Answer:
Amount of fuel used per year is supposed to be 34150 KJ/kg
Answer:
A) 0.03382 kg/s
B) 7.0372 Kw
C) 4.3982
D) 0.7396 kw
Explanation:
Given data:
Evaporator at 60 C
Space temperature = 25 C
power consumed by compressor = 1.6 kw
T1( evaporator temperature ) = 12°C
attached below is the detailed solution
Answer:
Vab = 80V
Explanation:
The only current flowing in the circuit is supplied by the 100 V source. Its only load is the 40+60 ohm series circuit attached, so the current in that loop is (100V)/(40+60Ω) = 1A. That means V1 = (1A)(60Ω) = 60V.
Vab will be the sum of voltages around the right-side "loop" between terminals 'a' and 'b'. It is (working clockwise from terminal 'b') ...
Vab = -10V +60V +(0A×10Ω) +30V
Vab = 80V
Answer:
ΔT= 11.94 °C
Explanation:
Given that
mass of water = 10 kh
Time t= 15 min
Heat lot from water = 400 KJ
Heat input to the water = 1 KW
Heat input the water= 1 x 15 x 60
=900 KJ
By heat balancing
Heat supply - heat rejected = Heat gain by water
As we know that heat capacity of water


Now by putting the values
900 - 400 = 10 x 4.187 x ΔT
So rise in temperature of water ΔT= 11.94 °C