Answer:
Correct option a) True.
Explanation:
It is true since the Vickers hardness value refers to the force applied in a 136 ° diamond tip penetrator divided by the surface of the groove produced in the material, the lower the impression made on this greater the value will be end of the Vickers measurement and greater its hardness.
The equation to determine the Vickers hardness value will be:
Hv= ((1.854 × P)/(d²)) (kg/mm²)
Therefore a value of 220 Vickers refers to a harder material than another value of 180 Vickers.
15- T
16- F
17- T
I think this is correct
Answer:
a)
, b)
, c) 
Explanation:
a) The tank can be modelled by the Principle of Mass Conservation:

The mass flow rate exiting the tank is:



b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:


Properties of water are obtained from tables:



The specific enthalpy at outlet is:


c) After a quick interpolation from data availables on water tables, the final temperature is:

Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit