Thermodynamics Solution

P 3000atm D  0.17in A S 4 D 2 ˜  A 0.023 in2 F P A˜ g 32.174 ft sec2 mass F g  mass 1000.7 lb_m Ans. 1.7 Pabs = U g˜ h˜ Patm U 13.535 gm cm3 ˜  g 9.832 m s2 ˜  h  56.38cm

P_atm 101.78kPa P_abs  U g˜ h˜ P_atm P_abs 176.808 kPa Ans.

1.8 U 13.535 gm cm3 ˜  g 32.243 ft s2 ˜  h  25.62in

P_atm 29.86in_Hg P_abs  U g˜ h˜ P_atm P_abs 27.22 psia Ans.

Chapter 1 - Section A - Mathcad Solutions

1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C).

Guess solution: t 0

Given t= 1.8t32 Find t() 40 Ans.

1.5 By definition: P F A

= F = mass g˜ Note: Pressures are in gauge pressure. P  3000bar D  4mm A S 4 D 2 ˜  A 12.566 mm2 F  P A˜ g 9.807m s2 mass F g  mass 384.4 kg Ans. 1.6 By definition: P F A = F = mass g˜

F_Mars  K x˜ F_Mars 4u103m K g_Mars FMars mass  g_Mars 0.01m K kg Ans. 1.12 Given: z P d d U  g˜ = and: U M P˜ R T˜ = Substituting: z P d d M P˜ R T˜  ˜g =

Separating variables and integrating: P_sea P_Denver P 1 P ´ µ µ ¶ d 0 z_Denver z M g˜ R T˜

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 ´ µ µ ¶ d =

After integrating: ln PDenver P_sea

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M  ˜g R T˜ ˜zDenver =

Taking the exponential of both sides

and rearranging: _P Denver Psea e M  ˜g R T˜ ˜zDenver §¨ © ·¹ ˜ = P_sea  1atm M 29 gm mol  g 9.8m s2  1.10 Assume the following: U 13.5 gm

cm3  g 9.8m s2  P  400bar h P U g˜  h 302.3 m Ans.

1.11 The force on a spring is described by: F = K_s x where K_s is the spring constant. First calculate K based on the earth measurement then g_Mars based on spring measurement on Mars.

On Earth: F ₌ mass g˜ ₌ K x˜ mass  0.40kg g 9.81m s2  x  1.08cm F  mass g˜ F 3.924 N K_s F x  K_s 363.333N m On Mars: x 0.40cm

Ans. w_moon  M g˜ _moon w_moon 18.767 lbf Ans.

1.14 cost_bulb 5.00dollars 1000hr ˜10

hr day

 cost_elec 0.1dollars

kW hr˜ ˜10 hr day˜ W70 

cost_bulb 18.262dollars

yr costelec 25.567

dollars yr cost_total  cost_bulbcost_elec cost_total 43.829dollars

yr Ans. 1.15 D  1.25ft mass  250lb_m g 32.169ft s2  R 82.06cm 3 atm ˜ mol K˜  T  (10273.15)K z_Denver 1 mi˜ M g˜ R T˜ ˜zDenver 0.194 P_Denver P_sea e M  ˜g R T˜ ˜zDenver §¨ © ·¹ ˜

 P_Denver 0.823 atm Ans.

P_Denver 0.834 bar Ans.

1.13 The same proportionality applies as in Pb. 1.11. g_earth 32.186 ft s2 ˜  g_moon 5.32 ft s2 ˜  'l_moon 18.76

'l_earth 'l_moon gearth g_moon ˜

 'l_earth 113.498

Ans. (b) P_abs F

A

 P_abs 110.054 kPa Ans.

(c) 'l 0.83m Work F 'l˜ Work 15.848 kJ Ans. 'E_P  mass g˜' l˜ 'E_P 1.222 kJ Ans. 1.18 mass 1250kg u 40m s  E_K 1 2mass u 2 ˜  E_K 1000 kJ Ans.

Work E_K Work 1000 kJ Ans.

1.19 Wdot mass g˜' h˜ time ˜0.91˜0.92 = Wdot 200W g 9.8m s2  'h  50m P_atm 30.12in_Hg A S 4 D 2 ˜  A 1.227 ft2

(a) F P_atm˜Amass g˜ F 2.8642u103lb_f Ans.

(b) P_abs F A

 P_abs 16.208 psia Ans.

(c) 'l 1.7ft Work F 'l˜ Work 4.8691u103ft lb˜ _f Ans. 'P_E  mass g˜' l˜ 'P_E 424.9 ft lb˜ _f Ans. 1.16 D  0.47m mass  150kg g 9.813m s2  P_atm 101.57kPa A S 4 D 2 ˜  A 0.173 m2

mdot Wdot g 'h˜ ˜0.91˜0.92  mdot 0.488kg s Ans. 1.22 a) cost_coal 25.00 ton 29 MJ kg ˜  cost_coal 0.95 GJ1 cost_gasoline 2.00 gal 37 GJ m3 ˜  cost_gasoline 14.28 GJ1 cost_electricity 0.1000 kW hr˜  cost_electricity 27.778 GJ1

b) The electrical energy can directly be converted to other forms of energy whereas the coal and gasoline would typically need to be converted to heat and then into some other form of energy before being useful.

The obvious advantage of coal is that it is cheap if it is used as a heat source. Otherwise it is messy to handle and bulky for tranport and storage.

Gasoline is an important transportation fuel. It is more convenient to transport and store than coal. It can be used to generate electricity by burning it but the efficiency is limited. However, fuel cells are currently being developed which will allow for the conversion of gasoline to electricity by chemical means, a more efficient process.

Electricity has the most uses though it is expensive. It is easy to transport but expensive to store. As a transportation fuel it is clean but batteries to store it on-board have limited capacity and are heavy.

1.24 Use the Matcad genfit function to fit the data to Antoine's equation. The genfit function requires the first derivatives of the function with respect to the parameters being fitted.

Function being fit: f T A(  BC) e

A B T C  §¨ © ·¹ 

First derivative of the function with respect to parameter A

A f T A(  BC) d d exp A B TC 

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First derivative of the function with respect to parameter B

B f T A(  BC) d d 1  TC exp A B TC 

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First derivative of the function with respect to parameter C

C f T A(  BC) d d B TC ( )2 exp A B TC 

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˜ o t 18.5  9.5  0.2 11.8 23.1 32.7 44.4 52.1 63.3 75.5

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 Psat 3.18 5.48 9.45 16.9 28.2 41.9 66.6 89.5 129 187

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T  t273.15 lnPsat ln Psat( )

Array of functions used by Mathcad. In this case, a₀ = A, a₁ = B and a₂ = C.

Guess values of parameters

F T a(  ) exp a₀ a1 Ta₂ 

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exp a₀ a1 Ta₂ 

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1  Ta₂ exp a0 a₁ Ta₂ 

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˜ a₁ Ta₂



2 exp a0 a₁ Ta₂ 

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 guess 15 3000 50 

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Apply the genfit function A B C

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genfit T Psat(  guessF)  A B C

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13.421 2.29u103 69.053 

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Ans.

Compare fit with data.

240 260 280 300 320 340 360 0 50 100 150 200 Psat f T A(  BC) T

This is an open-ended problem. The strategy depends on age of the child, and on such unpredictable items as possible financial aid, monies earned by the child, and length of time spent in earning a degree.

c)

The salary of a Ph. D. engineer over this period increased at a rate of 5.5%, slightly higher than the rate of inflation.

i 5.511 % i Find i( ) C₂ C₁ (1i) t₂t₁ = Given C₂ 80000dollars yr  C₁ 16000dollars yr  t₂  2000 t₁  1970 b)

The increase in price of gasoline over this period kept pace with the rate of inflation. C₂ 1.513dollars gal C₂  C₁˜(1i)t2t1 i 5% C₁ 0.35dollars gal  t₂  2000 t₁  1970 a) 1.25 T_nb273.15K 56.004 degC Ans. T_nb 329.154 K T_nb B A ln Psat kPa

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K ˜  Psat  1atm

t₁  20 degC˜ C_P 4.18 kJ kg degC˜ ˜  M_H2O 30 kg˜ t₂ t₁ 'U_total MH2O˜CP   t₂ 20.014 degC Ans.

(d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus

Q  'U_total Q 1.715kJ Ans.Ans.

(e) In all cases the total internal energy change of the universe is zero.

2.2 Similar to Pb. 2.1 with mass of water = 30 kg. Answers are: (a) W = 1.715 kJ

(b) Internal energy change of the water = 1.429 kJ

(c) Final temp. = 20.014 deg C (d) Q = -1.715 kJ

Chapter 2 - Section A - Mathcad Solutions

2.1 (a) M_wt 35 kg˜ g 9.8 m s2 ˜

 'z  5 m˜

Work M_wt˜' zg˜ Work 1.715 kJ Ans.

(b) 'U_total Work 'U_total 1.715 kJ Ans.

(c) By Eqs. (2.14) and (2.21): dUd PV( )₌ C_P˜dT Since P is constant, this can be written:

M_H2O˜C_P˜dT ₌ M_H2O˜dUM_H2O˜ dVP˜

Q₃₄  800J W₃₄ 300J

'Ut₃₄ Q₃₄W₃₄ 'Ut₃₄ 500J Ans.

Step 1 to 2 to 3 to 4 to 1: Since 'Ut is a state function, 'Ut for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the

'Ut values for all of the steps must sum to zero.

'Ut₄₁ 4700J 'Ut₂₃ ''Ut₁₂'Ut₃₄ Ut₄₁

'Ut₂₃ 4000J Ans.

Step 2 to 3: 'Ut₂₃ 4u103J Q₂₃  3800J

W₂₃ 'Ut₂₃Q₂₃ W₂₃ 200J Ans.

For a series of steps, the total work done is the sum of the work done for each step.

W₁₂₃₄₁  1400J

2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor.

i 9.7amp E 110V Wdot_mech  1.25hp Wdot_elect  iE˜ Wdot_elect 1.067u103W

Qdot Wdot_electWdot_mech Qdot 134.875 W Ans.

2.5 Eq. (2.3): 'Ut ₌ QW

Step 1 to 2: 'Ut₁₂  200J W₁₂  6000J

Q₁₂ 'Ut₁₂W₁₂ Q₁₂ 5.8u103J Ans.

'U  12˜kJ Q  'U Q  kJ12 Ans.

2.13Subscripts: c, casting; w, water; t, tank. Then m_c˜'U_cm_w˜'U_wm_t˜'U_{t =} 0

Let C represent specific heat, C ₌ C_{P =} C_V Then by Eq. (2.18) m_c˜'C_c˜ t_cm_w˜'C_w˜ t_wm_t˜'C_t˜ t_{t =} 0 m_c  2 kg˜ m_w  40 kg˜ m_t  5 kg˜ C_c 0.50 kJ kg degC˜ ˜  C_t 0.5 kJ kg degC˜ ˜  C_w 4.18 kJ kg degC˜ ˜ 

t_c  500 degC˜ t₁  25 degC˜ t₂ 30 degC˜ (guess) Given m_c˜C_c˜



t₂t_{c =}



m_w˜C_wm_t˜C_t ˜



t₂t₁

t₂  Find t



₂ t₂ 27.78 degC Ans. W₄₁ W₁₂₃₄₁W₁₂W₂₃W₃₄ W₄₁ 4.5u103J Ans. Step 4 to 1: 'Ut₄₁  4700J W₄₁ 4.5u103J

Q₄₁ 'Ut₄₁W₄₁ Q₄₁ 200 J Ans. Note: Q_{12341 =} W₁₂₃₄₁

2.11 The enthalpy change of the water = work done.

M  20 kg˜ C_P 4.18 kJ kg degC˜ ˜  't  10 degC˜ Wdot 0.25 kW˜ 'W M C˜'P˜ t Wdot  'W 0.929 hr Ans. 2.12 Q  7.5 kJ˜ 'U  12˜kJ W 'U Q W 19.5kJ Ans.

A 3.142 m2 mdot U u˜ A˜ mdot 1.571u104kg

s

Wdot mdot g˜' z˜ Wdot 7.697u103kW Ans.

2.18 (a) U1 762.0 kJ kg ˜  P1 1002.7 kPa˜ V1 1.128 cm3 gm ˜  H₁  U₁P₁˜V₁ H₁ 763.131kJ kg Ans. (b) U₂ 2784.4 kJ kg ˜  P₂  1500 kPa˜ V₂ 169.7 cm 3 gm ˜  H₂  U₂P₂˜V₂ 'U  U₂U₁ 'H  H₂H₁ 'U 2022.4kJ kg Ans. 'H 2275.8 kJ kg Ans. 2.15 mass  1 kg˜ C_V 4.18 kJ kg K˜ 

(a) 'T  1K 'Ut  mass C˜'_V˜ T 'Ut 4.18 kJ Ans.

(b) g 9.8m s2  'E_P 'Ut 'z 'EP mass g˜  _'z 426.531 m Ans. (c) 'E_K 'Ut u 'E_K 1 2˜mass  u 91.433m s Ans. 2.17 'z  50m U 1000kg m3  u 5m s  D  2m A S 4D 2 

mdot C˜ _p˜



T₃T₁ mdot₂˜C_P˜



T₃T_{2 =} Qdot T₃˜C_P˜



mdot₁mdot_{2 =} Qdotmdot₁˜C_P˜T₁mdot₂˜C_P˜T₂ mdot₁ 1.0kg s  T₁ 25degC mdot₂ 0.8kg s  T₂  75degC C_P 4.18 kJ kg K˜  Qdot 30kJ s 

T₃ Qdotmdot1˜CP˜T1mdot2˜CP˜T2 mdot₁mdot₂



˜C_P

 T₃ 43.235 degC Ans.

2.25By Eq. (2.32a): 'H 'u

2 2  ₌ 0 'H ₌ C_P˜'T By continuity, incompressibility u2 u1 A₁ A₂ ˜ = C_P 4.18 kJ kg degC˜ ˜  2.22 D₁  2.5cm u₁ 2m s  D₂  5cm

(a) For an incompressible fluid, U=constant. By a mass balance, mdot = constant = u₁A₁U = u₂A₂U

u₂ u₁ D1 D₂

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2 ˜  u₂ 0.5m s Ans. (b) 'E_K 1 2u2 2 1 2u1 2   'E_K 1.875 J kg Ans.

2.23 Energy balance: mdot₃˜H₃



mdot₁˜H₁mdot₂˜H_{2 =} Qdot Mass balance: mdot₃mdot₁mdot_{2 =} 0

Therefore: mdot₁˜



H₃H₁ mdot₂˜



H₃H_{2 =} Qdot or

u₂ 3.5m s

 molwt 29 kg

kmol 

Wsdot  98.8kW ndot 50kmol hr  C_P 7 2˜R  'H  C_P˜



T₂T₁ 'H 6.402u103 kJ kmol By Eq. (2.30): Qdot 'H u2 2 2 u₁2 2 

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˜ndotWsdot  Qdot 9.904kW Ans. 2.27By Eq. (2.32b): 'H 'u 2 2 g˜ _c  = also V₂ V₁ T₂ T₁ P₁ P₂ ˜ = By continunity, constant area u2 u1 V₂ V₁ ˜ = u₂ u₁ T2 T₁ ˜ P1 P₂ ˜ = 'u2 ₌ u₂2u₁2 'u2 u₁2 A1 A₂

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˜ = 'u2 u₁2 D1 D₂

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˜ = SI units: u₁ 14 m s ˜  D₁  2.5 cm˜ D₂  3.8 cm˜ 'T u1 2 2 C˜ _P 1 D₁ D₂

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˜  'T 0.019 degC Ans. D₂  7.5cm 'T u1 2 2 C˜ _P 1 D₁ D₂

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˜  'T 0.023 degC Ans.

Maximum T change occurrs for infinite D2: D₂  f cm˜ 'T u1 2 2 C˜ _P 1 D₁ D₂

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˜  'T 0.023 degC Ans. 2.26 T1 300K T2 520K u1 10 m s 

H₂ 2726.5 kJ kg ˜  By Eq. (2.32a): Q H₂H₁ u2 2 u₁2  2   Q 2411.6kJ kg Ans. 2.29 u1 30 m s ˜  H₁ 3112.5 kJ kg ˜  H₂ 2945.7 kJ kg ˜  u₂ 500 m s ˜  (guess)

By Eq. (2.32a): Given H₂H₁ u1 2 u₂2  2 = u₂  Find u



₂ u₂ 578.36m s Ans. D₁  5 cm˜ V₁ 388.61 cm 3 gm ˜  V₂ 667.75 cm 3 gm ˜  'H ₌ C_P˜'T 7 2˜R˜



T2T1 = 'u2 u₁2 T2 T₁ P₁ P₂ ˜

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˜ = P₁ 100 psi˜ P₂ 20 psi˜ u₁ 20 ft s ˜  T₁  579.67 rankine˜ R 3.407 ft lb˜ f

mol rankine˜ molwt 28 gm mol  T₂  578 rankine˜ (guess) Given 7 2˜R˜



T2T1 u₁2 2  T2 T₁ P₁ P₂ ˜

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₂ T₂ 578.9 rankine Ans. 119.15 degF˜ ( ) 2.28 u1 3 m s ˜  u₂ 200 m s ˜  H₁ 334.9 kJ kg ˜ 

By Eq. (2.23): Q  n C˜ _P˜



t₂t₁ Q 18.62kJ Ans. 2.31 (a) t₁  70 degF˜ t₂  350 degF˜ n 3 mol˜

C_V 5 BTU

mol degF˜ ˜

 By Eq. (2.19):

Q  n C˜ _V˜



t₂t₁ Q 4200 BTU Ans. Take account of the heat capacity of the vessel:

m_v  200 lb˜ _m c_v 0.12 BTU lb_m˜degF ˜



Q 



m_v˜c_vn C˜ _V ˜



t₂t₁ Q 10920 BTU Ans. (b) t₁  400 degF˜ t₂  150 degF˜ n  4 mol˜ Continuity: D2 D1

u₁˜V₂ u₂˜V₁ ˜

 D₂ 1.493 cm Ans.

2.30 (a) t₁  30 degC˜ t₂ 250 degC˜ n  3 mol˜

C_V 20.8 J

mol degC˜ ˜



By Eq. (2.19): Q  n C˜ _V˜



t₂t₁ Q 13.728 kJ Ans. Take into account the heat capacity of the vessel; then

m_v  100 kg˜ c_v 0.5 kJ kg degC˜ ˜



Q 



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t₂t₁ Q 11014 kJ Ans. (b) t₁ 200 degC˜ t₂ 40 degC˜ n 4 mol˜

C_P 29.1 joule mol degC˜ ˜

Wdot W_s˜mdot Wdot 39.52 hp Ans. 2.34 H₁ 307 BTU lb_m ˜  H₂ 330 BTU lb_m ˜  u₁ 20 ft s ˜  molwt 44 gm mol ˜  V₁ 9.25 ft 3 lb_m ˜  V₂ 0.28 ft 3 lb_m ˜  D₁ 4 in˜ D₂ 1 in˜ mdot S 4 D1 2 ˜ ˜u₁ V₁  mdot 679.263lb hr u₂ mdot V2 S 4 D2 2 ˜ ˜  u₂ 9.686 ft sec Ws 5360 BTU lbmol ˜  Eq. (2.32a): Q H₂H₁ u2 2 u₁2  2  Ws molwt   Q 98.82BTU lb_m C_P 7 BTU mol degF˜ ˜  By Eq. (2.23): Q n C˜ _P˜



t₂t₁ Q 7000BTU Ans. 2.33 H₁ 1322.6 BTU lb_m ˜  H₂ 1148.6 BTU lb_m ˜  u₁ 10 ft s ˜  V₁ 3.058 ft 3 lb_m ˜  V₂ 78.14 ft 3 lb_m ˜  D₁  3 in˜ D₂ 10 in˜ mdot 3.463u104 lb sec mdot S 4 D1 2 ˜ ˜u₁ V₁  u₂ mdot V2 S 4 D2 2 ˜ ˜  u₂ 22.997 ft sec Eq. (2.32a): W_s H₂H₁ u2 2 u₁2  2   W_s 173.99BTU lb

'H 17.4 kJ mol Ans. Q  n 'H˜ Q 602.08 kJ Ans. 'U QW n  'U 12.41 kJ mol Ans.

2.37 Work exactly like Ex. 2.10: 2 steps, (a) & (b). A value is required for PV/T, namely R.

T₁  293.15 K˜ T₂  333.15 K˜ R 8.314 J

mol K˜ _{P1 1000 kPa˜ P2  100 kPa˜} (a) Cool at const V1 to P2

(b) Heat at const P2 to T2 CP 7 2˜R  C_V 5 2˜R  T_a2 T₁ P2 P₁ ˜  T_a2 29.315 K

Qdot mdot Q˜ Qdot 67128BTU

hr Ans. 2.36 T₁  300 K˜ P  1 bar˜ n 1 kg˜ 28.9 gm mol ˜  n 34.602 mol V₁ 83.14 bar cm 3 ˜ mol K˜ ˜ T1 P ˜  V₁ 24942cm 3 mol W n V₁ V₂ V P ´ µ ¶ d ˜ = ₌ n P˜ ˜



V₁V_{2 =} n P˜ ˜



V₁3 V˜ ₁ Whence W   Pn˜ 2˜ V˜ ₁ W 172.61kJ Ans. Given: T₂ T₁ V2 V₁ ˜ = ₌ T₁˜3 Whence T₂  3 T˜ ₁ C_P 29 joule mol K˜ ˜  'H  C_P˜



T₂T₁

Re 22133 55333 110667 276667

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Re D U˜ u˜ P o   u 1 1 5 5

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cm  Note:HD = H/D in this solution HD  0.0001 P 9.0 10˜ 4 kg m s˜  U 996 kg m3  2.39 Ans. 'H 1.164 kJ mol 'H  'H_a'H_b Ans. 'U 0.831 kJ mol 'U  'U_a'U_b 'U_b 6.315u103 J mol 'U_b 'H_bP₂˜



V₂V₁ 'H_a 7.677u103 J mol 'H_a 'U_aV₁˜



P₂P₁ V₂ 0.028 m 3 mol V₂ R T˜ 2 P₂  V₁ 2.437u103 m 3 mol V₁ R T˜ 1 P₁  'U_a 5.484u103 J mol 'U_a C_V˜'T_a 'H_b 8.841u103 J mol 'H_b C_P˜'T_b 'T_a 263.835K 'T_a T_a2T₁ 'T_b 303.835 K 'T_b T₂T_a2

Ans. Cost 799924 dollars Cost 15200 Wdot kW

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0.573 ˜  Wdot 1.009u103kW Wdot mdot H˜



₂H₁

Assume that the compressor is adiabatic (Qdot = 0). Neglect changes in KE and PE. H₂ 536.9 kJ kg ˜  H₁ 761.1kJ kg  mdot 4.5kg s  2.42 Ans. 'P'L 0.632  0.206  11.254  3.88 

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kPa m 'P'L 2 D ˜U ˜ ufF 2 ˜

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o   Ans. mdot 0.313 1.956 1.565 9.778

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kg s mdot U u˜ S 4 ˜ D2

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o   f_F 0.00635 0.00517 0.00452 0.0039

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f_F 0.3305 ln 0.27 HD˜ 7 Re

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0.9 

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a bit of algebra leads to Work c P₁ P₂ P P Pb ´ µ µ ¶ d ˜  Work 0.516 J gm Ans. Alternatively, formal integration leads to

Work c P₂P₁ b ln P2b P1b

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˜ 

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˜  Work 0.516 J gm Ans. 3.5 N ₌ ab P˜ a 3.9 10˜ 6˜atm1 b 0.1˜109˜atm2 P₁ 1 atm˜ P₂ 3000 atm˜ V  1 ft˜ 3 (assume const.)

Combine Eqs. (1.3) and (3.3) for const. T:

Work V P₁ P₂ P ab P˜ ( )P˜ ´ µ ¶ d ˜

 Work 16.65 atm ft˜ 3 Ans.

Chapter 3 - Section A - Mathcad Solutions

3.1 E 1 U T U d d

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˜ = N 1 U P U d d

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˜ = P T

At constant T, the 2nd equation can be written: dU U = N dP˜ ln U₂ U₁

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= N' P˜ N 44.1810 6  ˜ ˜bar1  U₂₌ 1.01˜U₁ 'P ln 1.01( ) N

 'P 225.2 bar P₂₌ 226.2 bar˜ Ans.

3.4 b  2700 bar˜ c 0.125 cm 3 gm ˜  P₁ 1 bar˜ P₂  500 bar˜ Since Work V₁ V₂ V P ´ µ ¶ d  =

P₂ 1 bar˜ T₁ 600 K˜ C_P 7 2˜R

 C_V 5

2˜R 

(a) Constant V: W₌ 0 and 'U ₌ Q ₌ C_V˜'T T₂ T₁ P2 P1 ˜  'T  T₂T₁ 'T 525K 'U  C_V˜'T Q and 'U 10.91 kJ mol Ans. 'H  C_P˜'T 'H 15.28 kJ mol Ans. (b) Constant T: 'U ₌ 'H ₌ 0 and Q ₌ W Work R T˜ ₁ln P2 P₁

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˜  Q and Work 10.37 kJ mol Ans. (c) Adiabatic: Q ₌ 0 and 'U ₌ W ₌ C_V˜'T 3.6 E  1.2 10˜ 3˜degC1 C_P 0.84 kJ kg degC˜ ˜  M 5 kg˜ V₁ 1 1590 m3 kg ˜

 P  1 bar˜ t₁  0 degC˜ t₂  20 degC˜ With beta independent of T and with P=constant,

dV

V = E dT˜ V2  V1˜exp E t

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'V  V2V1 'V_total  M 'V˜ 'V_total 7.638u105m3 Ans.

Work ' VP˜ _total (Const. P) Work 7.638joule Ans. Q  M C˜ _P˜



t₂t₁ Q 84 kJ Ans.

'H_total Q 'H_total 84 kJ Ans.

'U_total QWork 'U_total 83.99 kJ Ans.

3.8

Step 41: Adiabatic T₄ T₁ P4 P₁

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R C_P ˜  T₄ 378.831 K 'U₄₁  C_V˜



T₁T₄ 'U₄₁ 4.597u103 J mol 'H₄₁ C_P˜



T₁T_{4 'H} 41 6.436 10 3 u J mol Q₄₁ 0 J mol  Q₄₁ 0 J mol W₄₁ 'U₄₁ W₄₁ 4.597u103 J mol P₂ 3bar T₂  600K V₂ R T˜ 2 P₂  V₂ 0.017 m 3 mol Step 12: Isothermal 'U₁₂ 0 J mol  'U₁₂ 0 J mol 'H₁₂ 0 J mol ˜  _'H 12 0 J mol J CP C_V  T₂ T₁ P2 P₁

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J 1 J ˜  T₂ 331.227 K 'T  T₂T₁ 'U  C_V˜'T 'H  C_P˜'T W and 'U 5.586 kJ mol Ans. 'H 7.821 kJ mol Ans. 3.9 P₄ 2bar C_P 7 2R  CV 5 2R  P₁ 10bar T₁  600K V₁ R T˜ 1 P₁  V₁ 4.988u103 m 3 mol

Step 34: Isobaric 'U₃₄  C_V˜



T₄T₃ 'U₃₄ 439.997 J mol 'H₃₄  C_P˜



T₄T₃ 'H₃₄ 615.996 J mol Q₃₄  C_P˜



T₄T₃ Q₃₄ 615.996 J mol W₃₄ R˜



T₄T₃ W₃₄ 175.999 J mol 3.10 For all parts of this problem: T_{2 =} T₁ and

'U ₌ 'H ₌ 0 Also Q ₌ Work and all that remains is

to calculate Work. Symbol V is used for total volume in this problem. P₁ 1 bar˜ P₂  12 bar˜ V₁ 12 m˜ 3 V₂ 1 m˜ 3 Q₁₂  TR˜ ₁ln P2 P₁

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˜  Q₁₂ 6.006u103 J mol W₁₂ Q₁₂ W₁₂ 6.006u103 J mol P₃ 2bar V₃  V₂ T₃ P3˜V3 R  T₃ 400 K Step 23: Isochoric 'U₂₃ C_V˜



T₃T₂ 'U₂₃ 4.157u103 J mol 'H₂₃ C_P˜



T₃T₂ 'H₂₃ 5.82u103 J mol Q₂₃  C_V˜



T₃T₂ Q₂₃ 4.157u103 J mol W₂₃ 0 J mol  W₂₃ 0 J mol P₄ 2 bar T₄ 378.831 K V₄ R T˜ 4 P₄  V₄ 0.016 m 3 mol

P_i P₁ V1 V₂

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J ˜  (intermediate P) P_i 62.898 bar W₁ P_i˜V₂P₁˜V₁ J 1  W₁ 7635 kJ

Step 2: No work. Work W₁ Work 7635 kJ Ans.

(d) Step 1: heat at const V₁ to P₂ W₁₌ 0 Step 2: cool at const P₂ to V₂

W₂ P₂˜



V₂V₁ Work W₂ Work 13200 kJ Ans. (e) Step 1: cool at const P₁ to V₂

W₁ P₁˜



V₂V₁ W₁ 1100 kJ (a) Work n R˜ T˜ ln P₂ P₁

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˜ = Work P₁˜V₁ ln P₂ P₁

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˜  Work 2982 kJ Ans. (b) Step 1: adiabatic compression to P₂

J 5 3  V_i V₁ P₁ P2

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1 J ˜  (intermediate V) V_i 2.702 m3 W₁ P₂˜V_iP₁˜V₁ J 1  W₁ 3063 kJ

Step 2: cool at const P₂ to V₂

W₂ P₂˜



V₂V_i W₂ 2042 kJ Work W₁W₂

Work 5106 kJ Ans. (c) Step 1: adiabatic compression to V₂

P₁ 100 kPa˜ P₂ 500 kPa˜ T₁ 303.15 K˜ C_P 7 2˜R  C_V 5 2˜R  J CP C_V 

Adiabatic compression from point 1 to point 2:

Q₁₂ 0 kJ mol ˜  'U₁₂₌ W₁₂₌ C_V˜'T₁₂ T₂ T₁ P2 P₁

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J 1 J ˜  'U₁₂  C_V˜



T₂T₁ 'H₁₂  C_P˜



T₂T₁ W₁₂ 'U₁₂ 'U₁₂ 3.679 kJ mol 'H12 5.15 kJ mol W12 3.679 kJ mol Ans. Cool at P₂ from point 2 to point 3:

T₃  T₁ 'H₂₃ C_P˜



T₃T₂ Q₂₃ 'H₂₃ 'U₂₃ C_V˜



T₃T₂ W₂₃ 'U₂₃Q₂₃

Step 2: heat at const V₂ to P₂ W₂₌ 0

Ans.

Work W₁ Work 1100 kJ

3.17(a) No work is done; no heat is transferred.

'Ut ₌ 'T ₌ 0 T_{2 =} T_{1 =} 100 degC˜ Not reversible (b) The gas is returned to its initial state by isothermal compression.

Work n R˜ T˜ ln V1 V₂

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˜ = but n R˜ T˜ ₌ P₂˜V₂ V₁  4 m˜ 3 V₂ 4 3 m 3 ˜  P₂ 6 bar˜ Work P₂˜V₂ ln V1 V₂

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˜  Work 878.9 kJ Ans. 3.18 (a)

Work 1.094 kJ mol

(b) If each step that is 80% efficient accomplishes the same change of state, all property values are unchanged, and the delta H and delta U values are the same as in part (a). However, the Q and W values change.

Step 12: W₁₂ W12 0.8  W₁₂ 4.598 kJ mol Q₁₂  'U₁₂W₁₂ Q₁₂ 0.92 kJ mol Step 23: W₂₃ W23 0.8  W₂₃ 1.839 kJ mol Q₂₃  'U₂₃W₂₃ Q₂₃ 5.518 kJ mol Step 31: W₃₁ W₃₁˜0.8 W₃₁ 3.245 kJ mol Q₃₁  W₃₁ Q₃₁ 3.245 kJ mol 'H₂₃ 5.15 kJ mol 'U23 3.679 kJ mol Ans. Q₂₃ 5.15 kJ mol W23 1.471 kJ mol Ans.

Isothermal expansion from point 3 to point 1:

'U₃₁₌ 'H₃₁₌ 0 P₃  P₂ W₃₁ R T˜ ₃ ln P1 P₃

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˜  Q₃₁  W₃₁ W₃₁ 4.056 kJ mol Q31 4.056 kJ mol Ans.

FOR THE CYCLE: 'U ₌ 'H ₌ 0

Q  Q₁₂Q₂₃Q₃₁ Work W₁₂W₂₃W₃₁

Q 1.094 kJ mol

(b) Adiabatic: P₂ P₁ V1 V₂

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J ˜  T₂ T₁ P2 P₁ ˜ V2 V₁ ˜  T₂ 208.96 K P₂ 69.65 kPa Ans. Work P2˜V2P1˜V1 J 1  Work 994.4kJ Ans,

(c) Restrained adiabatic: Work₌ 'U ₌ 'P_ext˜ V

P_ext  100 kPa˜ Work P_ext˜



V₂V₁ Work 400kJ Ans. n P1˜V1 R T˜ ₁  'U ₌ n C˜'_V˜ T T₂ Work n C˜ V T₁   T₂ 442.71 K Ans. P₂ P₁ V1 V2 ˜ T2 T1 ˜  P₂ 147.57 kPa Ans.

FOR THE CYCLE:

Q  Q₁₂Q₂₃Q₃₁ Work W₁₂W₂₃W₃₁

Q 3.192 kJ

mol Work 3.192

kJ mol

3.19Here, V represents total volume.

P₁  1000 kPa˜ V₁ 1 m˜ 3 V₂ 5 V˜ ₁ T₁ 600 K˜ C_P 21 joule mol K˜ ˜  C_V  C_PR J CP C_V 

(a) Isothermal: Work n R˜ T˜ ₁ln V1 V₂

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˜ = P₂ P₁ V1 V₂ ˜  T₂ T₁ T₂ 600 K P₂ 200 kPa Ans. Work P₁˜V₁ ln V1 V₂

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˜  Work 1609kJ Ans.

W₂₃ 0 kJ mol ˜  'U₂₃  C_V˜



T₃T₂ Q₂₃  'U₂₃ 'H₂₃ C_P˜



T₃T₂ Q₂₃ 2.079 kJ mol 'U23 2.079 kJ mol 'H23 2.91 kJ mol

Process: Work W₁₂W₂₃ Work 2.502 kJ

mol Ans. Q  Q₁₂Q₂₃ Q 0.424 kJ mol Ans. 'H  'H₁₂'H₂₃ 'H 2.91 kJ mol Ans. 'U  'U₁₂'U₂₃ 'U 2.079 kJ mol Ans. 3.20 T₁  423.15 K˜ P₁  8bar˜ P₃  3 bar˜ C_P 7 2˜R  C_V 5 2˜R  T₂ T₁ T₃ 323.15 K˜ Step 12: 'H₁₂ 0 kJ mol ˜  'U₁₂ 0 kJ mol ˜  If r V1 V₂ = V₁ V₃ = Then r T1 T₃ P₃ P₁ ˜  W₁₂ R T˜ ₁˜ln r() W12 2.502 kJ mol Q12 W12 Q12 2.502 kJ mol Step 23:

P₁ 1 bar˜ P₃ 10 bar˜ 'U  C_V˜



T₃T₁ 'H  C_P˜



T₃T₁ 'U 2.079 kJ mol Ans. 'H 2.91 kJ mol Ans.

Each part consists of two steps, 12 & 23. (a) T₂  T₃ P₂ P₁ T2 T₁ ˜  W₂₃ R T˜ ₂ln P3 P₂

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˜  Work W₂₃ Work 6.762 kJ mol Ans. Q  'U Work Q 4.684 kJ mol Ans.

3.21 By Eq. (2.32a), unit-mass basis: molwt 28 gm mol  'H 1 2 'u 2 ˜  ₌ 0 But 'H ₌ C_P˜'T Whence 'T u2 2 u₁2 



 2 C˜ _P = C_P 7 2 R molwt ˜  u₁ 2.5 m s ˜  u₂ 50 m s ˜  t₁ 150 degC˜ t₂ t₁ u2 2 u₁2  2 C˜ _P   t₂ 148.8 degC Ans. 3.22 C_P 7 2˜R  C_V 5 2˜R  T₁ 303.15 K˜ T₃ 403.15 K˜

Q₂₃  'H₂₃ 'U₂₃  C_V˜



T₃T₂ W₂₃ 'U₂₃Q₂₃ Work W₁₂W₂₃ Work 4.972 kJ mol Ans. Q  'U Work Q 2.894 kJ mol Ans.

For the second set of heat-capacity values, answers are (kJ/mol): 'U ₌ 1.247 'U ₌ 2.079 (a) Work₌ 6.762 Q ₌ 5.515 (b) Work₌ 6.886 Q ₌ 5.639 (c) Work₌ 4.972 Q ₌ 3.725 (b) P₂ P₁ T₂  T₃ 'U₁₂ C_V˜



T₂T₁ 'H₁₂ C_P˜



T₂T₁ Q₁₂ 'H₁₂ W₁₂ 'U₁₂Q₁₂ W₁₂ 0.831 kJ mol W₂₃ R T˜ ₂ln P3 P₂

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˜  W₂₃ 7.718 kJ mol Work W₁₂W₂₃ Work 6.886 kJ mol Ans. Q  'U Work Q 4.808 kJ mol Ans. (c) T₂  T₁ P₂ P₃ W₁₂ R T˜ ₁ln P2 P₁

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˜  'H₂₃  C_P˜



T₃T₂

For the process: Work W₁₂W₂₃ Q  Q₁₂Q₂₃ Work 5.608 kJ mol Q 3.737 kJ mol Ans. 3.24 W₁₂₌ 0 Work₌ W₂₃₌ P₂



V₃V_{2 =} R˜



T₃T₂

But T_{3 =} T₁ So... Work₌ R T˜



₂T₁ Also W R T˜ ₁ln P P₁

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˜ = Therefore ln P P₁

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T₂T₁ T₁ = T₂ 350 K˜ T₁ 800 K˜ P₁ 4 bar˜ P P₁exp T2T1 T₁

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˜  P 2.279 bar Ans. 3.23 T₁ 303.15 K˜ T₂ T₁ T₃  393.15 K˜ P₁ 1 bar˜ P₃ 12 bar˜ C_P 7 2˜R  C_V 5 2˜R 

For the process: 'U  C_V˜



T₃T₁ 'H  CP˜



T3T1 'U 1.871 kJ mol 'H 2.619 kJ mol Ans. Step 12: P₂ P₃ T1 T₃ ˜  W₁₂ R T˜ ₁ ln P2 P₁

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˜  W₁₂ 5.608 kJ mol Q12  W12 Q12 5.608 kJ mol Step 23: W₂₃ 0 kJ mol ˜  Q₂₃  'U

T_B(final)₌ T_B n_{A =} n_B Since the total volume is constant,

2 n˜ _A˜ TR˜ ₁ P₁ n_A˜R˜



T_AT_B P2 = or 2 T˜ 1 P1 T_AT_B P2 = (1) (a) P₂ 1.25 atm˜ T_B T₁ P2 P₁

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J 1 J ˜  (2) T_A 2 T˜ ₁ P2 P₁ ˜ T_B  Q ₌ n_A˜



'U_A'U_B Define q Q nA = q C_V˜



T_AT_B2 T˜ ₁ (3) T_B 319.75 K T_A 430.25 K q 3.118 kJ mol Ans. 3.25 V_A  256 cm˜ 3 Define: 'P P₁ = r r 0.0639 Assume ideal gas; let V represent total volume:

P₁˜V_B= P₂˜



V_AV_B From this one finds: 'P P₁ V_A  V_AV_B = V_B VA˜(r1) r  V_B 3750.3 cm3 Ans. 3.26 T₁  300 K˜ P₁  1 atm˜ C_P 7 2˜R  C_V  C_PR J CP C_V 

The process occurring in section B is a reversible, adiabatic compression. Let P final( )₌ P₂ T_A(final) T_{= A}

T_A 2 T˜ ₁ P2 P₁ ˜ T_B  (1) T_A 469 K Ans. q C_V˜



T_AT_B2 T˜ ₁ q 4.032 kJ mol Ans. (d) Eliminate T_AT_B from Eqs. (1) & (3):

q 3 kJ mol ˜  P₂ q P˜ 1 2 T˜ 1˜CV P₁   P₂ 1.241 atm Ans. T_B T₁ P2 P1

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J 1 J ˜  (2) T_B 319.06 K Ans. T_A 2 T˜ ₁ P2 P₁ ˜ T_B  (1) T_A 425.28 K Ans.

(b) Combine Eqs. (1) & (2) to eliminate the ratio of pressures: T_A  425 K˜ (guess) T_B 300 K˜ Given T_B T₁ TATB 2 T˜ ₁

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J 1 J ˜ = T_B Find T



_B T_B 319.02 K Ans. P₂ P₁ TATB 2 T˜ 1

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˜  (1) P₂ 1.24 atm Ans. q C_V˜



T_AT_B2 T˜ ₁ q 2.993 kJ mol Ans. (c) T_B 325 K˜ By Eq. (2), P₂ P₁ TB T₁

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J J 1 ˜  P₂ 1.323 atm Ans.

Solve virial eqn. for final V. Guess: V₂ R T˜ P₂  Given P₂˜V₂ R T˜ 1 B V2  C V₂2  = V₂  Find V



₂ V₂ 241.33cm 3 mol Eliminate P from Eq. (1.3) by the virial equation:

Work  TR˜ V₁ V₂ V 1 B V  C V2 

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1 V ˜ ´ µ µ ¶ d ˜  _{Work 12.62} kJ mol Ans.

(b) Eliminate dV from Eq. (1.3) by the virial equation in P:

dV R T˜ 1 P2 C' 

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˜ ˜dP = W  TR˜ P₁ P₂ P 1  P C' P˜ 

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´ µ µ ¶ d ˜  W 12.596 kJ mol Ans. 3.30 B 242.5 cm 3 mol ˜  C 25200 cm 6 mol2 ˜  T  373.15 K˜ P₁ 1 bar˜ P₂ 55 bar˜ B' B R T˜  _{B' 7.817 10}3 u 1 bar C' C B 2  R2˜T2  _{C' 3.492 10}5 u 1 bar2 (a) Solve virial eqn. for initial V.

Guess: V₁ R T˜ P₁  Given P₁˜V₁ R T˜ 1 B V₁  C V₁2  = V₁ Find V



₁ V₁ 30780cm 3 mol

(b) B₀ 0.083 0.422 T_r1.6   B₀ 0.304 B₁ 0.139 0.172 T_r4.2   B₁ 2.262u103 Z 1



B₀Z B˜ ₁ Pr Tr ˜   Z 0.932 V Z R˜ T˜ P  V 1924cm 3 mol Ans. (c) For Redlich/Kwong EOS:

V  1 H  0 :  0.08664 <  0.42748 Table 3.1 D Tr( ) T_r0.5 Table 3.1 q T



_r <D T˜



r : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r T_r  Eq. (3.53)

Note: The answers to (a) & (b) differ because the relations between the two sets of parameters are exact only for infinite series.

3.32 Tc 282.3 K˜ T  298.15 K˜ Tr T T_c  T_r 1.056 P_c 50.4 bar˜ P  12 bar˜ P_r P Pc  P_r 0.238 Z  0.087 (guess) (a) B 140 cm 3 mol ˜  C 7200 cm 6 mol2 ˜  V R T˜ P  V 2066cm 3 mol Given P V˜ R T˜ 1 B V  C V2  = V  Find V( ) V 1919cm 3 mol Z P V˜ R T˜  Z 0.929 Ans.

Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.928 V Z R˜ T˜ P  V 1918cm 3 mol Ans. (e) For Peng/Robinson EOS:

V  1 2 H  1 2 :  0.07779 <  0.45724 Table 3.1 Table 3.1 D Tr Z



 1



0.374641.54226Z 0.26992Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r T_r  Eq. (3.53) Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.928 V Z R˜ T˜ P  V 1916.5cm 3 mol Ans. (d) For SRK EOS: V  1 H  0 :  0.08664 <  0.42748 Table 3.1 Table 3.1 D T



_rZ 1



0.4801.574Z 0.176Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r Tr  Eq. (3.53)

V 1791cm 3 mol Given P V˜ R T˜ 1 B V  C V2  = V  Find V( ) V 1625cm 3 mol Z P V˜ R T˜  Z 0.907 Ans. (b) B₀ 0.083 0.422 T_r1.6   B₀ 0.302 B₁ 0.139 0.172 T_r4.2   B₁ 3.517u103 Z 1



B₀Z B˜ ₁ Pr T_r ˜   Z 0.912 V Z R˜ T˜ P  V 1634cm 3 mol Ans. (c) For Redlich/Kwong EOS:

V  1 H  0 :  0.08664 <  0.42748 Table 3.1 Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.92 V Z R˜ T˜ P  V 1900.6cm 3 mol Ans. 3.33 T_c  305.3 K˜ T  323.15 K˜ T_r T T_c  T_r 1.058 P_c  48.72 bar˜ P  15 bar˜ P_r P P_c  P_r 0.308 Z  0.100 (guess) (a) B 156.7 cm 3 mol ˜  C 9650 cm 6 mol2 ˜  V R T˜ P 

Table 3.1 D T



_rZ 1



0.4801.574Z 0.176Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r Tr  Eq. (3.53) Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.907 V Z R˜ T˜ P  V 1624.8cm 3 mol Ans. (e) For Peng/Robinson EOS:

V  1 2 H  1 2 :  0.07779 <  0.45724 Table 3.1 D Tr( ) T_r0.5 Table 3.1 q T



_r <D T˜



r : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r T_r  Eq. (3.53) Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.906 V Z R˜ T˜ P  V 1622.7cm 3 mol Ans. (d) For SRK EOS: V  1 H  0 :  0.08664 <  0.42748 Table 3.1

P_c 37.6 bar˜ P 15 bar˜ P_r P P_c  P_r 0.399 Z  0.286 (guess) (a) B 194 cm 3 mol ˜  C 15300 cm 6 mol2 ˜  V R T˜ P  V 1930cm 3 mol Given P V˜ R T˜ 1 B V  C V2  = V  Find V( ) V 1722cm 3 mol Z P V˜ R T˜  Z 0.893 Ans. (b) B₀ 0.083 0.422 T_r1.6   B₀ 0.283 Table 3.1 D Tr Z



 1



0.374641.54226Z 0.26992Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r Tr  Eq. (3.53) Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



rPr



˜



ZVE T˜



rPr ˜  = Z  Find Z( ) Z 0.896 V Z R˜ T˜ P  V 1605.5cm 3 mol Ans. 3.34 T_c 318.7 K˜ T  348.15 K˜ T_r T T_c  T_r 1.092

Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.888 V Z R˜ T˜ P  V 1714.1cm 3 mol Ans. (d) For SRK EOS: V  1 H  0 :  0.08664 <  0.42748 Table 3.1 Table 3.1 D T



_rZ 1



0.4801.574Z 0.176Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r T_r  Eq. (3.53) B₁ 0.139 0.172 T_r4.2   B₁ 0.02 Z 1



B₀Z B˜ ₁ Pr T_r ˜   Z 0.899 V Z R˜ T˜ P  V 1734cm 3 mol Ans.

(c) For Redlich/Kwong EOS:

V  1 H  0 :  0.08664 <  0.42748 Table 3.1 D Tr( ) T_r0.5 Table 3.1 q T



_r <D T˜



r : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r T_r  Eq. (3.53) Calculate Z Guess:

Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.882 V Z R˜ T˜ P  V 1701.5cm 3 mol Ans. 3.35 T  523.15 K˜ P  1800 kPa˜ (a) B 152.5 cm 3 mol ˜  C 5800 cm 6 mol2 ˜  V R T˜ P  (guess) Given P V˜ R T˜ 1 B V  C V2  = V  Find V( ) Z P V˜ R T˜  V 2250cm 3 mol Z 0.931 Ans. Calculate Z Guess: Z  0.9 Given Eq. (3.52) Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.895 V Z R˜ T˜ P  V 1726.9cm 3 mol Ans. (e) For Peng/Robinson EOS:

V  1 2 H  1 2 :  0.07779 <  0.45724 Table 3.1 Table 3.1 D Tr Z



 1



0.374641.54226Z 0.26992Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r Tr  Eq. (3.53) Calculate Z

V 2252cm 3 mol Ans. 3.37 B 53.4 cm 3 mol ˜  C 2620 cm 6 mol2 ˜  D 5000 cm 9 mol3 ˜  n  mol T  273.15 K˜ Given P V˜ R T˜ 1 B V  C V2  D V3  = fP V( ) Find V( ) i 0 10 P_i 



101020 i˜ ˜bar V_i R T˜ P_i  (guess) Z_i fP



iVi ˜Pi R T˜  Eq. (3.12) Eq. (3.39) Z1_i 1 B P˜ i R T˜   Eq. (3.38) Z2_i 1 2 1 4 B P˜ _i R T˜    (b) T_c  647.1 K˜ P_c  220.55 bar˜ Z  0.345 T_r T T_c  P_r P P_c  B₀ 0.083 0.422 T_r1.6   T_r 0.808 P_r 0.082 B₀ 0.51 B₁ 0.139 0.172 T_r4.2   B₁ 0.281 Z 1



B₀Z B˜ ₁ Pr T_r ˜   V Z R˜ T˜ P  Z 0.939 V 2268cm 3 mol Ans. (c) Table F.2: molwt 18.015 gm mol ˜  V 124.99 cm 3 gm ˜ ˜molwt  or

Z_i 1 0.953 0.906 0.861 0.819 0.784 0.757 0.74 0.733 0.735 0.743 Z1_i 1 0.953 0.906 0.859 0.812 0.765 0.718 0.671 0.624 0.577 0.53 Z2_i 1 0.951 0.895 0.83 0.749 0.622 0.5+0.179i 0.5+0.281i 0.5+0.355i 0.5+0.416i 0.5+0.469i P_i -10 1·10 20 40 60 80 100 120 140 160 180 200 bar

Note that values of Z from Eq. (3.39) are not physically meaningful for pressures above 100 bar.

0 50 100 150 200 0.5 0.6 0.7 0.8 0.9 1 Z_i Z1_i Z2_i P_i˜bar1

Eq. (3.53)

Calculate Z for liquid by Eq. (3.56) Guess: Z  0.01 Given Z E T



_rP_r



ZHE T˜



_rP_r ˜



ZVE T˜



_rP_r 1E T



rPr Z q T



_r ˜E T



_rP_r

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˜  = Z  Find Z( )Z 0.057 V Z R˜ T˜ P  V 108.1cm 3 mol Ans. Calculate Z for vapor by Eq. (3.52) Guess: Z  0.9

Given Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr Z Z˜



E T



_rP_r ˜  = Z  Find Z( ) Z 0.789 V Z R˜ T˜ P  V 1499.2cm 3 mol Ans. 3.38 (a) Propane: T_c  369.8 K˜ P_c  42.48 bar˜ Z  0.152

T  313.15 K˜ P  13.71 bar˜ T_r T T_c  T_r 0.847 P_r P P_c  P_r 0.323

For Redlich/Kwong EOS:

V  1 H  0 :  0.08664 <  0.42748 Table 3.1 D Tr( ) T_r0.5 Table 3.1 q T



_r <D T˜



r : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r Tr 

Ans. V 1.538u103cm 3 mol V R T˜ P R B˜



0Z B˜ 1 T_c Pc ˜   B₁ 0.207 B₁ 0.139 0.172 T_r4.2   B₀ 0.468 B₀ 0.083 0.422 T_r1.6  

For saturated vapor, use Pitzer correlation:

Ans. V 94.17cm 3 mol V V_cZ_c



1 T r 0.2857 ª¬ º¼ ˜  Z_c 0.276 V_c 200.0 cm 3 mol ˜  T_r 0.847 T_r T T_c  Rackett equation for saturated liquid:

Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole.

R/K, Liq. R/K, Vap. Rackett Pitzer (a) 108.1 1499.2 94.2 1537.8 (b) 114.5 1174.7 98.1 1228.7 (c) 122.7 920.3 102.8 990.4 (d) 133.6 717.0 109.0 805.0 (e) 148.9 1516.2 125.4 1577.0 (f) 158.3 1216.1 130.7 1296.8 (g) 170.4 971.1 137.4 1074.0 (h) 187.1 768.8 146.4 896.0 (i) 153.2 1330.3 133.9 1405.7 (j) 164.2 1057.9 140.3 1154.3 (k) 179.1 835.3 148.6 955.4 (l) 201.4 645.8 160.6 795.8 (m) 61.7 1252.5 53.5 1276.9 (n) 64.1 1006.9 55.1 1038.5 (o) 66.9 814.5 57.0 853.4 (p) 70.3 661.2 59.1 707.8 (q) 64.4 1318.7 54.6 1319.0 (r) 67.4 1046.6 56.3 1057.2 (s) 70.8 835.6 58.3 856.4 (t) 74.8 669.5 60.6 700.5

Eq. (3.53)

Calculate Z for liquid by Eq. (3.56) Guess: Z  0.01 Given Z E T



_rP_r



ZHE T˜



_rP_r ˜



ZVE T˜



_rP_r 1E T



rPr Z q T



_r ˜E T



_rP_r

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˜  = Z  Find Z( ) Z 0.055 V Z R˜ T˜ P  V 104.7cm 3 mol Ans.

Calculate Z for vapor by Eq. (3.52) Guess: Z  0.9 Given Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.78 V Z R˜ T˜ P  V 1480.7cm 3 mol Ans. 3.39 (a) Propane T_c  369.8 K˜ P_c  42.48 bar˜ Z  0.152

T  (40273.15)K˜ T 313.15 K P  13.71 bar˜ T_r T T_c  T_r 0.847 P_r P P_c  P_r 0.323

From Table 3.1 for SRK:

V  1 H  0 :  0.08664 <  0.42748 D T



_rZ 1



0.4801.574Z 0.176Z2 1 T_r 1 2 

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2  q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r Tr 

Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole.

SRK, Liq. SRK, Vap. Rackett Pitzer (a) 104.7 1480.7 94.2 1537.8 (b) 110.6 1157.8 98.1 1228.7 (c) 118.2 904.9 102.8 990.4 (d) 128.5 703.3 109.0 805.0 (e) 142.1 1487.1 125.4 1577.0 (f) 150.7 1189.9 130.7 1296.8 (g) 161.8 947.8 137.4 1074.0 (h) 177.1 747.8 146.4 896.0 (i) 146.7 1305.3 133.9 1405.7 (j) 156.9 1035.2 140.3 1154.3 (k) 170.7 815.1 148.6 955.4 (l) 191.3 628.5 160.6 795.8 (m) 61.2 1248.9 53.5 1276.9 (n) 63.5 1003.2 55.1 1038.5 (o) 66.3 810.7 57.0 853.4 (p) 69.5 657.4 59.1 707.8 (q) 61.4 1296.8 54.6 1319.0 (r) 63.9 1026.3 56.3 1057.2 (s) 66.9 817.0 58.3 856.4 (t) 70.5 652.5 60.6 700.5

Eq. (3.53)

Calculate Z for liquid by Eq. (3.56) Guess: Z  0.01 Given Z E T



_rP_r



ZHE T˜



_rP_r ˜



ZVE T˜



_rP_r 1E T



rPr Z q T



_r ˜E T



_rP_r

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˜  = Z  Find Z( ) Z 0.049 V Z R˜ T˜ P  V 92.2cm 3 mol Ans.

Calculate Z for vapor by Eq. (3.52) Guess: Z  0.6 Given Z 1E T



_rP_r q T



_r ˜E T



_rP_r ZE T



rPr ZHE T˜



_rP_r



˜



ZVE T˜



_rP_r ˜  = Z  Find Z( ) Z 0.766 V Z R˜ T˜ P  V 1454.5cm 3 mol Ans. 3.40 (a) Propane T_c  369.8 K˜ P_c  42.48 bar˜ Z  0.152

T  (40273.15)K˜ T 313.15 K P  13.71 bar˜ T_r T T_c  T_r 0.847 P_r P Pc  P_r 0.323

From Table 3.1 for PR:

D T



_rZ 1



0.374641.54226Z 0.26992Z2 1 T_r 1 2 

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2  V  1 2 H  1 2 :  0.07779 <  0.45724 q T



_r <D T˜



rZ : T˜ _r  Eq. (3.54) E T



_rP_r : P˜ r T_r 

Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole.

PR, Liq. PR, Vap. Rackett Pitzer (a) 92.2 1454.5 94.2 1537.8 (b) 97.6 1131.8 98.1 1228.7 (c) 104.4 879.2 102.8 990.4 (d) 113.7 678.1 109.0 805.0 (e) 125.2 1453.5 125.4 1577.0 (f) 132.9 1156.3 130.7 1296.8 (g) 143.0 915.0 137.4 1074.0 (h) 157.1 715.8 146.4 896.0 (i) 129.4 1271.9 133.9 1405.7 (j) 138.6 1002.3 140.3 1154.3 (k) 151.2 782.8 148.6 955.4 (l) 170.2 597.3 160.6 795.8 (m) 54.0 1233.0 53.5 1276.9 (n) 56.0 987.3 55.1 1038.5 (o) 58.4 794.8 57.0 853.4 (p) 61.4 641.6 59.1 707.8 (q) 54.1 1280.2 54.6 1319.0 (r) 56.3 1009.7 56.3 1057.2 (s) 58.9 800.5 58.3 856.4 (t) 62.2 636.1 60.6 700.5

P_r 2.282 From Tables E.3 & E.4: Z₀ 0.482 Z₁ 0.126

Z  Z₀Z Z˜ ₁ Z 0.493 n P V˜ total Z R˜ T˜

 n 2171 mol

mass n molwt˜ mass 60.898 kg Ans.

3.42 Assume validity of Eq. (3.38).

P₁ 1bar T₁  300K V₁ 23000cm 3 mol  Z₁ P1˜V1 R T˜ ₁  Z₁ 0.922 B R T˜ 1 P₁ ˜



Z11  B 1.942u103cm 3 mol With this B, recalculate at P₂ P₂  5bar

Z₂ 1 B P˜ 2 R T˜ 1   Z₂ 0.611 V₂ R T˜ 1˜Z2 P₂  V₂ 3.046u103cm 3 mol Ans. 3.41 (a) For ethylene, molwt 28.054 gm

mol  T_c 282.3 K˜ P_c 50.40 bar˜ Z  0.087 T  328.15 K˜ P 35 bar˜ Tr T T_c  P_r P P_c  T_r 1.162 P_r 0.694

From Tables E.1 & E.2: Z₀ 0.838 Z₁  0.033

Z  Z₀Z Z˜ ₁ Z 0.841 n 18 kg˜ molwt  V_total Z n˜ R˜ T˜ P  V_total 0.421 m3 Ans. (b) T  323.15 K˜ P 115 bar˜ V_total  0.25 m˜ 3 T_r T T_c  T_r 1.145 P_r P P_c 

P 16 bar˜ T_c 369.8 K˜ P_c 42.48 bar˜ Z  0.152 V_c 200 cm 3 mol ˜  Z_c 0.276 molwt 44.097 gm mol  T_r T T_c  T_r 0.865 P_r P P_c  P_r 0.377 V_liq V_cZ_c



1 T r 0.2857 ª¬ º¼ ˜  V_liq 96.769cm 3 mol

V_tank  0.35 m˜ 3 m_liq 0.8 V˜ tank V_liq molwt  m_liq 127.594 kg Ans. B₀ 0.083 0.422 T_r1.6   B₀ 0.449 B₁ 0.139 0.172 T_r4.2   B₁ 0.177 3.43 T  753.15 K˜ T_c 513.9 K˜ T_r T T_c  T_r 1.466 P  6000 kPa˜ P_c 61.48 bar˜ P_r P P_c  P_r 0.976 Z  0.645 B₀ 0.083 0.422 T_r1.6   B₀ 0.146 B₁ 0.139 0.172 Tr4.2   B₁ 0.104 V R T˜ P



B0Z B˜ 1 ˜R T_c P_c ˜   V 989cm 3 mol Ans.

For an ideal gas: V R T˜ P

 V 1044cm

3 mol

V R T˜ P



B0Z B˜ 1 ˜R T_c P_c ˜   V 9.469u103cm 3 mol m_vap Vvap V molwt  m_vap 98.213 kg Ans. 3.46 (a) T  333.15 K˜ T_c 305.3 K˜ T_r T Tc  T_r 1.091 P  14000 kPa˜ P_c 48.72 bar˜ P_r P P_c  P_r 2.874 Z  0.100 V_total 0.15 m˜ 3 molwt 30.07 gm mol 

From tables E.3 & E.4: Z0  0.463 Z1  0.037 V_vap R T˜ P



B0Z B˜ 1 ˜R T_c Pc ˜   V_vap 1.318u103cm 3 mol m_vap 0.2 V˜ tank V_vap molwt  m_vap 2.341 kg Ans. 3.45 T  298.15 K˜ T_c  425.1 K˜ T_r T T_c  T_r 0.701 P  2.43 bar˜ P_c  37.96 bar˜ P_r P P_c  P_r 0.064 Z  0.200 V_vap 16 m˜ 3 molwt 58.123 gm mol ˜  B₀ 0.083 0.422 T_r1.6   B₀ 0.661 B₁ 0.139 0.172 T_r4.2   B₁ 0.624

Whence T  T_r˜T_c T 391.7 K or 118.5 degC˜ Ans. 3.47 Vtotal  0.15 m˜ 3 T  298.15 K˜ T_c 282.3 K˜ P_c 50.40 bar˜ Z  0.087 molwt 28.054 gm mol  V Vtotal 40 kg˜ molwt

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 P V˜ ₌ P_r˜P_c˜V ₌ Z R˜ T˜ or P_{r =} D Z˜ where D R T˜ P_c˜V  D 4.675 Whence P_{r =} 4.675 Z˜ at T_r T Tc  T_r 1.056 Z  Z₀Z Z˜ ₁ Z 0.459 V Z R˜ T˜ P  V 90.87cm 3 mol m_ethane Vtotal V molwt  m_ethane 49.64 kg Ans. (b) V Vtotal 40 kg˜  P  20000 kPa˜ P V˜ ₌ Z R˜ T˜ ₌ Z R˜ T˜ _r˜T_c or T_r D Z = where D P V˜ R T˜ _c  D 29.548mol kg Whence T_r 0.889 Z = at P_r P P_c  P_r 4.105

This equation giving T_r as a function of Z and Eq. (3.57) in conjunction with Tables E.3 & E.4 are two relations in the same variables which must be satisfied at the given reduced pressure. The intersection of these two relations can be found by one means or another to occur at about:

P_r1 P1 P_c

 P_r1 0.452

V_total  0.35 m˜ 3 Z  0.100

From Tables E.1 & E.2: Z₀  .8105 Z₁  0.0479

Z  Z₀Z Z˜ ₁ Z 0.806 V₁ Z R˜ T˜ 1 P₁  V₁ 908cm 3 mol T₂  493.15 K˜ T_r2 T2 T_c  T_r2 1.615

Assume Eq. (3.38) applies at the final state. B₀ 0.083 0.422 T_r21.6   B₀ 0.113 B₁ 0.139 0.172 T_r24.2   B₁ 0.116 P₂ R T˜ 2 V₁



B₀Z B˜ ₁ ˜R Tc P_c ˜   P₂ 42.68 bar Ans.

This equation giving P_r as a function of Z and Eq. (3.57) in conjunction with Tables E.3 & E.4 are two relations in the same variables which must be satisfied at the given reduced temperature. The intersection of these two relations can be found by one means or another to occur at about:

P_r  1.582 and Z  0.338 P  P_c˜P_r P 79.73 bar Ans.

3.48 mwater  15 kg˜ Vtotal  0.4 m˜ 3 V V_total m_water  V 26.667cm 3 gm Interpolate in Table F.2 at 400 degC to find: P ₌ 9920 kPa˜ Ans.

3.49 T₁  298.15 K˜ T_c  305.3 K˜ T_r1 T1 T_c

 T_r1 0.977

3.51 Basis: 1 mole of LIQUID nitrogen T_n 77.3 K˜ T_c 126.2 K˜ T_r Tn Tc  T_r 0.613 P 1 atm˜ P_c 34.0 bar˜ P_r P P_c  P_r 0.03 Z  0.038 molwt 28.014 gm mol ˜  V_liq 34.7 cm˜ 3 B₀ 0.083 0.422 T_r1.6   B₀ 0.842 B₁ 0.139 0.172 T_r4.2   B₁ 1.209 Z 1



B₀Z B˜ ₁ Pr T_r ˜   Z 0.957 3.50 T  303.15 K˜ T_c 304.2 K˜ T_r T Tc  T_r 0.997

V_total  0.5 m˜ 3 P_c 73.83 bar˜ Z  0.224 molwt 44.01 gm mol ˜  B₀ 0.083 0.422 T_r1.6   B₀ 0.341 B₁ 0.139 0.172 T_r4.2   B₁ 0.036 V Vtotal 10 kg˜ molwt

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 V 2.2u103cm 3 mol P R T˜ V



B₀Z B˜ ₁ ˜R Tc P_c ˜   P 10.863 bar Ans.