Mathcad Solutions for Pressure, Force, and Temperature Calculations

P 3000atm D 0.17in A S

4˜D² A 0.023 in² F P A˜ g 32.174 ft

sec²

mass F

g mass 1000.7 lb_m Ans.

1.7 P_abs = U˜g˜hP_atm U 13.535 gm

cm³

˜ g 9.832 m

s²

˜ h 56.38cm

P_atm 101.78kPa P_abs U˜g˜hP_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˜hP_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˜D² 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 F_Mars

mass g_Mars 0.01m K

kg Ans.

1.12 Given:

z d P d

U ˜g

= and: U M P˜ R T˜

= Substituting:

z d P d

M P˜ R T˜ ˜g

=

Separating variables and integrating:

P_sea P_Denver

1 P P

´µ µ¶

d

0 z_Denver

M g˜ z R T˜

§¨ ©

·

¹

´µ µ¶

= d

After integrating: ln P_Denver

P_sea

§ ¨

©

·

¹

M˜g

R T˜ ˜z_Denver

=

Taking the exponential of both sides

and rearranging: P_Denver P_sea e

M˜g

R T˜ ˜z_Denver

©§¨ ·

˜ 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˜70W cost_bulb 18.262dollars

yr cost_elec 25.567dollars 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 82.06cm³˜atm

mol K˜ T (10273.15)K z_Denver 1 mi˜ M g˜

R T˜ ˜z_Denver 0.194

P_Denver P_sea e

M˜g

R T˜ ˜z_Denver

§¨© ·

˜ 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 g_earth g_moon

˜ 'l_earth 113.498

M 'l_earth˜lb_m M 113.498 lb_m

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˜D² 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˜D² A 0.173 m² (a) F P_atm˜Amass g˜ F 1.909u10⁴N

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

§¨

©

·

o

¹

First derivative of the function with respect to parameter B

B

f T A( BC) d

d

1

TC exp A B TC

§¨

©

·

˜

¹

o

First derivative of the function with respect to parameter C

C

f T A( BC) d

d

B TC ( )²

exp A B TC

§¨

©

·

˜

¹

o

t

18.5 9.5

0.2 11.8 23.1 32.7 44.4 52.1 63.3 75.5

§¨ ¨

¨ ¨

¨ ¨

¨ ¨

¨ ¨

¨ ¨

¨ ©

·

¸ ¸

¸ ¸

¸ ¸

¸ ¸

¸ ¸

¸

¹

Psat

3.18 5.48 9.45 16.9 28.2 41.9 66.6 89.5 129 187

§¨ ¨

¨ ¨

¨ ¨

¨ ¨

¨ ¨

¨ ¨

¨ ©

·

¸ ¸

¸ ¸

¸ ¸

¸ ¸

¸ ¸

¸

¹

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

§

¨ ©

·

¹

exp a₀ a₁ Ta₂

§

¨ ©

·

¹

1

Ta₂ exp a₀ a₁ Ta₂

§

¨ ©

·

˜

¹

a₁ Ta₂

^{2 exp a0}

a₁ Ta₂

§

¨ ©

·

˜

¹ ª «

« «

« «

« «

« «

« ¬

º »

» »

» »

» »

» »

» ¼

guess

15 3000

50

§¨ ¨

¨©

·

¸

¹

Apply the genfit function

A B C

§¨ ¨

¨©

·

¸

¹

genfit T Psat( guessF)

A B C

§¨ ¨

¨©

·

¸

¹

13.421 2.29u10³

69.053

§ ¨

¨ ¨

©

·

¸

¹

Ans.

Compare fit with data.

240 260 280 300 320 340 360

0 50 100 150 200

Psat

f T A( BC)

T

To find the normal boiling point, find the value of T for which Psat = 1 atm.

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 C₁ 16000dollars yr

t₂ 2000 yr t₁ 1970

b)

The increase in price of gasoline over this period kept pace with the rate of inflation.

C₂ 1.513dollars C₂ C₁˜(1i)^t2t1 gal

i 5%

C₁ 0.35dollars t₂ 2000 gal

t₁ 1970 a)

1.25

T_nb273.15K 56.004 degC Ans.

T_nb 329.154 K

T_nb B

A ln Psat

§¨

kPa

©

·

¹

C

§ ¨

¨ ©

·

¹

˜K Psat 1atm

t₁ 20 degC˜ C_P 4.18 kJ kg degC˜

˜ M_H2O 30 kg˜

t₂ t₁

'U_total M_H2O˜C_P

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˜'g˜ z 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˜P˜dV

Take Cp and V constant and integrate: M_H2O˜C_P^˜'

t₂t₁ = ^Utotal

Q₃₄ 800J W₃₄ 300J

'Ut₃₄ Q₃₄W₃₄ 'Ut₃₄ 500J Ans.

Step 1 to 2 to 3 to 4 to 1: Since 'U^t is a state function, 'U^t for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the

'U^t 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): 'U^t = QW

Step 1 to 2: 'Ut₁₂ 200J W₁₂ 6000J

Q₁₂ 'Ut₁₂W₁₂ Q₁₂ 5.8u10³J Ans.

Step 3 to 4:

'U 12˜kJ Q 'U Q 12kJ 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 =

^mw˜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: Q₁₂₃₄₁ = 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 'UQ

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 cm³

˜ gm H₂ U₂P₂˜V₂ 'U U₂U₁ 'H H₂H₁ 'U 2022.4kJ

kg Ans. 'H 2275.8kJ

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 'E_P

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

4D²

mdot C˜ _p^˜

T₃T₁ ^mdot2˜CP˜

^T3T2 = ^Qdot T₃˜C_P^˜

mdot₁mdot₂ = ^{Qdotmdot1˜CP˜T1mdot2˜CP˜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˜C_P˜T₁mdot₂˜C_P˜T₂ mdot₁mdot₂

˜^CP T₃ 43.235 degC Ans.

2.25By Eq. (2.32a): 'H 'u²

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

§ ¨

©

·

¹

2

˜ u₂ 0.5m

s Ans.

(b) 'E_K 1

2u₂² 1 2u₁²

'E_K 1.875 J

kg Ans.

2.23 Energy balance: mdot₃˜H₃

mdot₁˜H₁mdot₂˜H₂ ^{= Qdot} Mass balance: mdot₃mdot₁mdot₂ = 0

Therefore: mdot₁^˜

H₃H₁ ^mdot2˜

^{H3H2 = 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 u₂² 2

u₁² 2

§ ¨

©

·

¹

^˜molwt

ª

« ¬

º »

¼

^{˜ndotWsdot Qdot 9.904kW Ans.}

2.27By Eq. (2.32b): 'H 'u²

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

§ ¨

©

·

¹

2 1

ª« «¬ º»

˜

»¼

= 'u² u₁² D₁

D₂

§ ¨

©

·

¹

4 1

ª« «¬ º»

˜

»¼

=

SI units: u₁ 14 m

˜ s D₁ 2.5 cm˜ D₂ 3.8 cm˜ 'T u₁²

2 C˜ _P 1 D₁ D₂

§ ¨

©

·

¹

4

ª«

«¬ º»

˜

»¼

'T 0.019 degC Ans.

D₂ 7.5cm

'T u₁²

2 C˜ _P 1 D₁ D₂

§ ¨

©

·

¹

4

ª«

«¬ º»

˜

»¼

'T 0.023 degC Ans.

Maximum T change occurrs for infinite D2:

D₂ f˜cm 'T u₁²

2 C˜ _P 1 D₁ D₂

§ ¨

©

·

¹

4

ª«

«¬ º»

˜

»¼

'T 0.023 degC Ans.

2.26 T₁ 300K T₂ 520K u₁ 10m s

H₂ 2726.5 kJ

˜kg

By Eq. (2.32a): Q H₂H₁ u₂²u₁²

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 u₁²u₂²

= 2 u₂ Find u

₂

u₂ 578.36m

s Ans.

D₁ 5 cm˜ V₁ 388.61 cm³

˜ gm V₂ 667.75 cm³

˜ gm 'H = C_P˜'T 7

2˜R^˜

T₂T₁ 'u² u₁² T₂ =

T₁ P₁ P₂

§

˜

¨ ©

·

¹

2 1

ª« «¬ º»

˜

»¼

=

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

§

˜

¨ ©

·

¹

2 1

ª« «¬ º»

˜

»¼

˜molwt

=

T₂ Find T

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

m_v˜c_vn C˜ _V ^˜

^{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 ft³

lb_m

˜ V₂ 0.28 ft³ lb_m

˜ D₁ 4 in˜ D₂ 1 in˜

mdot S

4˜D₁²˜u₁

V₁ mdot 679.263lb hr

u₂ mdot V₂ S 4˜D₂²

˜ u₂ 9.686 ft

sec W_s 5360 BTU lbmol

˜

Eq. (2.32a): Q H₂H₁ u₂²u₁²

2 W_s

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 ft³

lb_m

˜ V₂ 78.14 ft³ lb_m

˜ D₁ 3 in˜ D₂ 10 in˜

mdot 3.463u10⁴ lb mdot sec

S

4˜D₁²˜u₁ V₁

u₂ mdot V₂ S

4˜D₂²

˜ u₂ 22.997 ft

sec

Eq. (2.32a): W_s H₂H₁ u₂²u₁²

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 C_P 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 24942cm³ mol

W n

V₁ V₂

V P

´µ

¶ d

= ˜ = n P˜ ^˜

V₁V₂ = n P˜ ^˜

V₁3 V˜ ₁

Whence W n˜P˜2˜V₁ 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

§ ¨

¨ ¨

¨ ©

·

¸ ¸

¹

Re D˜U˜u P

o



u 1 1 5 5

§ ¨

¨ ¨

¨ ©

·

¸ ¸

¹

m D s

2 5 2 5

§ ¨

¨ ¨

¨ ©

·

¸ ¸

¹

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

'H 1.164 kJ Ans.

'H 'H_a'H_b mol

'U 0.831 kJ Ans.

'U 'U_a'U_b mol

'U_b 6.315u10³ J 'U_b 'H_bP₂^˜

V₂V₁ mol

'H_a 7.677u10³ J 'H_a 'U_aV₁^˜

P₂P₁ mol

V₂ 0.028 m³ V₂ R T˜ ₂ mol

P₂ V₁ 2.437u10³ m³

V₁ R T˜ ₁ mol P₁

'U_a 5.484u10³ J 'U_a C_V˜'T_a mol

'H_b 8.841u10³ J 'H_b C_P˜'T_b mol

'T_a 263.835K 'T_a T_a2T₁

'T_b 303.835 K 'T_b T₂T_a2

Cost 799924 dollars Ans.

Cost 15200 Wdot

§¨

kW

©

·

¹

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

mdot 4.5kg kg s 2.42

'P'L Ans.

0.632

0.206

11.254

3.88

§ ¨

¨ ¨

¨ ©

·

¸ ¸

¹

kPa 'P'L 2 m

D ˜U ˜f_F ˜u²

§¨ ©

·

¹

o



mdot Ans.

0.313 1.956 1.565 9.778

§ ¨

¨ ¨

¨ ©

·

¸ ¸

¹

kg mdot U˜u S s

˜4D²

§ ¨

©

·

¹

o



f_F

0.00635 0.00517 0.00452 0.0039

§ ¨

¨ ¨

¨ ©

·

¸ ¸

¹

f_F 0.3305 ln 0.27˜HD 7

§¨

Re

©

·

¹

0.9

ª

« ¬

º »

¼ ª «

¬

º »

¼

2

ª«

˜

«¬ º»

»¼

o



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

§ ¨

©

·

˜

¹

§

¨ ©

·

˜

¹

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

d U d

§ ¨

©

·

˜

¹

= N 1

U P d U d

§ ¨

©

·

˜

¹

=

P T

At constant T, the 2nd equation can be written:

dU

U = N˜dP ln U₂ U₁

§ ¨

©

·

¹

^{= 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 cm³

˜ 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

§ ¨

©

·

˜

¹

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

ª¬

E^˜

t₂t₁

º¼

^{'V V}2V₁ 'V_total M˜'V 'V_total 7.638u10⁵m³ Ans.

Work 'P˜ V_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 P₁ 8 bar˜

Step 41: Adiabatic T₄ T₁ P₄ P₁

§ ¨

©

·

¹

R C_P

˜ T₄ 378.831 K

'U₄₁ C_V^˜

T₁T₄ ^'U41 4.597u10³ J mol 'H₄₁ C_P^˜

T₁T_{4 'H}

41 6.436u10³ 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 m³ mol

Step 12: Isothermal 'U₁₂ 0 J

mol 'U₁₂ 0 J

mol

'H₁₂ 0 J

˜mol 'H₁₂ 0 J mol J C_P

C_V T₂ T₁ P₂ P₁

§ ¨

©

·

¹

J1 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 C_V 5

2R

P₁ 10bar T₁ 600K V₁ R T˜ ₁

P₁ V₁ 4.988u10³ m³ 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: T₂ = 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 R˜T₁ln P₂

P₁

¨ §

©

·

˜

¹

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 m³ mol

P_i P₁ V₁ V₂

§ ¨

©

·

¹

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˜ ˜Tln

P₂ P₁

§ ¨

©

·

˜

¹

= Work P₁˜V₁ ln P₂ P₁

§ ¨

©

·

˜

¹

Work 2982 kJ Ans.

(b) Step 1: adiabatic compression to P₂

J 5 3

V_i V₁ P₁ P₂

§ ¨

©

·

¹

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 C_P

C_V

Adiabatic compression from point 1 to point 2:

Q₁₂ 0 kJ

˜mol 'U₁₂= W₁₂= C_V˜'T₁₂ T₂ T₁ P₂ P₁

§ ¨

©

·

¹

J1 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

Work W₁ Work 1100 kJ Ans.

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

'U^t = 'T = 0 T₂ = T₁ = 100 degC˜ Not reversible (b) The gas is returned to its initial state by isothermal compression.

Work n R˜ ˜T ln V₁ V₂

§ ¨

©

·

˜

¹

= but n R˜ ˜T = P₂˜V₂

V₁ 4 m˜ ³ V₂ 4

3˜m³ P₂ 6 bar˜ Work P₂˜V₂ ln V₁

V₂

¨ §

©

·

˜

¹

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

¨ §

©

·

˜

¹

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

§ ¨

©

·

¹

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 C_P C_V

(a) Isothermal: Work n R˜ ˜T₁ln V₁ V₂

¨ §

©

·

˜

¹

= P₂ P₁ V₁

V₂

˜ T₂ T₁ T₂ 600 K P₂ 200 kPa Ans.

Work P₁˜V₁ ln V₁ V₂

¨ §

©

·

˜

¹

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

§ ¨

©

·

˜

¹

Work W₂₃ Work 6.762 kJ

mol Ans.

Q 'UWork Q 4.684 kJ

mol Ans.

3.21 By Eq. (2.32a), unit-mass basis: molwt 28 gm

mol 'H 1

2˜'u²

= 0

But 'H = C_P˜'T Whence 'T

u₂²u₁² 2 C˜ _P

=

C_P 7 2

R molwt

˜ u₁ 2.5 m

˜ s u₂ 50 m

˜ s t₁ 150 degC˜

t₂ t₁ u₂²u₁² 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 'UWork 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 'H}12

W₁₂ 'U₁₂Q₁₂ W₁₂ 0.831 kJ mol

W₂₃ R T˜ ₂ln P₃ P₂

§ ¨

©

·

˜

¹

W₂₃ 7.718 kJ

mol

Work W₁₂W₂₃ Work 6.886 kJ

mol Ans.

Q 'UWork Q 4.808 kJ

mol Ans.

(c) T₂ T₁ P₂ P₃ W₁₂ R T˜ ₁ln P₂

P₁

§ ¨

©

·

˜

¹

'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

V₃V₂ = R^˜

T₃T₂ But T₃ = T₁ So... Work= R T^˜

₂T₁ Also W R T˜ ₁ln P

P₁

§ ¨

©

·

˜

¹

= Therefore

ln P P₁

§ ¨

©

·

¹

T₂T₁ T₁

= T₂ 350 K˜ T₁ 800 K˜ P₁ 4 bar˜

P P₁exp T₂T₁ T₁

¨ §

©

·

˜

¹

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

§ ¨

©

·

˜

¹

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˜R˜T₁ 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

§ ¨

©

·

¹

J1 J

˜ (2)

T_A 2 T˜ ₁ P₂ P₁

˜ T_B Q ⁼ n_A^˜

'U_A'U_B

Define q Q

n_A

= 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 V_A˜(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 C_P 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<}