Answer:
There are 6 types of pressure control valves and their function is to regulate the pressure below a threshold level within safe limits and to maintain and control pressure of a particular circuit.
Explanation:
The six type of Pressure valve with their functions are given below:
a. Unloading Valve:
These type of pressure valve are used to pour fluid into the container at very low or no pressure.
b. Safety valve:
These are used when the pressure within the vessel is in excess as inside temperature is near about preset [point point then these valves are open to release the extra pressure and are closed once normal conditions are regained.
c. Pressure Reducing Valve:
These are basically used for the control of the pressure in downstream not exceeding the design limits.
d. Pressure Relief Valves:
These are basically used to limit and regulate the pressure of any system.
e. Counter Balance Valve:
These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.
f. Sequence Valve:
These are used to maintain sequence or order in the operations of two parts or branches.
Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Answer:
The differences are listed below
Explanation:
The differences between consolidation and compaction are as follows:
In compaction the mechanical pressure is used to compress the soil. In consolidation, there is an application of stead pressure.
In compaction, there is a dynamic load by rapid mechanical methods like tamping, rolling, etc. In consolidation, there is static and sustained pressure applied for a long time.
In compaction, the soil volume is reduced by removing air from the void. In consolidation, the soil volume is reduced by squeezing out water from the pores.
Compaction is used for sandy soil, consolidation on the other hand, is used for clay soil.
