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:
The answers are as follow:
a) 10 mm
b) 12.730 N/
c) 127.307 N/
d) 0.25
Explanation:
d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9
Force = F =1000 N
let us find initial area first, A1 = pi* = 78.55
using reduction in area formula : 0.9 = (A1 - A2 ) / A1
solving it will give, A2 = 0.1 A1 = 7.855
a) The specimen elongation is final length - initial length
50 - 40 = 10 mm
b) Engineering stress uses the original area for all stress calculations,
Engineering stress = force / original area = F / A1 = 1000 / 78.55
Engineering stress = 12.730
c) True stress uses instantaneous area during stress calculations,
True fracture stress = force / final area = F / A2 = 1000 / 7.855
True Fracture stress = 127.30
e) strain = change in length / original length
strain = 10 / 40 = 0.25