Answer:
We know that all petrol engines are works on Otto cycle.Otto cycle have four process out of four two are constant volume process and others two are isentropic processes.
There are lots of limitations for practical Otto cycle these are as follows
1.In practical cycle heat can not add at constant volume.
2.In practical cycle there is a gap between combustion of two set of fuel.
3.Lots of heat is dissipated by cylinders.
4.Valve opening and closing is not a sudden process it requires some time.
5.There is a limitations of cylinder material ,it means that temperature of cycle can not rise after a specified limit of material.
Due to these above limitations practical cycles have low efficiency as compare to ideal cycle.
Answer:
for a) F= 744.97 N
for a) F= 167.85 N
for a) F= 764.57 N
Explanation:
the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.
Then from steam tables
T= 120°C → P required=Pr= 198.67 kPa
then the pressure developed by the piston is
P = (m*g + F)/A
where m= mass of the piston ,g= gravity F= force required and A= area of the piston
then
Pr = P = (m*g + F)/A
F = Pr*A-m*g
since A= π/4*D²
F =π/4* Pr*D²-m*g
replacing values
F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²
F= 744.97 N
b) for T₂=80°C → Pr₂=47.41 kPa
F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²
F₂= 167.85 N
c) for m=0 (mass of the piston neglected) ,the force required is
F₃ =π/4*Pr*D² = π/4*198.67 *10³Pa*(0.07m)²= 764.57 N
F₃ =764.57 N
Answer:
C. Signal-to-quantization-noise ratio
Explanation:
Answer:
Explanation:
Given
Required
Plot a steam and leaf display for the given data
Start by categorizing the data by their tenth values:
The 0.3's is will be plotted as thus:
The 0.4's is as follows:
The 0.5's is as follows:
The 0.6's is as thus:
Lastly, the 0.7's is as thus:
The combined steam and leaf plot is:
