The displacement volume of an internal combustion engine is 2.2 liters. The processes within each cylinder of the engine are mod
eled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by p1 = 1.6 bar, T1 = 325 K and V1 = 2.36 liters.
1) If the cycle is executed 2100 times per min, determine:
a) the compression ratio.
b) the net work per cycle, in kJ.
c) the maximum temperature, in K.
d) the power developed by the engine, in kW.
e) and the thermal efficiency.
1) If the cycle is executed 2100 times per min, determine:
a) the compression ratio.
b) the net work per cycle, in kJ.
c) the maximum temperature, in K.
d) the power developed by the engine, in kW.
e) and the thermal efficiency.
1 answer:
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Answer:
ordinary bulb total cost is $39.54
fluorescent bulb total cost is $13.05
amount save = 39.54 - 13.05 = $26.49
resistance = 626.1 ohm
Explanation:
in the 1st part
bulb on time = 3 year = 4380 hours
life of bulb = 750 h
so number of bulb required =
number of bulb required = 6
cost of 6 bulb is = 6 × 0.75 = $4.5
so
cost of operation is = 100 × 4380 ×
cost of operation = $35.04
so total cost will be = $4.5 + $35.04 = $39.54
and
when compare with florescent bulb
time = 3 year = 4380 h
life of bulb = 10000 h
so number of bulb required =
number of bulb required = 0.43 = 1
cost of 6 bulb is = 1 × 5 = $5
so
cost of operation is = 23 × 4380 ×
cost of operation = $8.05
so total cost will be = $5 + $8.05 = $13.05
in part 2nd
total amount save while compare bulb is
amount save = 39.54 - 13.05 = $26.49
and in part 3rd
resistance of bulb is
resistance =
resistance =
resistance = 626.1 ohm