Peoples Physics Book Version-2

http://www.ck12.org Chapter 19. Thermodynamics and Heat Engines Version 2

Some Important Points

• In a practical heat engine, the change in internal energy must be zero over a complete cycle. Therefore, over
  a complete cycleW=∆Q.


• The work done by a gas during a portion of a cycle =P∆V, note∆Vcan be positive or negative.

Gas Heat Engines

• When gas pressure-forces are used to move an object then work is done on the object by the expanding gas.
  Work can be done on the gas in order to compress it.


• If you plot pressure on the vertical axis and volume on the horizontal axis (seeP−Vdiagrams in the last
  chapter), the work done in any complete cycle is the area enclosed by the graph. For a partial process, work is
  the area underneath the curve, orP∆V.

Question:A heat engine operates at a temperature of 650K. The work output is used to drive a pile driver, which is
a machine that picks things up and drops them. Heat is then exhausted into the atmosphere, which has a temperature
of 300K.

a) What is the ideal efficiency of this engine?

b.) The engine drives a 1200kg weight by lifting it 50min 2. 5 sec. What is the engine’s power output?

c) If the engine is operating at 50% of ideal efficiency, how much power is being consumed?

d) The fuel the engine uses is rated at 2. 7 × 106 J/kg. How many kg of fuel are used in one hour?

Answer:

a) We will plug the known values into the formula to get the ideal efficiency.

η= 1 −

Tcold

T

hot=1-300K650K=54%

b) To find the power of the engine, we will use the power equation and plug in the known values.

P=

W

t

=

F d

t

=

mad

t

=

1200kg× 9 .8m/s^2 ×50m

2 .5sec

=240kW

c) First, we know that it is operating at 50% of ideal efficiency. We also know that the max efficiency of this engine
is 54%. So the engine is actually operating at

. 5 ×54%=27%

of 100% efficiency. So 240kW is 27% of what?

. 27 x=240kW⇒x=
240kW
. 27

=890kW