Rank These Engines On The Basis Of The Work They Perform Per Cycle.

You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.

Forums
Homework Help
Introductory Physics Homework Help

Internal-Combustion Engine Prototypes Ranking Task?

Thread starter cupcake
Start date Nov 11, 2010

Homework Statement

Rank these engines on the basis of their designed power output and their thermal efficiency.

1) Qin = 3000 J, Qout = 1500 J, f=150 Hz.
2) Qin = 4000 J, Qout = 3000 J, f=50 Hz.
3) Qin = 1000 J, Qout = 200 J, f=200 Hz.
4) Qin = 2000 J, Qout = 500 J, f=50 Hz.
5) Qin = 2000 J, Qout = 1000 J, f=100 Hz.
6) Qin = 1500 J, Qout = 500 J, f=200 Hz.

Homework Equations

P=Work/t, and Work=Qin-Qout

The Attempt at a Solution

I calculated the power output using the above formulas, where t=1/f, but I don't know why my answers are wrong... this is mastering physics question :(

Answers and Replies

Homework Statement

Rank these engines on the basis of their designed power output and their thermal efficiency.

1) Qin = 3000 J, Qout = 1500 J, f=150 Hz.
2) Qin = 4000 J, Qout = 3000 J, f=50 Hz.
3) Qin = 1000 J, Qout = 200 J, f=200 Hz.
4) Qin = 2000 J, Qout = 500 J, f=50 Hz.
5) Qin = 2000 J, Qout = 1000 J, f=100 Hz.
6) Qin = 1500 J, Qout = 500 J, f=200 Hz.

Homework Equations

P=Work/t, and Work=Qin-Qout

The Attempt at a Solution

I calculated the power output using the above formulas, where t=1/f, but I don't know why my answers are wrong... this is mastering physics question :(

Perhaps you could show us your answers. What is your equation for thermal efficiency? Are you taking that into account?

so,

1) W1=1500 P1=W1/t= 225000
2) W2=1000 P2=W2/t= 50000
3) W3=800 P3=W3/t= 160000
4) W4=1500 P4= 75000
5) W5=1000 P5=100000
5) W6=1000 P6=200000

so, the rank from the largest to the smaller will be, P1 > P6 > P3 > P5 > P4 > P2

thermal efficiency formula is 1-Qin/Qout, that one is for the next question (arrange by basis of thermal efficiency)

so,
1) W1=1500 P1=W1/t= 225000
2) W2=1000 P2=W2/t= 50000
3) W3=800 P3=W3/t= 160000
4) W4=1500 P4= 75000
5) W5=1000 P5=100000
5) W6=1000 P6=200000

so, the rank from the largest to the smaller will be, P1 > P6 > P3 > P5 > P4 > P2

thermal efficiency formula is 1-Qin/Qout, that one is for the next question (arrange by basis of thermal efficiency)

Looks right. What is the problem?