Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
<em></em>
<em>attached below is the detailed solution </em>
Answer:
a)COP=5.01
b)
KW
c)COP=6.01
d)
Explanation:
Given
= -12°C,
=40°C
For refrigeration
We know that Carnot cycle is an ideal cycle that have all reversible process.
So COP of refrigeration is given as follows
,T in Kelvin.

a)COP=5.01
Given that refrigeration effect= 15 KW
We know that 
RE is the refrigeration effect
So
5.01=
b)
KW
For heat pump
So COP of heat pump is given as follows
,T in Kelvin.

c)COP=6.01
In heat pump
Heat rejection at high temperature=heat absorb at low temperature+work in put

Given that
KW
We know that 


d)
Answer:
What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should never cross each other they should never have only one measurement value
Answer:
Mechanical power of pump is 74.07%.
Explanation:
Power of motor = 15 KW
Efficiency of motor= 90%
So the actual power(P) supplied by motor = 0.9 x 15 KW
P=13.5 KW
Water flow rate = 50 L/s
Volume flow rate = 50 L/s
We know that

So

We know that pump is an open system and work input for open system can be calculated as
W=VΔP
ΔP is the pressure difference
V is the volume flow rate
So by putting the values
W=0.05 (300-100) (here ΔP=300 - 100=200 KPa)
W=10 KW
So mechanical power of pump
η =0.7407
Mechanical power of pump is 74.07%.