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:
A
Explanation:
Due to ethanol's lower energy content, FFVs operating on E85 get roughly 15% to 27% fewer miles per gallon than when operating on regular gasoline, depending on the ethanol content.
Answer:
Option A
Chemical engineering
Explanation:
Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.
Answer:
Electrical faults are also caused due to human errors such as selecting improper rating of equipment or devices, forgetting metallic or electrical conducting parts after servicing or maintenance, switching the circuit while it is under servicing, etc.
Explanation:
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = 
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ = 
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
