain size reduction
- It is based on the fact that dislocations will experience hindrances while trying to move from a grain into the next because of abrupt change in orientation of planes.
- Hindrances can be two types: forcible change of slip direction, and discontinuous slip plane.
- Smaller the grain size, often a dislocation encounters a hindrance. Yield strength of material will be increased.
- Yield strength is related to grain size (diameter, d ) as Hall Petch relation:

Strengthening by Grain size reduction (contd..)
- Grain size reduction improves not only strength, but also the toughness of many alloys.
- If d is average grain diameter,
is grain boundary area per unit volume,
is mean number of intercepts of grain boundaries per unit length of test line,
is number of grains per unit area on a polished surface:

- Grain size can also be measured by comparing the grains at a fixed magnification with standard grain size charts.
- Other method: Use of ASTM grain size number (Z). It is related to grain diameter, (in mm) as follows:

Solid solution strengthening
- Impure foreign atoms in a single phase material produces lattice strains which can anchor the dislocations.
- Effectiveness of this strengthening depends on two factors size difference and volume fraction of solute. Solute atoms interact with dislocations in many ways:
- elastic interaction
- modulus interaction
- stacking-fault interaction
- electrical interaction
- short-range order interaction
- long-range order interaction
Answer:
diameter= 1 in
Explanation:
Some authors have developed studies to find the recommended flow rate at the pump outlet, in order to avoid high pressure losses and cavitation.
Assuming that this pump carries water, the recommended speed is 2m / s, so knowing the flow rate = 20gpm we can find the diameter of the pipe.
Q=flow rate=20gpm=0.0012618m^3/s
V=speed=2m/s
A=Area
Q=VA
A=Q/V

Taking into account that the flow is through a circular pipe we can use the equation to find the area of a circle, and then find the diameter
![A=\frac{\pi D^2 }{4} \\D=\sqrt[2]{\frac{4A}{\pi } } =\sqrt{\frac{4(630.8)}{\pi } } =23.34mm](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20D%5E2%20%7D%7B4%7D%20%5C%5CD%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B4A%7D%7B%5Cpi%20%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B4%28630.8%29%7D%7B%5Cpi%20%7D%20%7D%20%3D23.34mm)
the result is 23.34mm, if we find the closest commercial diameter to this value, we find that it is 25.4mm = 1in
Answer:
The force will be "125 N".
Explanation:
The given values are:



As we know,
⇒ 
then,
⇒ 
⇒ 
Since,
The pressure on both the pistons are equal, then
⇒ 
or,
⇒ 
By substituting the values, we get
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 