Answer:
,
Explanation:
The drag force is equal to:
Where is the drag coefficient and
is the frontal area, respectively. The work loss due to drag forces is:
The reduction on amount of fuel is associated with the reduction in work loss:
Where and
are the original and the reduced frontal areas, respectively.
The change is work loss in a year is:
The change in chemical energy from gasoline is:
The changes in gasoline consumption is:
Lastly, the money saved is:
Answer:
The source temperature is 1248 R.
Explanation:
Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.
Given:
Temperature of the heat sink is 520 R.
Second law efficiency is 60%.
Actual thermal efficiency is 35%.
Calculation:
Step1
Reversible efficiency is calculated as follows:
Step2
Source temperature is calculated as follows:
T = 1248 R.
The heat engine is shown below:
Thus, the source temperature is 1248 R.
Answer:
Amount of air left in the cylinder=m=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2
Initial temperature=T=20 C
Final Volume==0.1
Using gas equation
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0