Answer:
Depends on the battery and the current type.
Is it AC or DC?
Explanation:
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Answer:

Explanation:
From the information given:
Life requirement = 40 kh = 40 
Speed (N) = 520 rev/min
Reliability goal
= 0.9
Radial load
= 2600 lbf
To find C10 value by using the formula:

where;


The Weibull parameters include:



∴
Using the above formula:


![C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}](https://tex.z-dn.net/?f=C_%7B10%7D%20%3D%203640%20%5Ctimes%20%5Cbigg%5B%5Cdfrac%7B1248%7D%7B0.9933481582%7D%5Cbigg%5D%5E%7B%5Cdfrac%7B3%7D%7B10%7D%7D)

Recall that:
1 kN = 225 lbf
∴


Find the solution in the attachments
Note: Question was incomplete, so the complete question is added in the attachments.
Answer:
the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW
the isentropic pump efficiency is 78%
Explanation:
Given that;
m = 1.2 kg/sec
T = 50 degree Celsius { Vf = 0.001012 m^3/kg}
P1 = 1.5 Mpa
P2 = 15 Mpa
W-actual = 21 kw
W reversible = m*Vf (p2 - p1)
= 1.2 * 0.001012 * ( 15*10^3 - 1.5*10^3)
= 1.2 * 0.001012 * 13500
= 16.39 kW
the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW
Isentropic Pump efficiency = W-reversible / W-actual
= 16.39 / 21 = 0.78
= 78%
the isentropic pump efficiency is 78%