where (liQc = (liQ + (liQ'

If we add Eqs.

(8.31)

and

(8.33),

we obtain

«(liQ

+

T (liQ') _{> , 0}

(8.34)

(8.35)

We now adjust the direction of operation and the size of the Carnot engine so that the

1 66 I ntroduction to the Second law of Thermodynamics

composite engine produces no work ; the work required to operate E' is supplied by the Carnot engine, or vice versa. Then, vv,:

=

0, and Eq.

(8.34)

becomes

(8.36)

Under what condition will the relations Eqs.

(8.35)

and

(8.36)

be compatible ?

Because each of the cyclic integrals can be considered as a sum of terms, we write Eqs.

(8.36)

and

(8.35)

in the forms

(8.37)

and

Ql + Q2 + Q3 + Q4 + . . . >

o.

Tl T2 T3 T4 ⁽

8.38

)

The sum on the left-hand side of Eq. ₍

8.37

) consists of a number of terms, some positive and some negative. But the positive ones just balance the negative ones, and the sum is zero. We have to find numbers (temperatures) such that by dividing each term in Eq. ₍

8.37

) by a proper number we can obtain a sum in which the positive terms predominate, and thus fulfill the requirement of the inequality

(8.38).

We can make the positive terms predominate if we divide the positive terms in Eq. ₍

8.37)

by small numbers and the negative terms by larger numbers. However, this means that we are associating positive values of Q with low temperatures and negative values with high temperatures. This implies that heat is extracted from reservoirs at low temperatures and rejected to reservoirs at higher temperatures in the operation of the composite engine. The composite engine is conse­

quently an impossible engine, and our assumption, Eq.

(8.33),

must be incorrect. It follows that for any engine E',

We distinguish two cases :

Case I. The engine E' is reversible.

�

o.

We have excluded the possibility expressed by (

8.33

)

.

If we assume that for E'

f

^{- < 0 dQ}_T ^' _'

(

8.39

)

then we can reverse this engine, which changes all the signs but not the magnitudes of the Q's. Then we have

f

^dQ-_T ^'

> 0

_'

and the proof is the same as before. This forces us to the conclusion that for any system

f =

0 (all reversible cycles).

Therefore every system has a state property

S,

the entropy, such that

dS =

^dQrev _{T .}

The study of the properties of the entropy will be undertaken in the next chapter.

(

8

.

4

0)

(

8.41

)

The C l a usius I n eq u a l ity 1 67

Case II. The engine E' is not reversible.

For any engine we have only the possibilities expressed by

(8.39).

We have shown that the equality holds for the reversible engine. Since the heat and work effects associated with an irreversible cycle are different from those associated with a reversible cycle, this implies that the value of

f IlQ/T

for an irreversible cycle is different from the value, zero, associated with the reversible cycle. We have shown that for any engine the value cannot be greater than zero ; consequently, it must be less than zero. Therefore for irreversible cycles we must have

(all irreversible cycles). (

8.42)

8 . 1 5 T H E C LA U S I U S I N E Q U A LITY

Consider the following cycle : A system is transformed irreversibly from state

1

to state

2,

then restored reversibly from state

2

to state

1.

The cyclic integral is

=

IlQirr

+

II IlQrev

< 0,

T T 2 T

and it is less than zero, by

(8.42),

since it is an irreversible cycle. Using the definition of

dS,

this relation becomes

I2 1lQirr

1

T 2

+

II dS

^< 0.

The limits can be interchanged on the second integral (but not on the first !) by changing the sign. Thus we have

or, by rearranging, we have

IlQirr T _{_} f2

₁

_dS

< 0,

dS > IlQirr.

If the change in state from state

1

to state

2

is an infinitesimal one, we have

dS > IlQirr T '

(8.43)

(8.44)

This is the Clausius inequality, which is a fundamental requirement for a real transfor­

mation. The inequality

(8.44)

enables us to decide whether or not some proposed trans­

formation will occur in nature. We will not ordinarily use

(8.44)

just as it stands but will manipulate it to express the inequality in terms of properties of the state of a system, rather than in terms of a path property such as

IlQirr'

The Clausius inequality can be applied directly to changes in an isolated system. For any change in state in an isolated system,

IlQirr

₌ 0. The inequality then becomes

dS >

0.

(8.45)

The requirement for a real transformation in an isolated system is that

dS

be positive ; the entropy must increase. Any natural change occurring within an isolated system is attended by an increase in entropy of the system. The entropy of an isolated system

1 68 I nt rod uction to the Second law of Thermodynamics

continues to increase so long as changes occur within it. When the changes cease, the system is in equilibrium and the entropy has reached a maximum value. Therefore the condition of equilibrium

in an isolated system

is that the entropy have a maximum value.

These, then, are also fundamental properties of the entropy: (1) the entropy of an isolated system is increased by any natural change which occurs within it ; and (2) the entropy of an isolated system has a maximum value at equilibrium. Changes in a non­

isolated system produce effects in the system and in the immediate surroundings. The system and immediate surroundings constitute a composite isolated system in which the entropy increases as natural changes occur within it. Thus in the universe the entropy increases continually as natural changes occur within it.

Clausius expressed the two laws of thermodynamics in the famous aphorism : " The energy of the universe is constant ; the entropy strives to reach a maximum."

8 . 1 6 C O N C L U S I O N

B y what may seem a rather long route, the existence of a property of a system-the entropy-has been demonstrated. The existence of this property is a consequence of the second law of thermodynamics. The zeroth law defined the temperature of a system ; the first law, the energy; and the second law, the entropy. Our interest in the second law stems from the fact that this law has something to say about the natural direction of a trans­

formation. It denies the possibility of constructing a machine that causes heat to flow from a cold to a hot reservoir without any other effect. In the same way, the second law can identify the natural direction of a chemical reaction. In some situations the second law declares that neither direction of the chemical reaction is natural ; the reaction must then be at equilibrium. The application of the second law to chemical reactions is the most fruitful approach to the subject of chemical equilibrium. Fortunately, this application is easy and is done without interminable combinations of cyclic engines.

Q U E S T I O N S

S.l Using the considerations of Section 7.6, how can the Kelvin statement Vf;.y ::;; 0 of Section 8.3 be amplified to (a) WCY = 0 in a reversible cycle and (b) Wcy < 0 in an irreversible cycle ?

8.2 Would the Carnot engine efficiency be increased more by (a) increasing Tl at fixed Tz or (b) decreasing Tz at fixed Tl ? Explain.

S.3 How can

S

^¢lQreviT vanish when integrated around a cycle while the cyclical integral of IIQrev remains finite?

8.4 Verify Eq. (8.43) [with Eq. (8.41)J by (a) evaluating

S

^r/lQirriT for the irreversible Joule expansion of an ideal gas from volume Vl to volume Vz (Fig. 7.7) ; and (b) evaluating

S

^I1Qrev/T for the iso­

thermal reversible expansion of the gas between the same volumes.

P R O B L E M S Conversion factors :

1 watt = 1 joule per second (1 W = 1 J/s) 1 horsepower = 746 watts (1 hp = 746 W)

8.1 a) Consider the impossible engine that is connected to only one heat reservoir and produces net work in the surroundings. Couple this impossible engine to an ordinary Carnot engine in such a way that the composite engine is the " stove-refrigerator."

Problems 1 69

b) Couple the " stove-refrigerator " to an ordinary Carnot engine in such a way that the composite engine produces work in an isothermal cycle.

S.2 What is the maximum possible efficiency of a heat engine that has a hot reservoir of water boiling under pressure at 125 DC and a cold reservoir at 25 °C?

8.3 The Chalk Point, Maryland, generating station is a modern steam generating plant supplying electrical power to the Washington, D.C., and surrounding Maryland areas. Units One and Two have a gross generating capacity of 710 MW. The steam pressure is 3600 Ibs/in2 = 25 MPa and the superheater outlet temperature is 540 DC (1000 DF). The condensate temperature is at 30 DC (86 oF).

a) What is the Carnot efficiency of the engine ?

b) If the efficiency of the boiler is 91.2 %; the overall efficiency of the turbine, which includes the Carnot efficiency and its mechanical efficiency, is 46.7 % ; and the efficiency of the generator is 98.4 %, what is the overall efficiency of the unit ? (Note : Another 5 % of the total must be subtracted to account for other plant losses.)

c) One ofthe coal burning units produces 355 MW. How many metric tons (1 metric ton = 1 Mg) of coal/hr are required to fuel this unit at its peak output if the heat of combustion of the coal is 29.0 MJjkg?

d) How much heat per minute is rejected to the 30 DC reservoir in the operation of the unit in (c) ? e) If 250,000 gallons/minute of water pass through the condenser, what is the temperature rise of

the water ? Cp = 4.1 8 J/K g ; 1 gallon = 3.79 litres ; density = 1.0 kg/L.

(Data courtesy of William Herrmann, the Potomac Electric Power Company.)

8.4 a) Liquid helium boils at about 4 K, and liquid hydrogen boils at about 20 K. What is the efficiency of a reversible engine operating between heat reservoirs at these temperatures ? b) If we wanted the same efficiency as in (a) for an engine with a cold reservoir at ordinary

temperature, 300 K, what must the temperature of the hot reservoir be ?

S.5 The solar energy fiux is about 4 J /cm2 min. In a nonfocusing collector the temperature can reach a value of about 90 DC. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25 DC, calculate the area of collector needed if the heat engine is to produce 1 horsepower. Assume that the engine operates at maximum efficiency.

8.6 A refrigerator IS operated by a t-hp motor. If the interior of the box is to be maintained at - 20 DC against a maximum exterior temperature of 35 DC, what is the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume that the coefficient of performance is 75 % of the value for a reversible engine.

S.7 Suppose an electrical motor supplies the work to operate a Carnot refrigerator. If the heat leak into the box is 1200 J/s and the interior of the box is to be maintained at - 10 DC while the exterior is at 30 °C, what size motor (in horsepower) must be used if the motor runs continuously? Assume

that the efficiencies involved have their largest possible values.

8.8 Suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0 DC. Liquid water is taken in at 0 DC and converted to ice at 0 DC. To convert 1 g of ice to 1 g liquid, A.Hfus = 334 J/g are required. If the temperature outside the box is 20 DC, what mass of ice can be produced in one minute by a t-hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible values.

8.9 Under 1 atm pressure, helium boils at 4.216 K. The heat of vaporization is 84 J/mo!. What size motor (in horsepower) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.216 K to liquid at 4.216 K in one minute ? Assume that the ambient temperature is 300 K and that the coefficient of performance of the refrigerator is 50 % of the maximum possible.

S.lO A 0. 1 horsepower motor is used to run a Carnot refrigerator. If the motor runs continuously, what will be the temperature reached inside the box if the heat leak into the box is 500 J/s and the outside temperature is 20 DC? Assume that the machine performs with maximum efficiency.

1 70 I ntroduction to the Second law of Thermodynamics

8 . 1 1 If a heat pump is to provide a temperature of 21 °C inside the house from an exterior reservoir at 1 °C, calculate the maximum value for the coefficient of performance. If the cold end of the heat pump is made as a solar collector, what must the area of the collector be if the 1 °C tempera­

ture is maintained while pumping 2 kJ/s into the house as heat ? Assume that the solar fiux is 40 kJ m- 2 min - 1.

8.12 If a fossil fuel power plant operating between 540 °C and 50 °C provides the electrical power to run a heat pump that works between 25 °C and 5 °C, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler ?

a) Assume that the efficiencies are equal to the theoretical maximum values.

b) Assume that the power plant efficiency is 70 % of maximum and that the coefficient of per­

formance of the heat pump is 10 % of maximum.

c) If a furnace can use 80 % of the energy in fossil fuel to heat the house, would it be more economical in terms of overall fossil fuel consumption to use a heat pump or a furnace ? Do the calculations for cases (a) and (b).

8.13 A 23 600 BTU/hr air-conditioning unit has an energy efficiency ratio (EER) of 7.5. The EER is defined as the number of BTU /hr extracted from the room divided by the power consumption of the unit in watts (1 BTU = 1 .055 kJ).

a) What is the actual coefficient of performance of this refrigerator?

b) If the outside temperature is 32 °C and the inside temperature is 22 DC, what percent of the theoretical maximum value is the coefficient of performance ?

8.14 The standard temperatures for evaluating the performance of heat pumps for high temperatures are 70 of for the inside temperature and 47 of for the outside temperature. For low-temperature heating the standard temperatures are 70 of and 17 of. Calculate the theoretical coefficient of pe�formance for the heat pump under both these conditions. The values achieved by commercial machines range from 1.0[sic] to 2.4 for low-temperature heating and from 1.7 to 3.2 for high­

temperature heating.

8.15 The standard conditions for evaluating air conditioners are 80 OF interior temperature and 95 OF exterior temperature. Calculate the theoretical coefficient of performance under these conditions.

What value of EER does this coefficient of performance translate to ? (EER is defined in Problem 8.13.) Note: The EER values for commercial machines range from fl. ? :, .0 12.80.

8.16 a) Suppose we choose the efficiency of a reversible engine as the thermometric property for a thermodynamic temperature scale. Let the cold reservoir have a fixed temperature. Measure the efficiency of the engine with the hot reservoir at the ice point, 0 degrees, and with the hot reservoir at the steam point, 100 degrees. What is the relation between temperatures, t, on this scale and the usual thermodynamic temperatures T ?

b ) Suppose the hot reservoir has a fixed temperature and we define the temperature scale by measuring efficiency with the cold reservoir at the steam point and at the ice point. Find the relation between t and T for this case. (Choose 100 degrees between the ice point and the steam point.)

8.17 Consider the following cycle using 1 mol of an ideal gas, initially at 25 °C and 1 atm pressure.

Step 1. Isothermal expansion against zero pressure to double the volume (Joule expansion).

Step 2. Isothermal, reversible compression from t atm to 1 atm.

a) Calculate the value of � (liQ/T ; note that the sign conforms with Eq. (8.42).

b) Calculate I1S for step 2.

c) Realizing that for the cycle, I1Scycle = 0, find I1S for step 1.

d) Show that I1S for step 1 is not equal to the Q for step 1 divided by T.

In document Physical Chemistry 3th Castellan (Page 195-200)