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Chapter 2 Solutions Engineering and Chemical Thermodynamics 2e

Wyatt Tenhaeff Milo Koretsky School of Chemical, Biological, and Environmental Engineering Oregon State University [email protected]

2.1 Gas B will have a higher final temperature: Conceptually, the heat capacity is the amount of energy a gas needs to take in to change its temperature. The less energy it needs (lower heat capacity), the greater the temperature change. Alternatively, we can see it from the mathematics. Since Q  n cv dT and Q and n are identical the gas with a smaller heat capacity will have a greater temperature change.

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2.2 (a) process 1 will have a greater final temperature because all the energy supplied by the heat results in increasing the temperature – alternatively in process 2 some of the energy is used to do work as the piston expands. Mathematically, this results in cP being greater then cv. (b) It is convenient to use the property enthalpy for process 2 because h  q in a closed system  h  at constant pressure and cP    .  T  P

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2.3 The temperature will decrease. Temperature decreases as the pressure decreases (we can see from the steam tables)

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2.4 First, the steady-state energy balance across the valve shows that the enthalpy of the exit state, 2, equals the enthalpy of the inlet state, 1, i.e., it is an isenthalpic process; second, the enthalpy change of an ideal gas depends on temperature only. Therefore, since the enthalpy does not change, neither does the temperature.

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2.5 The temperature will decrease because work is done by the system and that energy comes from the molecular kinetic energy of the air

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2.6 The temperature will increase because flow work is done by the gas

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2.7 C>D>A>B Processes C and D will both result in an increase in temperature because work is being done on the system. Process C is irreversible, so it will result in more work and increase the temperature more than the reversible process D. Processes A and B will both result in a decrease in temperature because work is being done by the system. The irreversible process A does less work than the reversible process B, resulting in a smaller decrease in temperature.

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2.8 (a) steady state, open system with two streams in and one stream out, negligible change in kinetic and potential energy. (b) constant heat capacity (c) constant heat capacity and cP = cv, i.e., a liquid or a solid. (d) ideal gas (e) this equation always applies, it is the definition of enthalpy (f) reversible process (g) closed system with negligible macroscopic kinetic and potential energy. (h) no energy transfer across system boundaries via heat, i.e., an adiabatic process

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2.9 (a) greater than. The change in internal energy remains the same and q is 0 (adiabatic), so all the energy must now come from work (b) less than. The final temperature is much lower so energy will need to be removed by heat (c) greater than. More work is required for irreversible compression than reversible (d) less than. Using results from part (c) and the fact that u is the same.

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2.11 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. compressed

uncompressed x

State 1

M

State 2

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:

E  Q  W  W 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:

W   F  dx   kxdx 

1 2 kx 2

Hence: E 

1 2 kx 2

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.

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2.12 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 U  W

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 2.11, 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.

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2.13 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.

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2.14

Apartment

Surr. HOT System

Fridge

Q

+ W -

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 U  Q  W The Q term represents the heat from outside passing through the apartment’s walls. The W term 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 W will 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, Wajar  Wshut

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, Qajar  Qshut  U ajar  Qajar  Wajar   U shut  Qshut  Wshut  Tajar  Tshut

The refrigerator door should remain closed.

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2.15 The two cases are depicted below. Heater on

Heater off

Surr.

Surr.

System

System

Heater

Q

Heater

Q

W

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 U  Q  W  0

or Q W

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.

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2.16 The gas leaving the tank does flow work as it exits the valve. This work decreases the internal energy of the gas – lowering the temperature. During this process, water from the atmosphere will become supersaturated and condense. When the temperature drops below the freezing point of water, the water forms a solid.

Attractive interactions between the compressed gas molecules can also contribute to this phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of these interactions in Chapter 4.

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2.17 (a) It takes more energy to raise the temperature of a gas in a constant pressure cylinder. In both cases the internal energy of the gas must be increased. In the constant pressure cylinder work, Pv work must also be supplied to expand the volume against the surrounding’s pressure. This is not required with a constant volume. (b) As you perspire, sweat evaporates from your body. This process requires latent heat which cools you. When the water content of the environment is greater, there is less evaporation; therefore, this effect is diminished and you do not feel as comfortable.

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2.18 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 m H 2 O  0.5 kg 

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 qˆ  hˆ Refer to the development of Equation 2.28 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  25 º C  298.15 K . In the drying process, the absorbed water is vaporized into steam. Therefore, the expression for heat is v l  hˆH qˆ  hˆH 2O 2O

ˆv where is hH 2 O is the specific enthalpy of water vapor at P 1.01 bar and T  298.15 K and hˆlH O is the specific enthalpy of liquid water at P 1.01 bar and T  298.15 K . A hypothetical 2

path must be used to calculate the change in enthalpy. Refer to the diagram below P

hˆ

liquid 1 atm

vapor

P = 1 [atm]  hˆ 3

h1ˆ Psat

3.17 kPa

hˆ2

liquid

vapor

By adding up each step of the hypothetical path, the expression for heat is qˆ   h1   h2   h3 l sat l  hˆ , 25 º C   hˆ



 25 º C, 1.01 bar  hˆ Hv, satO  25 º C  hˆHl, satO 25 º C  hˆHv O  25 º C, 1.01 bar  hˆ Hv ,sat O 25 º C 

H 2O

H 2O

2

2

2

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2

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 qˆ  hˆHv O 25 º C, 3.17 kPa  hˆHl O 25 º C, 3.17 kPa  2 2

From the steam tables:  kJ  hˆvH,sat  2547.2  2O  kg   kJ  hˆlH, sat  104.87   O 2  kg 

(sat. H2O vapor at 25 ºC) (sat. H2O liquid at 25 ºC)

which upon substitution gives  kJ  qˆ  2442.3    kg 

Therefore,   kJ   Q  0.5 kg  2442.3     1221.2 kJ     kg   

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, W  IVt  30 A208 V1200 s  7488 kJ

Therefore, the efficiency is  1221.2 kJ   Q   100  %   100 % 16.3% W   7488 kJ 



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.

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2.19 m3 kg kJ uˆ1  xuˆ1v  (1 x )uˆ1l  1,355 kg Since the container is rigid, the volume is constant: vˆ1  xvˆ1v  (1 x )vˆ1l  0.373

vˆ2  vˆ1  0.373

m3 kg

and w=0 We know T2 and vˆ2 , so we can get the other properties from the steam tables: P2 = 10 bar uˆ 2  3,192

kJ kg

Applying the first law, qˆ  uˆ 2  uˆ1  1,838

kJ kg

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2.21 (a)

(b) An outline of the solution process follows

 

  

m3 Determine initial molar volume: vˆ1  1.79 kg Determine initial temperature and internal energy from the steam tables since the initial kJ state ( vˆ1 v 1, P1) is known: T1  120 o C; uˆ3  2,537.3 kg Determine the final pressure from a force balance: P2 = 3 105 Pa J Determine the work from the area under the curve above: wˆ  6.17 105 Kg Use an energy balance to determine the heat transferred, obtaining values for state 2 from kJ the steam tables: qˆ   1,780 Kg

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2.22 The amount of work done at constant pressure can be calculated by applying Equation 2.28

H  Q Hence, H  Q  mhˆ 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:



m

V  vˆ



1L 0.001 m

3 

  L      1.0 kg   m 3    0.0010     kg    

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:  kJ   kJ   kJ  uˆ  uˆl ,2 at 100 o C  uˆ l,1 at 25 o C  419.02   104.87    314.15    kg   kg   kg 



 

  

Solve for heat:   kJ   Q  muˆ  1.0 kg  314.05     314.15 kJ    kg   

and heat rate:

Q Q   t

314.15 kJ   0.52 kW   60 s  10 min.   min 

This value is the equivalent of five strong light bulbs.

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2.23 (a) From Steam Tables:

 kJ  uˆ1  2967.8    kg   kJ  uˆ 2  2659.8   kg 

(100 kPa, 400 ºC) (50 kPa, 200 ºC)

 kJ  uˆ  uˆ 2  uˆ1  308.0    kg  (b) From the definition of the heat capacity: T2

T2

T1

T1

 cv dT   cP  RdT

 u  u2  u1 

From Appendix A.2 cP  R( A  BT  CT 2  DT 2  ET 3 ) T2





 u  R  A  BT  CT 2  DT  2  ET 3  1 dT T1

Integrating  1 1 B C E 4  u  R (A  1)(T 2  T1 )  (T22  T12 )  (T 23  T13)  D(  )  ( T2 4  T1 )  2 3 4 T2 T1  

The following values were found in Table A.2.1 A  3.470 B  1.45  10  3 C0 D  1.21 104 E0 Substituting these values and using

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 J  R  8.314   mol  K  T1  (400  273.15 K)  673.15 K T2  ( 200  273.15 K)  473.15 K

provides  J   u  5551    mol    kJ   J   1 [mol H 2O]  1000 g  1 kJ   uˆ    5551        308.1     mol   18.0148 [g H 2 O]   1 kg  1000 J    kg 

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.

%Difference 

 308   308.1 100 %  0.03%  308.0

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2.24 (a) Referring to the energy balance for closed systems where kinetic and potential energy are neglected, Equation 2.13a states

U  Q  W (b) Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal U  0

According to Equation 2.38  P   n1RT1 ln  2   P1

P W  nRT ln  2  P1

  

From the ideal gas law: n1RT1  P1V1 P  W  P1V1 ln 2   P1 

Substitution of the values from the problem statement yields







 3 3  5 bar  5 W  8  10 Pa 2.5 10 m ln   8 bar  W   940 J 

The energy balance is 0 J  Q W  Q  940 J  (c) Since the process is adiabatic

Q0 The energy balance reduces to U  W

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The system must do work on the surroundings to expand. Therefore, the work will be negative and U  0 T2

U  n  cv T  0 T1

 T2  T1 T2 will be less than 30 ºC

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2.25 (a) (i). 1

Path B

Pa A th

P [bar]

3

2

2 1

0.01

0.02 0.03 v [m3 /mol]

(ii). Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal u  0

Since that enthalpy is a function of temperature only for an ideal gas. Therefore, h  0

Performing an energy balance and neglecting potential and kinetic energy produces u  q  w  0 For an isothermal, adiabatic process, Equation 2.38 states P  W  nRT ln 2   P1 

or w

P W  RT ln  2 n  P1

  

Substituting the values from the problem statement gives

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  J   1 bar  w   8.314   (88  273.15) K  ln    mol  K    3 bar    J  w  3299   mol

Using the energy balance above  J  q  w  3299  mol 

(b) (i).

See path on diagram in part (a)

(ii). Since the overall process is isothermal and u and h are state functions u 0

h0

The definition of work is w    PE dv 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 w   PE  dv   PE (v 2  v1 )

The ideal gas law can be used to solve for v2 and v1  J  mol    8 .314   (88  273.1...