2. What is the Function of the Camshaft in an Internal Combustion Engine?
1 answer:
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Answer:
heat transfer rate is -15.71 kW
Explanation:
given data
Initial pressure = 4 bar
Final pressure = 12 bar
volumetric flow rate = 4 m³ / min
work input to the compressor = 60 kJ per kg
solution
we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is
h1 = 262.96 kJ/kg
v1 = 0.05397 m³/kg
h2 = 310.24 kJ/kg
and here mass balance equation will be
m1 = m2
and mass flow equation is express as
m1 = .......................1
m1 =
m1 = 1.2353 kg/s
and here energy balance equation is express as
0 = Qcv - Wcv + m × [ ( h1-h2) + + g (z1-z2) ] ....................2
so here Qcv will be
Qcv = m × [ ] ......................3
put here value and we get
Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]
Qcv = -15.7130 kW
so here heat transfer rate is -15.71 kW
Answer:
Check image, right on PLATO
Answer:
A. 72.53 mol/min
B. 76.2%
Explanation:
Please check attachment to see the diagram of the components I drew.
I first converted flow rate to a molar flow rate. I got 11.5mol/ min after converting. Check attachment
A.) When we create a balance on all these components
n1 + n2 + n3
n1 = n2 + 11.5
Creating a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.5mol/min(1.00)
When we put the first equation into the second one:
(n2+11.5mol/min)0.18 + n2(0.05) + 11.5mol/min(1.00)
= 0.18n2 + 2.07min/mol = 0.5n2 + 11.5mol/min
Collect like terms
0.18n2 - 0.05n2 = 11.5 - 2.07min/mol
= 0.13n2 = 9.43
n2 = 9.43/0.13
n2 = 72.53 mol/min
72.53 is the flow rate of the gas stream leaving the burner.
B.) n1 = n2 + 11.5mol/min
n1 = 72.53mol/min + 11.5mol/min
n1 = 84.93mol/min
Amount of hexane entering condenser
n1c6H14 = y1C6H14 = 0.18(84.03mol/min)
= 15.1mol/min
The percentage condensed
= (11.5mol/min)/(15.1mol/min) * 100
= 76.2% of hexane is recovered as liquid
