What's the real power if Vrms = 100V, Irms = 2A, and the circuit has a power factor of 0.8?
1 answer:
You might be interested in
Answer:
a) 1
b) 1813.96 MJ/kmol
c) 32.43 MJ/kg , 1980.39 MJ/Kmol
Explanation:
molar mass of ethanol (C2H5OH) = 46 g/mol
molar mass of octane (C8H18) = 114 g/mol
therefore the moles of ethanol and octane
ethanol = 0.85 / 46
octane = 0.15 / 114
a) determine the molar air-fuel ratio and air-fuel ratio by mass
attached below
mass of air / mass of fuel = 12.17 / 1 = 12.17
b ) Determine the lower heating value
LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol
LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol
LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol
c) Determine higher heating value ( HHV )
HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol
HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol
HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg
HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol
Answer:
Amount of air left in the cylinder=m=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2
Initial temperature=T=20 C
Final Volume==0.1
Using gas equation
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0