Answer:
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Explanation:
Load 1: Active power P_1 = 20 HP = 14.91kW;
Reactive power 

Load 2: Active power P_2 = 20 kW;
Reactive power Q2 = 0 since the load is purely resistive.
Load 3: Active power
= 0 due to purely capacitiveload
Reactive power
= -20 Var
a) since all three loads are connected in parallel therefore
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar
Since Q = 0, the power factor is unity.
Supply current per phase is given by


Answer:
12 N-m
Explanation:
The dc motor is operating at 24 V that is its terminal voltage V =24 V
Armature resistance = 0.2 ohm
No load speed = 240 radian /sec
For motor we know that as the motor is on no load so so
Power developed in the motor
Now we know that power = torque× angular speed
So
Answer:
A) i) 984.32 sec
ii) 272.497° C
B) It has an advantage
C) attached below
Explanation:
Given data :
P = 2700 Kg/m^3
c = 950 J/kg*k
k = 240 W/m*K
Temp at which gas enters the storage unit = 300° C
Ti ( initial temp of sphere ) = 25°C
convection heat transfer coefficient ( h ) = 75 W/m^2*k
<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>
First step determine the Biot Number
characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125
Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3
Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method
attached below is a detailed solution of the given problem
<u>B) The physical properties are copper</u>
Pcu = 8900kg/m^3)
Cp.cu = 380 J/kg.k
It has an advantage over Aluminum
C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>
Given that:
P = 2200 Kg/m^3
c = 840 J/kg*k
k = 1.4 W/m*K
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)