b) Find the mass flow rate at the exit in kg/s. m kg/s 0.01 kg/s) exit 3) (10% e
ID: 2996645 • Letter: B
Question
b) Find the mass flow rate at the exit in kg/s. m kg/s 0.01 kg/s) exit 3) (10% each) 0.02 kmolls of gaseous Hydrogen, H2(g), is combusted with theoretical air at 25 oC in an experimental motor. Assume complete combustion and that the exhaust is at 1.0 bar. Find the following: a) Air fuel ratio on a mass basis AF 34.5L (+0.1) °C 0.5 °C) b) The dew point of the exhaust Tdp 4) (10% each) Consider a gaseous mixture that is 70% N2 and 30% CO2 by volume. Assuming an ideal gas: a) Find the mass fraction of the N mfN2 0.1 b) Find the gas constant for the mixture: kJ/(kg K) 0.1 5) (10%) Estimate the boiling point temperature of ethanol at 200 kPa. 370.3 °C 1 °C) bp (10%) A carbon tax has been proposed for all CO2 producing fuels. Assume thatExplanation / Answer
Saturation Properties at P = 800 kPa are
Tsat = 170.4 deg
hf = 721.1 kJ/kg
hg = 2769 kJ/kg
SEPARATE STREM
T = 30 deg
Since Tsat > T, The state is compressed liquid
Therefore Poperties at P = 800 kPa and T = 30 deg are
h = 126.5 kJ/kg
STEAM
Properties of steam at P = 800 kPa and T = 200 Deg are
h = 2839 kJ/kg
Let
1 indicate steam inlet
2 indicate separate liquid stream
3 indicate the exit
we have
h1 = 2839 kJ/kg
m1 = 0.5 kg/s
h2 = 126.5
h3 = hf @ P= 800 kPa = 721.1 kJ/kg
m2 = ?
m3 = ?
From mass balance
m1 + m2 = m3 ----(1)
From Energy balance
m1*h1 + m2*h2 = m3*h3
0.5*2839 + m2*126.5 = m3*721.1 -----(2)
Substituting 1 in 2
we get
0.5*2839 + m2*126.5 = m1*721.1 + m2*721.1
0.5*2839 + m2*126.5 = 0.5*721.1 + m2*721.1
594.6*m2 = 1058.95
m2 = 1.78 kg/s
m3 = m1 + m2 = 2.28 kg/s
Exit mass flow rate is 2.28 kg/s
3)
H2 + 1/2*(O2+3.76*N2) = H2O + 1.88 N2
Moles of H2 = 0.02 kmols
Since the combustion is completed
Moles of O2 = 0.01 kmoles
Moles of Air = 0.0476 kmoles
Mass of air = 0.0476*28.9*1000 = 1375.64
AF Ratio = Mass of air/100 = 13.76
The exhaust Pressure is 1 bar. The exhaust gas is H2O
Dew point temperature is the saturation temperature at the exhaust pressure
Therefore
The saturation temp at P = 1 bar is
Tsat = 99.62 C
Tdp = 99.62 C
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