Answer: (a) 0,142 (b) 52.99 (c) 2.83 (d) 88.26
Explanation:
If the refrigarating capacity is 150kw
(a) the mass flow rate of refrigerant, in kilograms per second is 0.142
(b) the power input to the compressor, in kilowatts is 52.99
(c) the coefficient of performance is 2.83
(d) the isentropic compressor efficiency is 68.6 per cent
Answer:
there hope it can help.......
Answer:
Technician B only
Explanation:
During rotor reconditioning, which is the process also known as machining and sanding, where sanding is the involves the application of between 120 and 150 grit sandpaper while using a non-excessive force that is applied non-directionally for up to 60 seconds on each side such that the surface roughness meets OE standards. The rotors are then cleaned by washing after they are serviced before they can then be installed.
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as

Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that



Now the pressure of helium by using the Adiabatic pressure temperature is

Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that


Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
Answer:
Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.