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sladkih [1.3K]
3 years ago
6

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

st temperature = 430°C, 7.0 × 109 cal of fuel produces 1.4 × 1010 J of work in one hour.(a) What is the actual efficiency of this engine?(b) What is the Carnot efficiency of the engine?(c) What is the power output, in hp, of this engine?
Physics
1 answer:
suter [353]3 years ago
5 0

Answer:

actual efficiency is  47.78 %

Carnot efficiency  is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency  and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy  = 7.0 × 10^9 cal  (4.184 J/1 cal)  = 29.29 × 10^9 J

actual efficiency  =  1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency  =  0.4778

actual efficiency is  47.78 %

and

Carnot efficiency  is = 1 - ( 703 / 2173 )

so Carnot efficiency  is  = 0.67648

Carnot efficiency  is 67.65 %

and

power output  = work / time

power output  =  1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
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Answer:

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Explanation:

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x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

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g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

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t=1.597s

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Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

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V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
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