A heat engine operating at steady state delivers a power output of 15.5 hp. The engine receives energy by heat transfer in the amount of Qh = 30.5 kJ during each cycle of operation from a high temperature thermal reservoir at Th = 940 K. The system is executing 50 cycles per minute. Determine the amount of work that is delivered during each cycle in kilojoules

w_cycle = 13.68 KJ / cycle

Explanation:

Given:

- Steady power out-put W_out = 15.5 hp

- System executes 50 cycles/min

Find:

Determine the amount of work that is delivered during each cycle in kilo-joules

Solution:

- First we will convert the steady output into KW as follows:

                                 W_out = 15.5 hp

The conversion factor is 1.36 KW per hp.

                                 W_out = 15.5 [hp] * [KW] / 1.36 [hp]

                                 W_out = 11.4 [KW] or [KJ/s]

- Given that there are 50 cycles in 60 sec. We will use direct proportionality to calculate the number of cycles per second:

                       50 cycles   > 60 s

                          x cycles   > 1 s

                                 x = 50/60 = 0.8333 cycles /s

- Next we will compute the amount of work done per cycle:

                                 w_cycle = W_out / x

                                 w_cycle = 11.4 [KJ/s] * [s] / 0.8333 [cycles]

                                 w_cycle = 13.68 KJ / cycle

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

momentum of soft ball three times less than baskey ball