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Homework #06 Solutions Ch. 22: 1, 4, 7, 13, 18, 21, 23, 29, 33, 41, plus bonus problem 59 Chapter 22 P22.1 A heat engine takes in 360 J of energy from a hot reservoir and performs 25.0 J of work in each cycle. Find (a) the efficiency of the engine and (b) the energy expelled to the cold reservoir in each cycle.

SOLUTION: (a)

e

(b)

Qc

Weng Qh

Qh

25.0 J 360 J

Weng

0.069 4 or 6.94%

360 J 25.0 J

335 J

P22.4 A multicylinder gasoline engine in an airplane, operating at 2 500 rev/min, takes in energy 7.89 103 J and exhausts 4.58 103 J for each revolution of the crankshaft. (a) How many liters of fuel does it consume in 1.00 h of operation if the heat of combustion is 4.03 107 J/L? (b) What is the mechanical power output of the engine? Ignore friction and express the answer in horsepower. (c) What is the torque exerted by the crankshaft on the load? (d) What power must the exhaust and cooling system transfer out of the engine?

SOLUTION: (a)

The input energy each hour is 60 min 1.18 109 J h 1h 1L 29.4 L h 4.03 107 J

7.89 103 J revolution 2 500 rev min

implying fuel input 1.18 109 J h (b) have

Qh

Weng

Qc . For a continuous-transfer process we may divide by time to

Qh t Usefu l pow er ou tpu t

P P

(c)

Qc

t W eng

t

Q h Qc t t t 3 7.89 10 J 4.58 10 3 J 2 500 rev 1 m in revolu tion revolu tion 1 m in 60 s

1.38 10 5 W

P

eng

eng

Qc

(d)

eng

W eng

t

1 hp 746 W

1.38 10 5 J s 2 500 rev 60 s

4.58 103 J 2 500 rev revolution 60 s

1.38 10 5 W

185 hp

1 rev 2 rad

527 N m

1.91 105 W

P22.7 A refrigerator has a coefficient of performance of 3.00. The ice tray compartment is at –20.0°C, and the room temperature is 22.0°C. The refrigerator can convert 30.0 g of water at 22.0°C to 30.0 g of ice at –20.0°C each minute. What input power is required? Give your answer in watts.

SOLUTION: COP 3.00

Qc . Therefore, W W

Qc . 3.00

The heat removed each minute is QC t

0.030 0 kg 4 186 J kg C 22.0 C 0.030 0 kg 2 090 J kg C 20.0 C

or,

Qc t

0.030 0 kg 3.33 105 J kg 1.40 104 J min

233 J s

Thus, the work done per second is

P

233 J s 3.00

77.8 W .

P22.13 An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at 250°C, and the isothermal compression takes place at 50.0°C. The gas takes in 1 200 J of energy

from the hot reservoir during the isothermal expansion. Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in each cycle.

SOLUTION: Isothermal expansion at Th

523 K

Isothermal compression at

Tc

323 K

Gas absorbs 1 200 J during expansion.

(a)

Qc

(b)

Weng

Qh

Qh

Tc Th

1 200 J

Qc

323 523

741 J

1 200 741 J

459 J

P22.18 An electric generating station is designed to have an electric output power of 1.40 MW using a turbine with two-thirds the efficiency of the Carnot engine. The exhaust energy is transferred by heat into the cooling tower at 110˚C. (a) Find the rate at which the station exhausts energy by heat, as a function of the fuel combustion temperature Th. If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change? (b) Find the exhaust power for Th = 800˚C. (c) Find the value of Th for which the exhaust power would be only half as large as in part (b). (d) Find the value of Th for which the exhaust power would be on quarter as large as in part (b).

SOLUTION: (a) "The actual efficiency is two thirds the Carnot efficiency" reads as an equation Weng

Weng

Qh

Qc

W eng

T 2 1 c 3 Th

2 Th Tc 3 Th

.

All the T's represent absolute temperatures. Then Qc

W eng

W eng

1.5 Th Th Tc

Qc W eng

1.5 Th Th Tc

1

1.5 Th Th Tc Th Tc

Qc

Weng

0.5 Th Tc Th Tc

Weng 0.5 Th Tc t Th Tc

Qc t

1.40 MW

0.5 Th 383 K Th 383 K

The dominating Th in the bottom of this fraction means that the exhaust power decreases as the firebox temperature increases.

(b)

(c) We require

Qc t

1.40 MW

Qc t

1 2

0.5 Th 383 K 0.5(1073 K) 383 K 1.40 MW Th 383 K (1073 383) K

1.87 M W = 1.40 M W

0.5 Th 383 K

0.5 Th 383 K Th 383 K

0.666Th 255 K

Th

1.87 MW

0.5 Th 383 K Th 383 K 638 K/ 0.166

0.666

3.84 103 K

(d) The mnimum possible heat exhaust power is approached as the firebox temperature goes to infinity, and it is |Qc|/ t = 1.40 MW( 0.5/1) = 0.7 MW. The heat exhaust power cannot be as small as (1/4)(1.87 MW) = 0.466 MW. So no answer exists. The energy exhaust cannot be that small.

P22.21

An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running

in reverse. That is, energy Qc is taken in from a cold reservoir and energy Qh is rejected to a hot reservoir. (a) Show that the work that must be supplied to run the refrigerator or heat pump is

W =

Th – Tc Qc Tc

(b) Show that the coefficient of performance of the ideal refrigerator is

COP =

Tc Th – Tc

SOLUTION: (a)

For a complete cycle,

W

Qh

Qc

Qc

Qh Qc

Eint

0 and

1

The text shows that for a Carnot cycle (and only for a reversible cycle) Qh Th Qc

Tc

Therefore,

W

(b) COP

Qc

Th Tc Tc

We have the definition of the coefficient of performance for a refrigerator, Qc

W Using the result from part (a), this becomes

Tc

COP

Th Tc

P22.23 How much work does an ideal Carnot refrigerator require to remove 1.00 J of energy from helium at 4.00 K and reject this energy to a room-temperature (293-K) environment?

SOLUTION: COP

W

P22.29

Carnot refrig

Tc T

4.00 289

0.013 8

Qc W

72.2 J per 1 J energy removed by heat.

A gasoline engine has a compression ratio of 6.00 and uses a gas for which

= 1.40. (a) What is the efficiency of the engine if it operates in an idealized Otto cycle? (b) What If? If the actual efficiency is 15.0%, what fraction of the fuel is wasted as a result of friction and energy losses by heat that could by avoided in a reversible engine? (Assume complete combustion of the air-fuel mixture.)

SOLUTION: Compression ratio (a)

6.00 ,

1.40

Efficiency of an Otto-engine e 1 e 1

(b)

1 6.00

V2 V1

1

0.400

51.2%

If actual efficiency e 15.0% losses in system are e e

36.2%

P22.33 Calculate the change in entropy of 250 g of water heated slowly from 20.0°C to 80.0°C. You may use the result of Problem 31.

SOLUTION: f

S i

dQ T

Tf

Ti

mcdT T

mc ln

S 250 g 1.00 cal g C ln

P22.41 12? (b) a 7?

Tf Ti

353 293

46.6 cal K

195 J K

If you roll two dice, what is the total number of ways in which you can obtain (a) a

SOLUTION: (a)

A 12 can only be obtained one way, as 6 6

(b)

A 7 can be obtained six ways: 6 1 , 5 2 , 4 3 , 3 4 , 2 5 , 1 6

*P20.59 An idealized diesel engine operates in a cycle known as the air-standard diesel cycle, shown in Figure 22.59. Fuel is sprayed into the cylinder at the point of maximum compression, B. Combustion occurs during the expansion B C, which is modeled as an isobaric process. Show that the efficiency of an engine operating in this idealized diesel cycle is

1 TD TC

e 1

TA TB

SOLUTION: P22.59 .

The heat transfer over the paths CD and BA is zero since they are adiabatic. Over path BC: QBC

nCP TC TB

Over path DA: QDA

nCV TA TD

Therefore,

QDA

Qc

.

Qh

.

The efficiency is then

0

P Adiabatic Processes B

C

0

D

and

A

QBC

e 1

Qc Qh

1

1 TD TA e 1 TC TB

TD

Vi

3Vi

TA CV

TC TB CP

FIG. P22.59

V