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:
The settlement that is expected is 1.043 meters.
Explanation:
Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil
The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula
where
'H' is the initial depth of the layer
is the Compression index
is the inital void ratio
is the initial effective stress at the depth
is the change in the effective stress at the given depth
Applying the given values we get
Answer:
int()
Explanation:
float() is using decimals, so that can't be it, like float(input( "how much does this cost?"))
print() is used to print something, not a user asking, like print("hello")
string() means like a whole, like string( I am good)
By elimination, int() is correct.
Hope this helps!
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
