Answer:
a socket can be driven by a flexible head socket wrench, I believe
Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:

For the pipe 1, the flow velocity is:

Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m

The Reynold´s number is:


Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:

For the pipe 2, the flow velocity is:

The Reynold´s number is:


The head of pipe 1 is:

The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:

The required pumping power is:

Answer:
The maximum theoretical height that the pump can be placed above liquid level is 
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

(
stands here for density,
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:


This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:

We insert that into our last equation and get:

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
Answer:
b. 10A
Explanation:
Using the formula, E= k × r×I
200= 0.5 ×2000×0.02×I
200=20×I
Dividing with 20
I = 200/20= 10A
Answer:


Explanation:
= Area of section 1 = 
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = 
= Density of fluid = 
= Area of section 2 = 
Mass flow rate is given by

The mass flow rate through the pipe is 
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

The speed at section 2 is
.