2.(10 pts)A proposed engine cycle employs an ideal gas and consists of the following sequence of transformations; a) Isothermal
compression at 300 o K from a pressure of 1bar to a pressure of 30bar b) Constant pressure heating to a temperature of 1600 o K. c) Isothermal expansion at 1600 o K to the original pressure of 1 bar. d) Constant pressure cooling to a temperature of 300 o K to complete the cycle An ideal regenerator connects d) to b) so that the heat given up in d) is used for the heating in b). For an engine using a kilamole of gas find the net work in kJ and the thermal efficiency. You may assume C p
1 answer:
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Assumptions:
- Steady state.
- Air as working fluid.
- Ideal gas.
- Reversible process.
- Ideal Otto Cycle.
Explanation:
Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):
- Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
- Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).
- Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
- Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.
- Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
- Exhaust 1-0: the working fluid is vented to the atmosphere.
If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:
where:
Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.
Answer:
See image attached.
Answer:
a) 23.551 hp
b) 516.89 hp
Explanation:
<u>given:</u>
<u>required:</u>
the power in hp
<u>solution:</u>
.............(1)
by substituting in the equation (1)
=353.27 lbf
..........(2)
by substituting in the equation (2)
= 2769.29 lbf
power is defined by
P=F.V
353.27*36.67
=12954.411 lbf.ft/s
=12954.411*.001818
=23.551 hp
2769.29*102.67
= 284323 lbf.ft/s
= 284323*.001818
= 516.89 hp