Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin
etic and potential energy effects are negligible. Determine:
a. power input required [kJ/kg]
b. isentropic compressor efficiency
c. rate of entropy production per unit mass [kJ/kg K] in the compressor
a. power input required [kJ/kg]
b. isentropic compressor efficiency
c. rate of entropy production per unit mass [kJ/kg K] in the compressor
1 answer:
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Answer:
the net work per cycle Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
the mass in air ( lb) can be determined by using the formula:
where;
R = 53.3533 ft.lbf/lb.R°
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
At state of volume 2; the relative volume can be determined as:
The specific energy at
is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
To determine the relative volume at state 4; we have:
The specific energy at
is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
where ;
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp
Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
