Koretsky - Solucionário

Chapter 1 Solutions

Engineering and Chemical Thermodynamics

Wyatt Tenhaeff Milo Koretsky

Department of Chemical Engineering Oregon State University

1.2

An approximate solution can be found if we combine Equations 1.4 and 1.5:

molecular k e V m 2  2 1  molecular k e kT  2 3 m kT V  3  

Assume the temperature is 22 ºC. The mass of a single oxygen molecule ism5.141026kg. Substitute and solve:

 

m/s 6 . 487  V

The molecules are traveling really, fast (around the length of five football fields every second). Comment:

We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitative expression for this expression, we have:

dv v v kT m kT m dv v f 2 2 2 / 3 2 exp 2 4 ) (               

where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by integrating the expression above

 

m/s 449 8 ) ( ) ( 0 0   





  m kT dv v f vdv v f V  

1.3

Derive the following expressions by combining Equations 1.4 and 1.5:

a a m kT V2 3 b b m kT V2 3 Therefore, a b b a m m V V  2 2  

1.4

We have the following two points that relate the Reamur temperature scale to the Celsius scale:



0ºC,0ºReamur



and



100ºC,80ºReamur



Create an equation using the two points:



ºReamur



0.8T



ºCelsius



T  At 22 ºC, Reamur º 6 . 17  T

1.5

(a)

After a short time, the temperature gradient in the copper block is changing (unsteady state), so the system is not in equilibrium.

(b)

After a long time, the temperature gradient in the copper block will become constant (steady state), but because the temperature is not uniform everywhere, the system is not in equilibrium.

(c)

After a very long time, the temperature of the reservoirs will equilibrate; The system is then homogenous in temperature. The system is in thermal equilibrium.

1.6

We assume the temperature is constant at 0 ºC. The molecular weight of air is kg/mol 0.029 g/mol 29   MW

Find the pressure at the top of Mount Everest:











 









_                        K 73.15 2 K mol J 314 . 8 m 8848 m/s .81 9 kg/mol 029 . 0 exp atm 1 P kPa 4 . 3 3 atm 330 . 0   P

Interpolate steam table data:

C º 4 . 71  sat

T for Psat 33.4kPa

Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the temperature remains constant. In reality the temperature decreases with height as we go up the mountain. However, a solution in which T and P vary with height is not as straight-forward.

1.7

To solve these problems, the steam tables were used. The values given for each part constrain the water to a certain state. In most cases we can look at the saturated table, to determine the state.

(a) Subcooled liquid

Explanation: the saturation pressure at T = 170 [oC] is 0.79 [MPa] (see page 508); Since the pressure of this state, 10 [bar], is greater than the saturation

pressure, water is a liquid.

(b) Saturated vapor-liquid mixture

Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume of this state, 3 [m3/kg], is in between these values we have a saturated vapor-liquid mixture.

(c) Superheated vapor

Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since the volume of this state, 0.05 [m3/kg], is greater than this value, it is a vapor.

(d) Superheated vapor

Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5 [MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state, 7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go to the superheated water vapor tables for P = 500 [kPa], we see the state is constrained to T = 200 [oC].

1.8

From the steam tables in Appendix B.1:

         kg m 003155 . 0 ˆ 3 critical v



T 374.15ºC,P22.089MPa



At 10 bar, we find in the steam tables

         kg m 001127 . 0 ˆ 3 sat l v          kg m 19444 . 0 ˆ 3 sat v v

Because the total mass and volume of the closed, rigid system remain constant as the water condenses, we can develop the following expression:





sat v sat l critical v x v x vˆ  1 ˆ  ˆ

where x is the quality of the water. Substituting values and solving for the quality, we obtain 0105

. 0 

x or 1.05 %

1.9

The calculation methods will be shown for part (a), but not parts (b) and (c)

(a)

Use the following equation to estimate the specific volume:



1.9MPa,250ºC

 

ˆ1.8MPa,250ºC



0.5



ˆ



2.0MPa,250ºC

 

ˆ1.8MPa,250ºC





ˆ v v v

v   

Substituting data from the steam tables,





         kg m 11821 . 0 C º 250 MPa, 9 . 1 ˆ 3 v

From the NIST website:





         kg m 11791 . 0 C º 250 MPa, 9 . 1 ˆ 3 NIST v

Therefore, assuming the result from NIST is more accurate

% 254 . 0 % 100 ˆ ˆ ˆ Difference % _          NIST NIST v v v (b) Linear interpolation:





         kg m 13284 . 0 C º 300 MPa, 9 . 1 ˆ 3 v NIST website:





         kg m 13249 . 0 C º 300 MPa, 9 . 1 ˆ 3 NIST v Therefore, % 264 . 0 Difference %  (c) Linear interpolation:





         kg m 12406 . 0 C º 270 MPa, 9 . 1 ˆ 3 v

Note: Double interpolation is required to determine this value. First, find the molar volumes at 270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from the previous step to find the molar volume at 270 ºC and 1.9 MPa.

NIST website:





         kg m 12389 . 0 C º 270 MPa, 9 . 1 ˆ 3 NIST v Therefore, % 137 . 0 Difference % 

With regards to parts (a), (b), and (c), the values found using interpolation and the NIST website agree very well. The discrepancies will not significantly affect the accuracy of any subsequent calculations.

1.10

For saturated temperature data at 25 ºC in the steam tables,

              kg L 003 . 1 kg m 001003 . 0 ˆ 3 sat l v

Determine the mass:

 

kg 997 . 0 kg L 003 . 1 L 1 ˆ          _sat l v V m

Because molar volumes of liquids do not depend strongly on pressure, the mass of water at 25 ºC and atmospheric pressure in a one liter should approximately be equal to the mass calculated above unless the pressure is very, very large.

1.11

First, find the overall specific volume of the water in the container:

          kg m 2 . 0 kg 5 m 1 ˆ 3 3 v

Examining the data in the saturated steam tables, we find

sat v sat

l v v

vˆ  ˆ ˆ at Psat 2bar

Therefore, the system contains saturated water and the temperature is

C º 23 . 120  _Tsat T

Let m represent the mass of the water in the container that is liquid, and _l m represent the mass _v of the water in the container that is gas. These two masses are constrained as follows:

kgm_l m_v 5

We also have the extensive volume of the system equal the extensive volume of each phase

V v m v m_lˆ_lsat  _vˆ_vsat 



_0.001061m3_/kg

 

_{ 0.8857m}3_/kg



_1m3 v l m m

Solving these two equations simultaneously, we obtain

kgm_l 3.88 kgm_v 1.12 Thus, the quality is

224 . 0 kg 5 kg 12 . 1    m m x v

The internal energy relative to the reference state in the saturated steam table is sat v v sat l lu m u m U  ˆ  ˆ



kJ/kg



47 . 504 ˆsat  l u



kJ/kg



5 . 2529 ˆ_vsat  u Therefore,



3.88kg





504.47



kJ/kg



 

 1.12kg





2529.5



kJ/kg





 U kJ 4 . 4790  U

1.12

First, determine the total mass of water in the container. Since we know that 10 % of the mass is vapor, we can write the following expression

 

 



sat



v sat l v v m V  0.9 ˆ  0.1 ˆ

From the saturated steam tables for Psat 1MPa

         kg m 001127 . 0 ˆ 3 sat l v          kg m 1944 . 0 ˆ 3 sat v v Therefore,

 

 

                                    kg m 1944 . 0 1 . 0 kg m 001127 . 0 9 . 0 m 1 ˆ 1 . 0 ˆ 9 . 0 3 3 3 sat v sat l v v V m kg 9 . 48  m and

 

0.1 ˆ 0.1



48.9kg





0.1944m3/kg



 sat v v mv V 3 m 950 . 0  v V

1.13

In a spreadsheet, make two columns: one for the specific volume of water and another for the pressure. First, copy the specific volumes of liquid water from the steam tables and the

corresponding saturation pressures. After you have finished tabulating pressure/volume data for liquid water, list the specific volumes and saturation pressures of water vapor. Every data point is not required, but be sure to include extra points near the critical values. The data when plotted on a logarithmic scale should look like the following plot.

The Vapor-Liquid Dome for H2O

0.0010 0.0100 0.1000 1.0000 10.0000 100.0000 0.0001 0.001 0.01 0.1 1 10 100 1000 Specific Volume, v (m3_/kg) P ressu re , P (M P a

) Liquid Vapor-Liquid Vapor

1.14

The ideal gas law can be rewritten as

P RT v

For each part in the problem, the appropriate values are substituted into this equation where

       K mol bar L 08314 . 0 R

is used. The values are then converted to the units used in the steam table.

(a)                              kg m 70 . 1 L m 10 mol kg 018 . 0 mol L 7 . 30 ˆ mol L 7 . 30 3 3 3 MW v v v

For part (a), the calculation of the percent error will be demonstrated. The percent error will be based on percent error from steam table data, which should be more accurate than using the ideal gas law. The following equation is used:

% 100 ˆ ˆ ˆ Error % _         ST ST IG v v v

where v_IG is the value calculated using the ideal gas law and v_ST is the value from the steam table. % 62 . 1 % kg m 6729 . 1 kg m 6729 . 1 kg m 70 . 1 Error % 3 3 3                                           

This result suggests that you would introduce an error of around 1.6% if you characterize boiling water at atmospheric pressure as an ideal gas.

(b)               kg m 57 . 3 ˆ mol L 3 . 64 3 v v  Error % 0.126%

For a given pressure, when the temperature is raised the gas behaves more like an ideal gas.

(c)               kg m 0357 . 0 ˆ mol L 643 . 0 3 v v %Error = 8.87% (d)               kg m 0588 . 0 ˆ mol L 06 . 1 3 v v %Error = 0.823%

1.15

The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall – your answer should vary. The volume of the room is









3 3 m 4 . 54 ft 1920 ft 10 ft 16 ft 12    V

The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles using the ideal gas law:









295.15K



K mol J 314 . 8 m 4 . 54 Pa 10 01325 . 1 5 3               RT PV n mol 2246  n

Use the molecular weight of air to obtain the mass:



 

 2246.27mol



0.029kg/mol



65.2kg  MWn

m

1.16

First, find the total volume that one mole of gas occupies. Use the ideal gas law:









_                     mol m 0244 . 0 Pa 10 1 K 15 . 293 K mol J 314 . 8 ₃ 5 P RT v

Now calculate the volume occupied by the molecules:

                     3 3 4 molcule one of Volume mole per molecules of Number r N voccupied _A π





_   _         _  23 10 3 m 10 5 . 1 3 4 mol 1 molecules 10 022 . 6 π occupied v            mol m 10 51 . 8 3 6 occupied v

Determine the percentage of the total volume occupied by the molecules:

% 0349 . 0 % 100 Percentage            v voccupied

1.17

Water condenses on your walls when the water is in liquid-vapor equilibrium. To find the

maximum allowable density at 70 ºF, we need to find the smallest density of water vapor at 40 ºF which satisfies equilibrium conditions. In other words, we must find the saturation density of water vapor at 40 ºF. From the steam tables,

         kg m 74 . 153 ˆ 3 sat v

v (sat. water vapor at 40 ºF = 4.44 ºC)

We must correct the specific volume for the day-time temperature of 70 ºF (21.1 ºC ) with the ideal gas law.

3 . 294 K 294.3 6 . 277 K 6 . 277 , ˆ ˆ T v T v_vsat 





                           kg m 163 kg m 74 . 153 K 6 . 277 K 3 . 294 ˆ ˆ 3 3 K 6 . 277 , 6 . 277 3 . 294 K 294.3 sat v v T T v Therefore,           3 3 m kg 10 13 . 6 ˆ 1 ˆ sat v v 

If the density at 70 ºF is greater than or equal to6.13103

 

kg/m3 , the density at night when the temperature is 40 ºF will be greater than the saturation density, so water will condense onto the wall. However, if the density is less than 6.13103

 

kg/m3 , then saturation conditions will not be obtained, and water will not condense onto the walls.

1.18

We consider the air inside the soccer ball as the system. We can answer this question by looking at the ideal gas law:

RT

Pv

If we assume it is a closed system, it will have the same number of moles of air in the winter as the summer. However, it is colder in the winter (T is lower), so the ideal gas law tells us that Pv will be lower and the balls will be under inflated.

Alternatively, we can argue that the higher pressure inside the ball causes air to leak out over time. Thus we have an open system and the number of moles decrease with time – leading to the under inflation.

1.19

First, write an equation for the volume of the system in its initial state:

V v m v m_lˆ_lsat  _vˆ_vsat  Substitute m_l  1



x



mandm_v xm:







x v xv



V m 1 ˆ_lsat  ˆ_vsat 

At the critical point

V v

mˆcritical 

Since the mass and volume don’t change





sat v sat l critical v x v x vˆ  1 ˆ  ˆ

From the steam tables in Appendix B.1:

         kg m 003155 . 0 ˆ 3 critical v



T 374.15ºC,P22.089MPa



At 0.1 MPa, we find in the steam tables

         kg m 001043 . 0 ˆ 3 sat l v          kg m 6940 . 1 ˆ 3 sat v v

Therefore, solving for the quality, we get 00125

. 0 

x or 0.125 %

1.20

As defined in the problem statement, the relative humidity can be calculated as follows

saturation at water of mass water of mass Humidity Relative 

We can obtain the saturation pressure at each temperature from the steam tables. At 10 [oC], the saturation pressure of water is 1.22 [kPa]. This value is proportional to the mass of water in the vapor at saturation. For 90% relative humidity, the partial pressure of water in the vapor is:

[kPa]pWater 0.91.221.10

This value represents the vapor pressure of water in the air. At 30 [oC], the saturation pressure of water is 4.25 [kPa]. For 590% relative humidity, the partial pressure of water in the vapor is:

[kPa]pWater 0.55.632.12

Since the total pressure in each case is the same (atmospheric), the partial pressure is

proportional to the mass of water in the vapor. We conclude there is about twice the amount of water in the air in the latter case.

1.21

(a)

When you have extensive variables, you do not need to know how many moles of each substance are present. The volumes can simply be summed.

b

a V

V V₁  

(b)

The molar volume, v1, is equal to the total volume divided by the total number of moles.

b a b a tot n n V V n V v     1 1

We can rewrite the above equation to include molar volumes for species a and b.

b b a b a b a a b a b b a a _v n n n v n n n n n v n v n v        1 b b a av x v x v₁   (c)

The relationship developed in Part (a) holds true for all extensive variables.

b a K K K₁   (d) b b a ak x k x k₁  

Chapter 2 Solutions

Engineering and Chemical Thermodynamics

Wyatt Tenhaeff Milo Koretsky

Department of Chemical Engineering Oregon State University

2.1

There are many possible solutions to this problem. Assumptions must be made to solve the problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the towel when you dry yourself. In other words, let

 

kg 5 . 0 2O  H m

Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy balance and neglecting potential and kinetic energy effects reveals

h qˆˆ

Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find the minimum energy required for drying the towel, assume that the temperature of the towel remains constant at T 25ºC298.15K. In the drying process, the absorbed water is vaporized into steam. Therefore, the expression for heat is

l O H v O H h h q 2 2 ˆ ˆ ˆ  where is h_Hv _O 2

ˆ _{is the specific enthalpy of water vapor at P1.01barand T 298.15K and} l

O H

h

2

ˆ _{is the specific enthalpy of liquid water at P1.01barand T 298.15K. A hypothetical} path must be used to calculate the change in enthalpy. Refer to the diagram below

P = 1 [atm] liquid vapor liquid vapor h ˆ 1 P h ˆ 2 h ˆ 3 _h ˆ Psat 1 atm 3.17 kPa

By adding up each step of the hypothetical path, the expression for heat is



































ˆ 25ºC,1.01bar ˆ 25ºC



C º 25 ˆ C º 25 ˆ bar 1.01 C, º 25 ˆ C º 25 ˆ ˆ , , , , 3 2 1 2 2 2 2 2 2 sat v O H v O H sat l O H sat v O H l O H sat l O H h h h h h h h h h q            

However, the calculation of heat can be simplified by treating the water vapor as an ideal gas, which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the hypothetical path containing the pressure change of the liquid can be neglected. This leaves



25ºC,3.17kPa



ˆ



25ºC,3.17kPa



ˆ ˆ 2 2 l O H v O H h h q 

From the steam tables:

       kg kJ 2 . 2547 ˆ , 2 sat v O H h (sat. H2O vapor at 25 ºC)        kg kJ 87 . 104 ˆ , 2 sat l O H h (sat. H2O liquid at 25 ºC)

which upon substitution gives

       kg kJ 3 . 2442 ˆ q Therefore,

 





1221.2

 

kJ kg kJ 442.3 2 kg 5 . 0 _             Q

To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20 minutes (1200 s) to dry the towel. From the definition of electrical work,

 



30 A





208

 

V





1200

 

s



7488

 

kJ 

 IVt

W

Therefore, the efficiency is

 

 

100 % 16.3% kJ 7488 kJ 221.2 1 % 100 _       _        _  W Q 

There are a number of ways to improve the drying process. A few are listed below.  Dry the towel outside in the sun.

 Use a smaller volume dryer so that less air needs to be heated.

 Dry more than one towel at a time since one towel can’t absorb all of the available heat. With more towels, more of the heat will be utilized.

2.3

In answering this question, we must distinguish between potential energy and internal energy. The potential energy of a system is the energy the macroscopic system, as a whole, contains relative to position. The internal energy represents the energy of the individual atoms and molecules in the system, which can have contributions from both molecular kinetic energy and molecular potential energy. Consider the compression of a spring from an initial uncompressed state as shown below.

M

State 1 State 2

x

uncompressed compressed

Since it requires energy to compress the spring, we know that some kind of energy must be stored within the spring. Since this change in energy can be attributed to a change of the macroscopic position of the system and is not related to changes on the molecular scale, we determine the form of energy to be potential energy. In this case, the spring’s tendency to restore its original shape is the driving force that is analogous to the gravity for gravitational potential energy.

This argument can be enhanced by the form of the expression that the increased energy takes. If we consider the spring as the system, the energy it acquires in a reversible, compression from its initial uncompressed state may be obtained from an energy balance. Assuming the process is adiabatic, we obtain: W W Q E   

We have left the energy in terms of the total energy, E. The work can be obtained by integrating the force over the distance of the compression:

2 2 1 kx kxdx dx F W 



 



 Hence: 2 2 1 kx E 

We see that the increase in energy depends on macroscopic position through the term x.

It should be noted that there is a school of thought that assigns this increased energy to internal energy. This approach is all right as long as it is consistently done throughout the energy balances on systems containing springs.

2.4

For the first situation, let the rubber band represent the system. In the second situation, the gas is the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the energy balance becomes

W

U 



Since work must be done on the rubber band to stretch it, the value of the work is positive. From the energy balance, the change in internal energy is positive, which means that the temperature of the system rises.

When a gas expands in a piston-cylinder assembly, the system must do work to expand against the piston and atmosphere. Therefore, the value of work is negative, so the change in internal energy is negative. Hence, the temperature decreases.

In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the rubber band goes to increase its potential energy; however, a part of it goes into stretching the polymer molecules which make up the rubber band, and the qualitative argument given above still is valid.

2.5

To explain this phenomenon, you must realize that the water droplet is heated from the bottom. At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The water vapor forms an insulation layer between the skillet and the water droplet. At low temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures.

2.6 Apartment System Fridge + -W Q HOT Surr.

If the entire apartment is treated as the system, then only the energy flowing across the apartment boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the refrigerator is not explicitly accounted for in the energy balance because it is within the system. By neglecting kinetic and potential energy effects, the energy balance becomes

W Q

U  



The Q term represents the heat from outside passing through the apartment’s walls. The Wterm represents the electrical energy that must be supplied to operate the refrigerator.

To determine whether opening the refrigerator door is a good idea, the energy balance with the door open should be compared to the energy balance with the door closed. In both situations, Q is approximately the same. However, the values of Wwill be different. With the door open, more electrical energy must be supplied to the refrigerator to compensate for heat loss to the apartment interior. Therefore,

shut ajar W

W 

where the subscript “ajar” refers the situation where the door is open and the subscript “shut” refers to the situation where the door is closed. Since,

shut ajar Q Q  shut shut shut ajar ajar ajar Q W U Q W U       shut ajar T T   

2.7

The two cases are depicted below.

System Heater Q Surr. W System Heater Q Surr.

Heater off Heater on

Let’s consider the property changes in your house between the following states. State 1, when you leave in the morning, and state, the state of your home after you have returned home and heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2, the state of the system is the same and U must be zero, so

0    U Q W or W Q 

where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that must be delivered to the heater. The case where more heat escapes will require more work and result in higher energy bills. When the heater is on during the day, the temperature in the system is greater than when it is left off. Since heat transfer is driven by difference in temperature, the heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off when you are gone.

2.8

The amount of work done at constant pressure can be calculated by applying Equation 2.57

Q H   Hence, h m Q H   ˆ 

where the specific internal energy is used in anticipation of obtaining data from the steam tables. The mass can be found from the known volume, as follows:

 

 

 

kg 0 . 1 kg m 0.0010 L m 001 . 0 L 1 ˆ 3 3                                    v V m

As in Example 2.2, we use values from the saturated steam tables at the same temperature for subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1:

 







 



                        kg kJ 15 . 314 kg kJ 87 . 104 kg kJ 02 . 419 C 25 at ˆ C 100 at ˆ ˆ u_l,2 o u_l,1 o u

Solve for heat:

 





314.15

 

kJ kg kJ 05 . 314 kg 0 . 1 ˆ _              m u Q

and heat rate:

 









 

0.52

 

kW min s 60 min. 0 1 kJ 15 . 314          t Q Q

2.9

(a)

From Steam Tables:

       kg kJ 8 . 2967 ˆ₁ u (100 kPa, 400 ºC)        kg kJ 8 . 2659 ˆ₂ u (50 kPa, 200 ºC)            kg kJ 0 . 308 ˆ ˆ ˆ u₂ u₁ u (b)

From Equations 2.53 and 2.63









      2 1 2 1 1 2 T T P T T vdT c R dT c u u u

From Appendix A.2

) (A BT CT2 DT 2 ET3 R c_P      







       2  1 1 3 2 2 T T dT ET DT CT BT A R u Integrating                   ( ) 4 ) 1 1 ( ) ( 3 ) ( 2 ) )( 1 ( ₂4 ₁4 1 2 3 1 3 2 2 1 2 2 1 2 T T E T T D T T C T T B T T A R u

The following values were found in Table A.2.1

0 10 21 . 1 0 10 45 . 1 470 . 3 4 3         E D C B A

K 473.15 K) 15 . 273 200 ( K 673.15 K) 15 . 273 400 ( K mol J 314 . 8 2 1             T T R provides                                               kg kJ 1 . 308 J 1000 kJ 1 kg 1 g 1000 O] H g [ 0148 . 18 O] H [mol 1 mol J 5551 ˆ mol J 5551 2 2 u u

The values in parts (a) and (b) agree very well. The answer from part (a) will serve as the basis for calculating the percent difference since steam table data should be more accurate.

 

% 03 . 0 % 100 0 . 308 1 . 308 308 %        Difference

2.10 (a)

Referring to the energy balance for closed systems where kinetic and potential energy are neglected, Equation 2.30 states

W Q

U  



(b)

Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal 0  U According to Equation 2.77               1 2 1 1 1 2 _ln ln P P RT n P P nRT W

From the ideal gas law:

1 1 1 1RT PV n         1 2 1 1 ln P P V P W

Substitution of the values from the problem statement yields







 

J 940 bar 8 bar 5 ln m 10 5 . 2 Pa 10 8 5 3 3             W W

The energy balance is

 

J 940 J 0     Q W Q (c)

Since the process is adiabatic 0

 Q

The energy balance reduces to

W

U 

The system must do work on the surroundings to expand. Therefore, the work will be negative and 1 2 0 0 2 1 T T T c n U U v T T        



2.11 (a) (i). 2 P [bar] v [m3/mol] 1 2 3 0.01 0.02 0.03 1 P ath A Path B (ii).

Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal

0  u

Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore, 0

 h

Performing an energy balance and neglecting potential and kinetic energy produces 0

  

u q w

For an isothermal, adiabatic process, Equation 2.77 states

       1 2 ln P P nRT W or         1 2 ln P P RT n W w





                   bar 3 bar 1 ln K ) 15 . 273 88 ( K mol J 8.314 w       mol J 3299 w

Using the energy balance above

       mol J 3299 w q (b)

(i). See path on diagram in part (a) (ii).

Since the overall process is isothermal and u and h are state functions 0  u 0  h

The definition of work is





 P dv

w _E

During the constant volume part of the process, no work is done. The work must be solved for the constant pressure step. Since it is constant pressure, the above equation simplifies to

) (v₂ v₁ P dv P w _E



 _E 

The ideal gas law can be used to solve for v2and v1









                                            mol m 010 . 0 Pa 10 3 K ) 15 . 273 88 ( K mol J 314 . 8 mol m 030 . 0 Pa 10 1 K ) 15 . 273 88 ( K mol J 314 . 8 3 5 1 1 1 3 5 2 2 2 P RT v P RT v

                            mol m 010 . 0 mol m 030 . 0 Pa) 10 3 ( 3 3 5 w       mol J 6000 w

Performing an energy balance and neglecting potential and kinetic energy results in 0    u q w         mol J 6000 w q

2.12

First, perform an energy balance. No work is done, and the kinetic and potential energies can be neglected. The energy balance reduces to

Q U  

We can use Equation 2.53 to get



 2 1 T T vdT c n Q

which can be rewritten as



 2 1 T T PdT c n Q

since the aluminum is a solid. Using the atomic mass of aluminum we find

mol 3 . 185 mol kg 0.02698 kg 5 _      n

Upon substitution of known values and heat capacity data from Table A.2.3, we get





_



_{ } 



            K 15 . 323 K 15 . 294 3 10 49 . 1 486 . 2 K mol J 314 . 8 mol 3 . 185 T dT Q

 

kJ 61 . 131  Q

2.13

First, start with the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes

W Q

U  



The value of the work will be used to obtain the final temperature. The definition of work (Equation 2.7) is



  2 1 V V EdV P W

Since the piston expands at constant pressure, the above relationship becomes



V2 V1



P

W  _E 

From the steam tables

         kg m 02641 . 0 ˆ 3 1 v (10 MPa, 400 ºC)

 

3 3 1 1 1 0.07923 m kg m 02641 . 0 kg) 3 ( ˆ                   m v V

Now V₂and v₂are found as follows

 

3 6 3 1 2 0.4536 m Pa 10 0 . 2 J 748740 m 07923 . 0        E P W V V

 

 

__         kg m 1512 . 0 kg 3 m 4536 . 0 ˆ 3 3 2 2 2 m V v

Since vˆ₂andP₂are known, state 2 is constrained. From the steam tables:

 

ºC 400 2  T                 kg m 0.1512 bar, 20 3

       kg kJ 2 . 2945 ˆ₂ u                 kg m 0.1512 bar, 20 3        kg kJ 4 . 2832 ˆ₁ u



100bar,400ºC









 



338.4

 

kJ kg kJ 4 . 2832 kg kJ 2 . 2945 kg 3 ˆ ˆ_{2 1} 1                       U m u u

Substituting the values of Uand Winto the energy equation allows calculation of Q W U Q 

 



748740 J



1.09 10

 

J [J] 338400     6  Q

2.14

In a reversible process, the system is never out of equilibrium by more than an infinitesimal amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of 1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible. To solve for the final temperature of the system, the energy balance will be written. The piston-cylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore, potential and kinetic energy effects can be neglected. The energy balance simplifies to

W

U 



Conservation of mass requires

2 1 n

n 

Let nn₁ n₂

The above energy balance can be rewritten as





 2 1 2 1 V V E T T vdT P dV c n

Since c and _v P_Eare constant:



T2 T1



P



V2 V1



nc_v   _E 

2

V and T₁can be rewritten using the ideal gas law

2 2 2 P nRT V  nR V P T₁ 1 1

Substituting these expressions into the energy balance, realizing that P_E P₂, and simplifying the equation gives

nR V P P T 2 7 2 5 1 1 2 2       _ 

 

 

 

 

           K mol bar L 08314 . 0 mol 0 . 1 L 10 bar 1 bar 2 1 2 1 R n V P P results in

 

K 206 2  T

To find the value for work, the energy balance can be used



T2 T1



nc U

W   _v 

Before the work can be calculated, T₁must be calculated

 







 



 





241

 

K K mol bar L 08314 . 0 mol 1 L 10 bar 2 1 1 1                nR V P T

Using the values shown above

 

J 727  

2.15

The maximum work can be obtained through a reversible expansion of the gas in the piston. Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston assembly is well-insulated, so the heat transfer contribution to the energy balance can be neglected, in addition to potential and kinetic energy effects. The energy balance reduces to

W

U 



In this problem, the process is a reversible, adiabatic expansion. For this type of process, Equation 2.90 states



2 2 1 1



1 1 V P V P k W   

From the problem statement (refer to problem 2.13),

 

 

 

bar 1 L 10 bar 2 2 1 1    P V P

To calculate W, V₂ must be found. For adiabatic, reversible processes, the following relationship (Equation 2.89) holds:

const PVk 

where k is defined in the text. Therefore,

k k V P P V 1 1 2 1 2        Noting that 5 7   v P c c

k and substituting the proper values provides

 

L 4 . 16 2  V

Now all of the needed values are available for calculating the work.



L bar



900

 

J

9  

 

W

From the above energy balance,

 

J 900   U

The change in internal energy can also be written according to Equation 2.53:



  2 1 T T vdT c n U

Since c is constant, the integrated form of the above expression is _v



_{2 1}



2 5 T T R n U         

Using the ideal gas law and knowledge of P₁ and V₁,

 

K 6 . 240 1 T and

 

K 3 . 197 2  T

The temperature is lower because more work is performed during the reversible expansion. Review the energy balance. As more work is performed, the cooler the gas will become.

2.16

Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible. Furthermore, no work is done on the system and potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes

0 ˆ u or 1 2 ˆ ˆ u u 

From the steam tables

       kg kJ 2 . 2619 ˆ₁ u (200 bar, 400 ºC)         kg kJ 2 . 2619 ˆ₂ u

The values of uˆ₂and P₂constrain the system. The temperature can be found from the steam tables using linear interpolation:

C º 5 . 327 2  T

 

_             kg kJ 2 . 2619 ˆ , bar 100 u₂

Also at this state,

         kg m 02012 . 0 ˆ 3 2 v Therefore,

 





3

 

3 2 0.020 m kg m 02012 . 0 kg 0 . 1                    mv V_vessel

2.17

Let the entire tank represent the system. Since no heat or work crosses the system boundaries, and potential and kinetic energies effects are neglected, the energy balance is

0  u

Since the tank contains an ideal gas

K 300 0 1 2 1 2     T T T T

The final pressure can be found using a combination of the ideal gas law and conservation of mass. 2 2 2 1 1 1 V P T V P T  We also know 1 2 2V V  Therefore, bar 5 2 1 2   P P

2.18 (a)

First, as always, simplify the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance is

W Q

U  



Since, this system contains water, we can the use the steam tables. Enough thermodynamic properties are known to constrain the initial state, but only one thermodynamic property is known for the final state: the pressure. Therefore, the pressure-volume relationship will be used to find the specific volume of the final state. Since the specific volume is equal to the molar volume multiplied by the molecular weight and the molecular weight is constant, the given expression can be written

const v

Pˆ1.5 

This equation can be used to solve for vˆ₂.

5 . 1 1 5 . 1 1 2 1 2 ˆ ˆ _             v P P v Using

 

 

 

 

bar 100 kg m 1 . 0 kg 10 m 1.0 ˆ bar 20 2 3 3 1 1           P v P gives          kg m 0.0342 ˆ 3 2 v

Now that the final state is constrained, the steam tables can be used to find the specific internal energy and temperature.

 

        kg kJ 6 . 3094 ˆ K 7 . 524 2 2 u T



  2 1 V V EdV P W or



  2 1 ˆ ˆ ˆ ˆ v v Edv P w

Since the process is reversible, the external pressure must never differ from the internal pressure by more than an infinitesimal amount. Therefore, an expression for the pressure must be

developed. From the relationship in the problem statement,

const v

P v

Pˆ1.5  ₁ˆ₁1.5 

Therefore, the expression for work becomes





   2 1 2 1 ˆ ˆ 1.5 5 . 1 1 1 ˆ ˆ 1.5 5 . 1 1 1 _ˆ ˆ 1 ˆ ˆ ˆ ˆ ˆ v v v v v d v v P v d v v P w

Integration and substitution of proper values provides

               _  kg kJ 284 kg m bar 840 . 2 ˆ 3 w

 





2840

 

kJ kg kJ 284 kg 10 _             W

A graphical solution is given below:

2 P [bar] v [m3/kg] 20 60 100 0.02 0.06 0.10 1 Work

To solve for Q , Umust first be found, then the energy balance can be used.







 



4918

 

kJ kg kJ 8 . 2602 kg kJ 6 . 3094 kg 10 ˆ ˆ_{2 1 }                      U m u u

Now Q can be found,

 

kJ 2840

 

kJ 2078

 

kJ 4918       U W Q (b)

Since the final state is the same as in Part (a), Uremains the same because it is a state function. The energy balance is also the same, but the calculation of work changes. The pressure from the weight of the large block and the piston must equal the final pressure of the system since

mechanical equilibrium is reached. The calculation of work becomes:



  2 1 ˆ ˆ ˆ v v E dv mP W

All of the values are known since they are the same as in Part (a), but the following relationship should be noted

2

P P_E 

Substituting the appropriate values results in

 

kJ 6580 

W

Again we can represent this process graphically:

2 P [bar] v [m3/kg] 20 60 100 0.02 0.06 0.10 1 Work

Now Q can be solved.

 

kJ 6580

 

kJ 1662

 

kJ 4918       U W Q (c)

This part asks us to design a process based on what we learned in Parts (a) and (b). Indeed, as is characteristic of design problems there are many possible alternative solutions. We first refer to the energy balance. The value of heat transfer will be zero when

W

U 



For the same initial and final states as in Parts (a) and (b),

 

kJ 4918    U W

There are many processed we can construct that give this value of work. We show two alternatives which we could use:

Design 1:

If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a reversible compression followed by an irreversible compression. Let the subscript “i" represent the intermediate state where the process switches from a reversible process to an irreversible process. The equation for the work then becomes

 

J 4918000 ˆ ˆ 1 ˆ ) ˆ ˆ ( ˆ ˆ 5 . 1 5 . 1 1 1 2 2 1             

_

i v v i dv v v P v v P m W

Substituting in known values (be sure to use consistent units) allows calculation of vˆ : _i

         kg m 0781 . 0 ˆ 3 i v

The pressure can be calculated for this state using the expression from part (a) and substituting the necessary values.

 

bar 0 . 29 5 . 1 1 1         i i i P v v P P

Now that both P and _i vˆ are known, the process can be plotted on a P-v graph, as follows: _i 2 P [bar] v [m3/kg] 20 60 100 0.02 0.06 0.10 1 i Work Design 2:

In an alternative design, we can use two irreversible processes. First we drop an intermediate weight on the piston to compress it to an intermediate state. This step is followed by a step similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this case, we again must find the intermediate state. Writing the equation for work:



 (ˆ ˆ1) 2(ˆ2 ˆ )



4918000

 

J m Pi vi v P v vi

W

However, we again have the relationship: 5 . 1 1 1        i i v v P P

Substitution gives one equation with one unknown vi:

 

J 4918000 ) ˆ ˆ ( ) ˆ ˆ ( _{1 2 2} 5 . 1 1 1                     i i i v v P v v v v P m W

There are two possible values vi to the above equation. Solution A:        kg m 043 . 0 ˆ 3 i v

which gives

 

bar 8 . 70  i P

This solution is graphically shown below:

2 P [bar] v [m3/kg] 20 60 100 0.02 0.06 0.10 1 i Work Solution B        kg m 0762 . 0 ˆ 3 i v which gives

 

bar 0 . 30  i P

This solution is graphically shown below:

2 P [bar] v [m3/kg] 20 60 100 0.02 0.06 0.10 1 i Work

2.19

(a)

Force balance to find k:

Piston

Fspring=kx Fatm=PatmA

Fgas=PgasA Fmass=mg

P_gas  P_atmmg_A  kx_A since V=Ax

P_gas  P_atmmg_A  kV A2

since V is negative. Now solve:







 





₂



2 3 2 2 5 5 m 1 . 0 m 02 . 0 m 1 . 0 m/s 81 . 9 kg 2040 Pa 10 1 Pa 10 2    k        m N 10 01 . 5 4 k

Work can be found graphically (see P-V plot) or analytically as follows: Substituting the expression in the force balance above:

W_A  P_atm mg_A kV A2     dV

 





           _  f i f i V V V V atm A d V A V k dV A mg P W 2



















       _      2 0 m 02 . 0 m 1 . 0 N/m 10 01 . 5 m 05 . 0 03 . 0 Pa 10 00 . 3 2 2 3 2 2 4 2 5 A W

 

J 10 5 3   A W V [m3] 1 2 3 0.01 0.03 0.05 Patm mg A kV A2 -W 10squares_square0.5kJ =5 kJ Work P [bar] 2 1 = (b)

You need to find how far the spring extends in the intermediate (int) position. Assume

PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer slightly). Since you know P and V for each state in Part (a), you can calculate n.

P_i P_f  V_i V_f    n or 105 Pa 2105 Pa  0.03 m3 0.05 m3    n  n  1.35 Now using the force balance

P_gas,int  P_atm  mg_A kV

A2

with the above equation yields:

P_int  P_{i V}Vi

int

 

n

 P_atm mg_A k V



int Vi



A2

This last equality represents 1 equation. and 1 unknown (we know k), which gives

3 int 0.0385m

V

W_B  P_gas Vi Vint  dV  P_gas Vint V_f  dV

expanding as in Part (a)

W_B  2105 N m2    Vi Vint  dV 5106 N m5V V_i V_int  d

 

V  3105 N m2    Vint V_f  dV  5106 N m5V V_int V_if  d

 

V Therefore,

 

kJ 85 . 3   B W

just like we got graphically.

V [m3] 1 2 3 0.01 0.03 0.05 Patm mg A P [bar] 2 1 Work mg A (c)

The least amount of work is required by adding differential amounts of mass to the piston. This is a reversible compression. For our assumption that PVn = const, we have the following

expression:





  f i f i V V V V gas rev C dV V const dV P W 35 . 1 ,



_{1 10}5_Pa



_0.05m3



1.35 _{1.75 10}3 .    const Therefore, W_C,rev  1.75103 .05 .03  dV V1.35   1.75103 0.35 V 0.35 .05 .03

 

J 2800   V [m3] 1 2 3 0.01 0.03 0.05 Patm P [bar] 2 1 Workrev

2.20

Before this problem is solved, a few words must be said about the notation used. The system was initially broken up into two parts: the constant volume container and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” denotes the final state.

To begin the solution, the mass of water present in each part of the system will be calculated. The mass will be conserved during the expansion process. Since the water in the rigid tank is saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is constrained to a specific state. From the steam tables,

 

kPa 8 . 1553 kg kJ 3 . 2595 ˆ kg kJ 64 . 850 ˆ kg kJ 12736 . 0 ˆ kg kJ 001156 . 0 ˆ , 1 , 1 , 1 , 1                              sat v i l i v i l i P u u v v (Sat. water at 200 ºC)

Knowledge of the quality of the water and the overall volume of the rigid container can be used to calculate the mass present in the container.

 

 

v i l i m v v m V₁0.05 ₁ ˆ_1, 0.95 ₁ ˆ_1,

Using the values from the steam table and V₁ 0.5

 

m3 provides

 

kg 13 . 4 1 m

Using the water quality specification,

 

 

kg 207 . 0 05 . 0 kg 92 . 3 95 . 0 1 1 1 1     m m m m l v

                kg kJ 9 . 2638 ˆ kg m 35202 . 0 ˆ , 2 3 , 2 i i u v (600 kPa, 200 ºC)

Enough information is available to calculate the mass of water in the piston assembly.

 

kg 284 . 0 ˆ₂ 2 2   v V m

Now that the initial state has been characterized, the final state of the system must be determined. It helps to consider what physically happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of the entire system will match the surroundings’. In other words,

        kg kJ 9 . 2638 ˆ ˆ_f u_2,i u (600 kPa, 200 ºC)

Thus, the change in internal energy is given by

l i l i v i v i i f l i v i m u m u m u m u m m U ( ₂ _1,  _1, )ˆ  ₂ˆ_2,  _1, ˆ_1,  _1, ˆ_1, 

Substituting the appropriate values reveals

 

kJ 0 . 541  U

To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa (weight of the piston was assumed negligible). From Equation 2.7,

) ( _{f i} E V V E dV P V V P W f i     



where

 

 

     

3 3 3 3 , 2 , 1 , 1 2 m 6 . 0 m 5 . 0 m 0.1 m 55 . 1 ˆ ) ( Pa 600000         i i l i v i f E V v m m m V P

Note: vˆ_2,i was used to calculate V_f because the temperature and pressure are the same for the final state of the entire system and the initial state of the piston-cylinder assembly. The value of W can now be evaluated.

 

kJ 570  

W

The energy balance is used to obtain Q.

 

kJ



570

 

kJ



1111

 

kJ 0 . 541        U W Q

2.21

A sketch of the process follows:

10 cm 50 cm A B p1,B = 20 bar T1,B = 250 oC p1,A = 10 bar T1,A = 700 oC H2O H2O system boundary Process A B p2 T2 H2O system boundary p2 T2 H2O

The initial states are constrained. Using the steam tables, we get the following:

State 1,A State 1,B

p 10 [bar] 20 [bar] T 700 [oC] 250 [oC] v 0.44779 [m3/kg] 0.11144 [m3/kg] u 3475.35 [kJ/kg] 2679.58 [kJ/kg] V 0.01 m3 0.05 m3 m V v 0.11 [kg] 0.090 [kg]

All the properties in the final state are equal. We need two properties to constrain the system: We can find the specific volume since we know the total volume and the mass:

 

 

__           kg m 30 . 0 kg 0.20 m 06 . 0 3 3 , 1 , 1 , 1 , 1 2 B A B A m m V V v

We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is rigid, W=0. An energy balance gives:

U  Q  W  0 Thus, U₂U₁ m_1,Au_{1, A} m_1,Bu_1,B or            kg kJ 3121 , 1 , 1 , 1 , 1 , 1 , 1 2 2 2 B A B B A A m m u m u m m U u

We have constrained the system with u2 and v2, and can find the other properties from the steam

Tables. Very close to

T2 = 500 [oC] and P2 = 1200 [kPa]