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%
Answer:
Flush the surface with clean water
Explanation:
An example can be demonstrated when cleaning an aircraft's transparent plastic windshield. The first step is to flush it with clean water to help remove all grit and sand that might be on the surface. Once the surface is free of any sand and grit capable of scratching the plastic surface, it can then be washed and then rinsed.
Answer:
The road will narrow
P.S. please add more information when asking a question
The examples of engineering controls is Biohazard waste containers and Spill clean up kits.
What is engineering controls?
An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.
An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.
Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.
Therefore, the Option C and D is correct.
