Q#20.1
A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle.
(a) How much heat must be supplied to the engine in each cycle?
(b) What is the thermal efficiency of the engine?
Answer:
For a heat engine,
W = $|Q_H|-|Q_C|$; e = $\frac{W}{Q_H}$; $Q_H$ > 0, $Q_C$ < 0.
Give, W = 2200 J and $|Q_C|$ = 4300 J
(a) $|Q_H|= W+|Q_C|$
$|Q_H|= 2200 J + 4300 J = 6500 J
(b) the thermal efficiency of the engine is
e = $\frac{2200 \ J}{6500 \ J}$ = 0.34
e = 34%
Since the engine operates on a cycle, the net Q equal the net W. But to calculate the efficiency we use the heat energy input, $Q_H$.
Q#20.2
An aircraft engine takes in 9000 J of heat and discards 6400 J each cycle.
(a) What is the mechanical work output of the engine during one cycle?
(b) What is the thermal efficiency of the engine?
Answer:
For a heat engine,
W = $|Q_H|-|Q_C|$; e = $\frac{W}{Q_H}$; $Q_H$ > 0, $Q_C$ < 0.
Give, $Q_H$ = 9000 J and $|Q_C|$ = 6400 J
(a) W = $|Q_H|-|Q_C|$
W = 9000 J - 6400 J = 2600 J
(b) the thermal efficiency of the engine is
e = $\frac{2600 \ J}{9000 \ J}$ = 0.29
e = 29%
Since the engine operates on a cycle, the net Q equal the net W. But to calculate the efficiency we use the heat energy input, $Q_H$.
Q#20.3 . A Gasoline Engine. A gasoline engine takes in 1.61 $\times 10^4 \ J$ of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 $\times 10^4 \ J/s$.
(a) What is the thermal efficiency?
(b) How much heat is discarded in each cycle?
(c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?
Answer:
The problem deals with a heat engine
Given, W = +3700 J and $Q_H$ = +16,100 J
Use Eq. e = $\frac{W}{Q_H}$ to calculate the efficiency e and
W = $|Q_H|-|Q_C|$ to calculate $Q_C$.
Power = $\frac{W}{t}$
(a) the thermal efficiency is
e = $\frac{3700 \ J}{16,100 \ J}$ = 0.23
e = 23%
(b) $|Q_C|=|Q_H|-W$
Heat discarded is
$|Q_C|=16,100 J - 3700 J = 12,400 J
(c) $Q_H$ is supplied by burning fuel; $Q_H$ = m$L_c$ where $L_c$ is the heat of combustion.
$m = \frac{Q_H}{L_c}=\frac{16,100 \ J}{4.60 \times 10^4 \ J/s}$
m = 0.350 g
(d) ) W = 3700 J per cycle
In t = 1.00 s. the engine goes through 60.0 cycles.
P = W/t = 60.0(3700 J)/1.00 s = 222 kW or
P = (2.22 $\times 10^5 \ W$)(1 hp/746 W)
P = 298 hp
$Q_C$ = −12,400 J. In one cycle $Q_{tot} = Q_C + Q_H$ = +3700 J. This equals $W_{tot}$ for one cycle.
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