Chapter 20
Page 1
CHAPTER 20 – Second Law of Thermodynamics; Heat Engines
1.
For the heat input, we have
Q
H
=
Q
L
+
W
= 8500 J + 2700 J = 11,200 J.
We find the efficiency from
e
=
W
/
Q
H
= (2700 J)/(11,200 J) = 0.24 =
24%
.
2.
We find the efficiency from
e
=
W
/
Q
H
= (8200 J)/(18.0 kcal)(4186 J/kcal) = 0.109 =
10.9%
.
3.
We find the rate of heat input from the efficiency:
e
=
W
/
Q
H
= (
W
/
t
)/(
Q
H
/
t
);
0.38 = (500 MW)/(
Q
H
/
t
), which gives
Q
H
/
t
= 1316 MW.
We find the rate of heat discharge from
Q
H
/
t
= (
Q
L
/
t
) + (
W
/
t
);
1316 MW = (
Q
L
/
t
) + 500 MW, which gives
Q
L
/
t
=
816 MW
.
4.
(
a
) We find the rate at which work is done from
P
=
W
/
t
= (180 J/cycle/cyl)(4 cyl)(25 cycles/s) =
1.8
×
10
4
J/s
.
(
b
) We find the heat input rate from
e
=
W
/
Q
H
= (
W
/
t
)/(
Q
H
/
t
);
0.25 = (1.8
×
10
4
J/s)/(
Q
H
/
t
), which gives
Q
H
/
t
=
7.2
×
10
4
J/s
.
(
c
)
We find the time to use one gallon from
t
=
E
/(
Q
H
/
t
) = (130
×
10
6
J/gal)(1 gal)/(7.2
×
10
4
J/s) = 1.81
×
10
3
s =
30 min
.
5.
We use the units to help us find the rate of heat input to the engine from the burning of the gasoline:
Q
H
/
t
= [(3.0
×
10
4
kcal/gal)/(38 km/gal)](90 km/h)(4186 J/kcal) = 2.97
×
10
8
J/h.
The horsepower is the work done by the engine.
We find the efficiency from
e
=
W
/
Q
H
= (20 hp)(746 W/hp)(3600 s/h)/(2.97
×
10
8
J/h) = 0.18 =
18%
.
6.
(
a
) Work is positive for an expansion.
Because the
work done is
represented by the area under the
PV
curve,
⏐
W
bc
⏐
>
⏐
W
ac
⏐
.
The net work is the sum of the works done in each leg.
For a
positive net work,
W
bc
> 0, so we have an expansion from b to c,
and the path must be
clockwise.
(
b
) We find the ratio of volumes from the ideal gas equation:
(
P
c
/
P
a
)/(
V
c
/
V
a
) =
T
c
/
T
a
;
(1)(
V
c
/
V
a
) =
T
c
/
T
a
, so
V
c
/
V
a
=
V
c
/
V
b
=
T
c
/
T
a
=
T
b
/
T
a
.
The work done during the isothermal expansion is
W
bc
=
nRT
b
ln(
V
c
/
V
b
) =
nRT
b
ln(
T
b
/
T
a
)
= (1.0 mol)(8.315 J/mol
·
K)(423 K) ln(423 K/273 K) = 1.54
×
10
3
J.
The work done during the constant pressure compression is
W
ca
=
P
a
(
V
a
–
V
c
) =
P
a
V
a
[1 – (
V
c
/
V
a
)] =
nRT
a
[1 – (
V
c
/
V
a
)]
= (1.0 mol)(8.315 J/mol
·
K)(273 K)[1 – (423 K/273 K)] = – 1.25
×
10
3
J.
For the isothermal expansion ?
U
bc
= 0, so
Q
bc
=
W
bc
= 1.54
×
10
3
J.
For the constant volume compression
W
ab
= 0, so
Q
ab
= ?
U
ab
=
nC
V
(
T
b
–
T
a
) = (1.0 mol)
*
(8.315 J/mol
·
K)(423 K – 273 K) = 1.87
×
10
3
J.
The efficiency is
e
=
W
net
/
Q
added
= (
W
bc
+
W
ca
)/(
Q
bc
+
Q
ab
)
= (1.54
×
10
3
J – 1.25
×
10
3
J)/(1.54
×
10
3
J + 1.87
×
10
3
J) = 0.085 =
8.5%
.
P
V
b
a
0
c
T
b
=
T
c

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