Answer:
a) 229.4281 hp.
b) 262.15 ft3/min.
Explanation:
Given data:
P1 = 14.2 psi
T1 = 60°F = 520° R
P2 = 120 psi
T2 = 500°F = 960° R
volumetric flow rate ( Av1 ) = 1200 ft^3 /min = 20 ft^3 / sec
attached below is the detailed solution
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Neither of the two technicians (Technician A and Technician B) is correct.
<h3>What is an
engine vacuum?</h3>
An engine vacuum can be defined as a type of engine which is designed and developed to derive its force from air pressure that's being pushed against one side of the piston of an automobile, while having a partial vacuum on the other side.
In this scenario, we can infer and logically conclude that neither of the two technicians (Technician A and Technician B) is correct because engine vacuum is high when the engine is operating under light loads and vice-versa.
Read more on engine vacuum here: brainly.com/question/14602340
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It would be 72cm bc u need to add up all the line in the back to
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:
![\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%280.3%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Ccdot%20%281.20%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%29%5Ccdot%20%2827.778%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%5Ccdot%20%5B%281.85%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%20-%20%281.50%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%5D%5Ccdot%20%2825%5Ctimes%2010%5E%7B6%7D%5C%2Cm%29)


The change in chemical energy from gasoline is:



The changes in gasoline consumption is:





Lastly, the money saved is:

