A particular heat engine has a mechanical power output of 4.50 kW and an efficiency of 23.0%. The engine expels 6.61 10

Post by **answerhappygod** » Mon May 09, 2022 5:36 pm

A particular heat engine has a mechanical power output of 4.50 kW and an efficiency of 23.0%. The engine expels 6.61 103 J of exhaust energy in each cycle.(a) Find the energy taken in during each cycle.(b) Find the time interval for each cycle.
A particular heat engine has a mechanical power output of 4.50 kW and an efficiency of 23.0%. The engine expels 6.61 103 J of exhaust energy in each cycle.(a) Find the energy taken in during each cycle.(b) Find the time interval for each cycle.
(a) Find the energy taken in during each cycle.(b) Find the time interval for each cycle.
Part 1 of 4 - ConceptualizeVisualize the input energy dividing up into work output and wasted output. The exhaust by heat must be the wasted part of the energy input, so the input will be on the order of 10000 J in each cycle. If each cycle took one second, the mechanical output would be around 3000 J each second. A cycle must take less than a second for the useful output to be 5000 J/s.
Part 1 of 4 - Conceptualize
Part 1 of 4 - Conceptualize
Visualize the input energy dividing up into work output and wasted output. The exhaust by heat must be the wasted part of the energy input, so the input will be on the order of 10000 J in each cycle. If each cycle took one second, the mechanical output would be around 3000 J each second. A cycle must take less than a second for the useful output to be 5000 J/s.
Visualize the input energy dividing up into work output and wasted output. The exhaust by heat must be the wasted part of the energy input, so the input will be on the order of 10000 J in each cycle. If each cycle took one second, the mechanical output would be around 3000 J each second. A cycle must take less than a second for the useful output to be 5000 J/s.
Part 2 of 4 - CategorizeWe will use the first law of thermodynamics and the definition of efficiency to solve for the input energy per cycle. We will find the time interval for each cycle from the definition of power.
Part 2 of 4 - Categorize
Part 2 of 4 - Categorize
We will use the first law of thermodynamics and the definition of efficiency to solve for the input energy per cycle. We will find the time interval for each cycle from the definition of power.
We will use the first law of thermodynamics and the definition of efficiency to solve for the input energy per cycle. We will find the time interval for each cycle from the definition of power.

: A Particular Heat Engine Has A Mechanical Power Output Of 4 50 Kw And An Efficiency Of 23 0 The Engine Expels 6 61 10 1 (20.99 KiB) Viewed 21 times

: A Particular Heat Engine Has A Mechanical Power Output Of 4 50 Kw And An Efficiency Of 23 0 The Engine Expels 6 61 10 2 (11.73 KiB) Viewed 21 times

: A Particular Heat Engine Has A Mechanical Power Output Of 4 50 Kw And An Efficiency Of 23 0 The Engine Expels 6 61 10 3 (12.94 KiB) Viewed 21 times

Tutorial Exercise A particular heat engine has a mechanical power output of 4.50 kW and an efficiency of 23.0%. The engine expels 6.61 * 10° 3 of exhaust energy in each cycle. (a) Find the energy taken in during each cycle. (b) Find the time interval for each cycle. Part 1 of 4 . Conceptualize Visualize the input energy dividing up into work output and wasted output. The exhaust by heat must be the wasted part of the energy input, so the input will be on the order of 10000) in each cycle. If each cycle took one second, the mechanical output would be around 3000 each second. A cycle must take less than a second for the useful output to be 5000/s. Part 2 of 4 . Categorize We will use the first law of thermodynamics and the definition of efficiency to solve for the input energy per cycle. We will find the time interval for each cycle from the definition of power.
Part 3 of 4 - Analyze (a) From the definition of efficiency, we have 11-10 Q 10,1 191 0.23 Solving for I, the energy taken in during each cycle, we have 1961 19,1 - 6610 6610 -0.23 0.23 = 8.6 8.58 kl. Screenshot
Part 4 of 4 - Analyze (b) From the definition of power, we have Weng At and solving for the time interval, gives Ar- Weng 6.61003 Your response differs from the correct answer by more than 100%. 3/ Submit Skip you cannot come back) Need Help?