AUTONOMÍA PEDAGÓGICA

EXERCISE 5.1

Basis: 10 000 Lph of thermic fluid Energy balance:

This change in energy is responsible for rise in temperature of the thermic fluid.

Rise in temperature, T = (232.823 ´ 3600)/(7500 ´ 2.68)

= 41.7 K or 41.7°C Outlet temperature of thermic fluid = 473.15 + 41.7

= 514.85 K (241.7ºC) Ans.

EXERCISE 5.2

Basis: 1 kmol oxygen T₁ = 350 K and T₂ = 1500 K Ref. 7:

5

H₁ =

Based on absolute enthalpies (Table – 5.22),

H₃ = 49 273 – [(11 603 – 8597)(350 – 298.15)/(400 – 298.15) + 8597]

= 25.7725 (1000 – 300) + 57 8938

Use of absolute enthalpies (Table 5.22):

H = – 314 357 – (– 347 921)

= 33 564 kJ/kmol Ans. (d)

EXERCISE 5.4

Basis: 1 kmol dry gas mixture from absorber T₁ = 343 K and T₂ = 618 K

Component kmol C ° equation constants_mp

n_i a_i × n_i b_i × n_i ´ 10³ c_i × n_i ´ 10⁶ d_i × n_i ´ 10⁹

CH₄ 0.0025 0.0481 0.1303 0.0299 – 0.0283

CO 0.0038 0.1103 – 0.0107 0.0443 – 0.0179

CO₂ 0.001 0.0214 0.0643 – 0.0411 0.0098

H₂ 0.7462 21.3492 0.7607 – 0.1101 0.5738

N₂ 0.2435 7.2054 – 1.2518 3.2100 – 1.2097

Ar 0.003 0.0623 — — —

H₂O 0.0126 0.4094 0.0010 0.1665 – 0.0573

Total 1.0126 29.2061 – 0.2840 3.2995 – 0.7296 For 100 kmol dry gas mixture,

heat exchanged Q = 100 [29.2061 (618.15 – 343.15)

– 0.2840 ´ 10^–3 (618.15² – 343.15²)/2 + 3.2995 ´ 10^–6 (618.15³ – 343.15³)/3

– 0.7296 ´ 10^–9 (618.15⁴ – 343.15⁴)/4

= 100 [8031.7 – 37.5 + 215.3 – 24.1]

= 818 540 kJ Ans. (a)

Use data of Table – 5.22.

n_i A = B =

Component kmol (H^o – H^o₀ + DH_f^o)₁ n_i × A (Hº – H₀^o + DH^o_f)₂ n_i × B

CH₄ 0.0025 – 55 311 – 138 – 42 622 – 107

CO 0.0038 – 103 981 – 395 – 95 802 – 364

CO 0.001 – 382 381 – 382 – 307 401 – 370

CO₂ 0.7462 + 9 709 + 7245 + 17 722 + 13 224 H₂ 0.2435 + 9 902 + 2411 + 18 029 + 4 390

Ar 0.003 — — — —

H₂O 0.0126 – 27 751 – 2870 – 218 133 – 2748

Total 1.0126 + 5871 + 14 025

Enthalpy change for 100 kmol dry gas mixture,

Q = 100 (14 025 – 5871) +100 ´ 0.003 ´ 20.7723 (618.15 – 343.15)

= 815 400 + 1714

= 817 114 kJ (including that of argon) Ans. (b) At 343.15 K (70ºC), p_w = 31.162 kPa (Ref. Table 6.13)

Mole fraction of water vapour in gas mixture = 0.0126/1.0126

= 0.012 44 If total system pressure is p,

31.162 = 0.012 44 p

or p = 2505 kPa a

º 2.505 MPa a or 25.05 bar a Ans. (c)

EXERCISE 5.5

Basis: 5000 kg/h hot oil

Heat load of oil = 5000 ´ 2.51 (423.15 – 338.15)

= 1066 750 kJ/h º 296.32 kW Heat gained by water = 10000 ´ 4.1868 (T – 294.15)

Where T = outlet temperature of cooling water Heat loss = Heat gain

10 000 ´ 4.1868 (T – 294.15) = 1066 750

T = 319.63 K (46.5ºC) Ans.

EXERCISE 5.6

Basis: 1 kg Diphyl DT Q =

Mean heat capacity = (2.03 + 2.206)/2 = 2.118 kJ/(kg×K) Q = 2.118 (533.15 – 453.15) = 169.4 kJ/kg Error = (169.4 – 200.9) 100/200.9

= – 15.7% Ans.(b)

Solving the equations, a¢ = – 1.1645, b = 0.0095 Basis: 5 kmol benzene – toluene mixture

(a) The mixture contains 3 kmol benzene and 2 kmol toluene.

Mass of benzene = 3 ´ 78 = 234 kg Mass of toluene = 2 ´ 92 = 184 kg Let T be the temperature of the mixture.

Heat gained by benzene = m C_l

= 234 [– 0.6051 (T – 303)] + 0 007 76 2

. (T² – 303²)

= 0.907 92 T² – 141.593 T – 40 452 Heat lost by toluene = m C

T

2 2

373

× ×

z

^{l dT}

= 184 [– 1.1645 (373 – T)] + (0.0095/2) (373² – T²)

= 416 77 + 214 268 T – 0.874 T² Heat loss by toluene = heat gain by benzene

0.907 92 T²= 141.593 T – 40 452 = 41 677 + 214.268 T – 0.874 T² 1.781 92 T²– 355.861 T – 82 129 = 0

Solving the quadratic equation, T = 336.6 K (63.45ºC) Ans. (a) (b) Use of data given in Table – 5.3 (Ref. 7).

T₁ = 303 K (30ºC)

Let the final temperature of the mixture after mixing is T K.

Heat gained by benzene = 3

303

z

T(– 7.2733 + 770.541 ´ 10^–3 T

– 1648.18 ´ 10^–6 T² + 1897.94 ´ 10^–9 T³)

= 3 [– 7.2733 (T – 303)

+ 770.541 ´ 10^–3 (T² – 303²)/2 – 1648.18 ´ 10^–3 (T⁹ – 303³)/3 + 1897.94 ´ 10^–3 (T⁴ – 303⁴)/4]

=1423.47 ´ 10^–9 T⁴ – 1624.17 ´ 10^–6 T³ + 1.1559 ´ 10^–3 T²

– 21.8199 T – 65 651.7 T₂ = 373 K (100ºC)

Heat given-up by toluene = 2

T

373

z

(1.8083 + 812.223 ´ 10^–3 T

– 1512.67 ´ 10^–6 T²+ 1630.01 ´ 10^–9 T³)

= 2 [1.8083(373 – T)

+ 812.223 ´ 10^–3 (373² – T²)/2 – 1512.67 ´ 10^–6 (373³ – T³)/3 + 1630.01 ´ 10^–9 (373⁴ – T⁴)/4]

= – 815 ´ 10^–9 T⁴ + 1008.4 ´ 10^–6 T³ – 0.8122 ´ 10^–3 T² – 3.6166 T + 77 795.4

Heat gained by benzene = heat given-up by toluene Equating the two and simplifying,

2238.47 ´ 10^–9 T⁴ – 2362.57 ´ 10^–6 T³ + 1.9681 T² – 18.2033 T = 143 447.1

The equation can be solved by trial and error method or by Mathcad.

Mixing temperature = 334.25 K (61.1ºC) Ans. (b)

EXERCISE 5.8

(a) From Table – 5.4 for acetic acid, log p = 4.682 06 – 1642.540

(b) From Table – 5.4, for sulphur trioxide, logp = 4.205 15 – 892.175

(a) From Table – 5.5, for acetone,

1

(b) From Table – 5.5, for carbon disulphide,

l_v= 26.74 ´ 0.821 97

= 21 98 kJ/mol Ans.

EXERCISE 5.10

From Table – 5.4, for chlorobenzene

log p = 4.110 83 – 1435.675

l_v= 8.31451 404.9 1.092(ln 4530 5.6182) (0.930 0.6396)

Watson equation:

Basis: 100 kg naphthalene

Total heat to be supplied consists of sensible heat, supplied to the solid from 303.15 K to its melting point (353.35 K), latent heat of fusion l_f at 353.35 K, sensible heat supply to liquid naphthalene from the melting point (353.35 K) to its normal boiling point (490 K) and the latent heat of vaporzation at 490 K.

Q₂= C

z

_s dT = 100 [– 0.092 (353.35 – 303.15) + (0.0046/2) (353.35² – 303.15²)]

= – 461.8 + 7579.9 = 7118.1 kJ Q₂= m × l_f = 100 ´ 150.7 = 15 070 kJ

For evaluation of Q₃, C₁ is required to be expressed in the polynomial form of temperature.

C_l = a + bT

1.738 = a + b ´ 353 (1)

2.135 = a + b ´ 473 (2)

Solving the equations, a = 0.57 and b = 0.003 31

Q₃ = m = 100 [0.57 (491 – 353) + (0.003 31/2)

z

^{Cl dT}

(491² – 353²)]

= 7866 + 19 276 = 27 142 kJ Q₄ = m ×l_v = 100 ´ 316.1 = 31 610 kJ Total heat load =

S

Q_i

= 7118.1 + 15 070 + 27 142 + 31 610 = 80 940.1 kJ

Latent heat of vaporisation of the eutectic mixture at 171 kPa a

= 278.0 kJ/kg (Ref. Table 5.6) Quantity of the eutectic mixture condensed

= 80 940.1/278.0

= 291.15 kg Ans.

EXERCISE 5.13

Basis: 10 000 kg/h superheated stem @ 0.44 MPa a and 543 K (270°C) is mixed with 7500 kg/h saturated steam at 0.44 MPa a.

From Table AIV.2 enthalpy of saturated steam = 2741.9 kJ/kg From Table AIV.3 enthalpy of superheated steam = 3004.5 kJ/kg Total enthalpy of 7500 kg/h saturated steam = 7500 ´ 2741.9

= 20 564 250 kJ/h º 5712.29 kW Total enthalpy of 10 000 kg/h superheated steam = 10 000 ´ 3004.5

= 30 045 000 kJ/h º 8345.83 kW Total enthalpy of mixed fluids = 5712.29 + 8345.83

= 14 058.12 kW º 50 609 250 kJ/h Specific enthalpy of mixture = 50 609 250

17 500 = 2891.96 kJ/kg From steam tables (AIV.3), temperature of mixed steam

= 489.15 K (216.0°C) at 0.44 MPa a which is superheated.

EXERCISE 5.14

Basis: 100 kg superheated steam @ 0.5 MPa a and 523 K (250°C) From Steam Tables AIV.3, specific enthalpy of superheated steam

= 2961.1 kJ/kg

From Steam Tables AIV.2, specific enthalpy of saturated steam at 0.5 MPa a

= 2747.5 kJ/kg at (T_S) saturation temp.

of 425 K (151.85°C).

Enthalpy of steam to be reduced = 2961.1 – 2747.5 = 213.6 kJ/kg Enthalpy of water to be sprayed = 125.66 kJ/kg

Let a be the quantity of water to be sprayed.

100 ´ 2961.1 + a ´ 125.66 = 2747.5 (100 + a)

or a = 8.15 kg Ans.

EXERCISE 5.15

Basis: 1 kg condensate

Operating pressure of dryer = 310 kPa a Operating pressure of flash vessel = 101.325 kPa a

At 310 kPa a, T_s= 407.8 K (134.65°C),

h = 566.23 kJ/kg and l_v = 2159.9 kJ/kg At 101.3 kPa a, T_s= 373.15 K (100°C),

h = 419.06 kJ/kg, l_v = 2256.9 kJ/kg and H = 2676.0 kJ/kg

Let the flash quantity be a kg per kg of condensate at 310 kPa.

(1 – a) 419.06 + 2676.0a = 566..23

a = 0.065 kg/kg condensate Enthalpy of saturated steam at 780 kPa a= 2767.5 kJ/kg

Let y and y¢ be the quantities of water to be sprayed before modification and after the modification per kg of saturated steam at 310 kPa, respectively.

y ´ 419.06 + (1 – y) 2767.5 = (566.23 + 2159.9) = 2726.13 y = 0.0176 kg/kg condensate After the modification,

y¢ ´ 419.06 + 0.065 ´ 2676.0 + (1 – y¢ – 0.065) 2767.5

= 2726.13

y¢ = 0.015 kg/kg condensate Make-up 0.8 MPa a steam = 1 – 0.065 – 0.015 = 0.92 kg Reduction in make-up steam = [(0.9824 - 0.92)/0.9824] 100

= 6.35% Ans.

EXERCISE 5.16

Basis: 1 kmol dry gas mixture. T₁ = 1473.15 K, T₂ = 573.15 K

Component kmol C° equation constants_mp n_i

a_i × n_i b_i × n_i ´ 10³ c_i × n_i ´ 10³ d_i × n_i ´ 10³

H₂ 0.557 15.936 0.568 – 0.082 0.428

CO 0.345 10.015 – 0.972 4.017 – 1.624

CO₂ 0.028 0.598 1.800 – 1.149 0.274

CH₄ 0.005 0.096 0.261 0.060 – 0.057

N₂ 0.065 1.923 – 0.334 0.857 – 0.323

H₂O 1.850 60.110 0.147 24.440 – 8.413

Total 2.850 88.678 1.470 28.143 – 9.715

Heat lost in WHB, DH = 88.678 (1473 – 573) + 1.470 ´ 10^–3 (1473² – 573²)/2 + 28.143 ´ 10^–6 (1473³ – 573³)/3 – 9.715 ´ 10^–9 (1473⁴ – 573⁴)/4

= 79 810.2 + 1353.4 + 28 216.9 – 11 172.1

= 98 208.4 kJ/kmol dry gas mixture º 9820 840 kJ/100 kmol dry gas mixture Steam pressure p = 4.0 MPa g = 4.1013 MPa a

T_s= 524.8 K (251.8°C) (Ref. A IV.2) Water enters WHB at 504.8 K (231.8°C), 20 K lower than 524.8 K.

Enthalpy of water at 504.8 K, h = 998.7 kJ/kg

Enthalpy of saturated steam at 4.1013 MPa a, H_s = 2800 kJ/kg Enthalpy supplied to water in WHB

= 2800 – 998.7

= 1801.3 kJ/kg Steam generated in WHB = 9820 840/1801.3

= 5452 kg per 100 kmol dry gas Ans.

EXERCISE 5.17

Basis: 100 kg saturated liquid ammonia at 705 kPa a Let y be the quantity of NH₃ vapours flashed.

Liquid NH₃ at 101.325 kPa a = 100 – y kg Heat balance:

Enthalpy of liquid NH₃ at 705 kPa a

= Enthalpy of liquid NH₃ at 101.3 kPa a + Enthalpy of flash vapours at 101.3 kPa a 100 ´ 265.56 = (100 – y) 49.1 + y ´ 1418.7 kJ/h (º 44.85 kW) Solving the equation, y = 15.8 kg or 15.8 % flashing Ans.

EXERCISE 5.18

Basis: 100 kg NH₃ vapours, entering 1st stage of compressor

Enthalpy H₁ = 100 ´ 1632.68 = 163 268 kJ/h (º 45.35 kW) of gas leaving 1st stage.

Enthalpy of 100 kg liquid NH₃ at 276 K = 100 ´ 213.92

= 21 392 kJ/h (º 5.94 kW) Heat, removed in flash cooler = 163 268 – 21 392

= 141 876 kJ/h (º 39.41 kW) Let y be the quantity of liquid NH₃, obtained from the condenser and is utilised in flash cooler. Out of this, a part will be flashed due to pressure reduction.

Another basis: 1 kg liquid NH₃ at 1930.3 kPa a and 321 K (48°C) Enthalpy of saturated liquid at 1930.3 kPa a

= Enthalpy of saturated liquid at 276 K (3°C) + Enthalpy of saturated vapours at 276 K (3°C)

100 ´ 431.07 = (100 – z) 213.92 + z ´ 1464.92 where z = amount of flash vapours, produced due to pressure reduction

z = 17.36 kg

Thus 17.36% liquid, entering the flash cooler (i.e. y kg) will be flashed.

Balance liq. NH₃, available for cooling 1st stage vapours

= (1 – 0.1734) y = 0.8266y kg/h This ammonia should cater for the heat load of 141 876 kJ/h.

0.8266 y ´ 1251.0 = 141 876

or y = 137.13 kg/h Ans. (a)

Heat load of condenser = 1663.91 – 431.07

= 169 059 kJ/h º 46.96 kW Cooling water to condenser = 169 059/[3600 ´ 4.1868 ´ (313 – 305)]

= 1.402 kg/s Ans. (b)

EXERCISE 5.19

Basis: 65 kW refrigeration load in chiller From Table 5.7,

Enthalpy of saturated R-134a liquid at 40°C (313.15 K) = 256.35 kJ/kg Saturation temperature of R-134a at 101.325 kPa = –26°C

Superheat = 10°C

Temperature of R-134a gas, leaving chiller = –26 + 10 = –16°C

Enthalpy of superheated R-134a gas at 101.325 kPa and –16°C = 391.0 kJ/kg (Ref. Table 5.71) Heat picked up by R-134a in chiller = 391.0 – 256.35

= 134.65 kJ/kg

Evaporation rate = 60 kJ× 1 kg s 134.65 kJ

= 0.4456 kg/s Ans.(a) With the incorporation of the economizer, R-134a gas heats upto 10°C at 101.325 kPa.

Heat picked up in the economizer = 412 – 391

= 21 kJ/kg

In exchange, R-134a liquid under saturation pressure (10.165 bar) at 40°C (313.15 K) is cooled.

Enthalpy of sub-cooled R-134a liquid = 256.35 – 21

= 235.35 kJ/kg Heat picked up in chiller = 391 – 235.35

= 155.65 kJ/kg Evaporation rate = 65

155.65 = 0.4176 kg/s with

economizer Ans. (b)

EXERCISE 5.20

(a) Locate a point representing 273 K (0°C) on p-H diagram for CO₂ (Fig. 5.11). Follow constant enthalpy line (i.e. vertical line) which inter-sects the pressure line representing 1.0 MPa g (= 1.101 MPa a).

Read dryness fraction of 0.277. Thus vapours produced by pressure

reduction is 27.7%. Ans.

(b) Enthalpy of CO₂ at 1.101 MPa a and 313 K(40°C) = 811 kJ/kg Enthalpy of vapour-liquid mixture at 1.101 MPa a = 500 kJ/kg Enthalpy required to produce CO₂ gas at 1.101 MPa a and 313 K

= 811.0 – 500 = 311 kJ/kg Ans.

EXERCISE 5.21

Basis: 1 kmol dry gas

T₁ = 403.15 K and T₂ = 313.15 K

Component kmol Molar mass n_i × M_i Critical Critical

y_i M_i pressure temperature

p_ci y_i ×p_ci T_ci y_i ×T_ci

MPa K

CO₂ 0.947 44.01 41.677 7.375 6.983 304.12 288.00

O₂ 0.008 31.9988 0.256 5.042 0.040 154.59 1.24

N₂ 0.030 28.0134 0.840 3.394 0.102 126.09 3.78

H₂ 0.015 2.016 0.030 1.297 0.019 33.2 0.48

Total 1.000 42.803 7.144 293.50

Average molar mass = 42.803 of incoming gas mixture Pseudo critical pressure p_c= 7.144 MPa

Pseudo critical temperature T_c= 293.5 K (20.35°C) New Basis: 100 kmol dry gas

Component kmol C° equation constants_mp n_i

a_i × n_i b_i × n_i ´ 10³ c_i × n_i ´ 10⁶ d_i × n_i ´ 10⁹

CO₂ 94.7 2023.313 6087.704 – 3887.492 928.051

O₂ 0.8 20.821 9.404 – 1.874 – 0.450

N₂ 3.0 88.773 – 15.423 39.549 – 14.904

H₂ 1.5 42.916 1.529 – 0.221 1.154

Total 100.0 2175.823 6083.214 – 3850.038 913.851 T_r1 = 403.15/293.5 = 1.374, T_r2= 313.15/293.5 = 1.067

Avg. T_r= (1.374 + 1.067)/2 = 1.220 p_r= 20/7.144 = 2.8

Fig. 5.1 does not provide value of C^R_mp for the required p_r and T_r. From Lee at el article (ref. 12).

mpR

C

R =

   

^{CmRp 0+w CmRp 1}

At p_r= 2.8 and T_r = 1.22

 

^{CmpR 0 = 5.681}

Acentric factor (w_i) for each component is to be considered and S (y_i× w_i) is required to be calculated for the mixture, However, since CO₂ is 94.7% in the mixture, its w_CO2 value is taken for calculation.

w = 0.225 for CO₂ (Ref. 3)

 

^{CmpR 1 = 7.222}

mpR

C

R = 5.681 + (0.225 ´ 7.222) = 7.306 C^R_mp= 7.306 ´ 8.314 51 = 60.746 C_mp = C^o_mp +C^R_mp=C^o_mp + 60.746 kJ/(kmol×K) Therefore, corrected heat capacity equation will become

C_mp= (2175.823 + 60.746 ´ 100)

+ 6083.214 ´ 10^–9 T – 3850.038 ´ 10^–3 T² + 913.851 ´ 10^–3 T³ Heat removed in water cooler

= 8250.423 (403.15 – 313.15) + 6083.214 ´ 10^–3 (403.15² – 313.15²)/2 – 3850.038 ´ 10^–6 (403.15³ – 313.15³)/3 + 913.851 ´ 10^–9

(403.15⁴ – 313.15⁴)/4

= 742 538 + 196 087 – 44 680 + 3838

= 897 783 kJ/h

Both are closely matching.

Since incoming gas mixture has 5.3 mole % inerts, liquefaction temperature will have to be found by trails such that at the liquefaction temperature, ratio of uncondensed CO₂ to inerts satisfy the vapour pressure data. At 40 bar a pressure, saturation temperature of CO₂ is +5.32°C (278.47 K). Presence of inerts will require liquefaction temperature to be lower than 278.47 K.

Trial 1: Assume liquefaction temperature of 273.15 K.

Consider enthalpy of liquid CO₂ = 0 kJ/kg at 273.15 K Heat given-up by gas from 313.15 K to 273.15 K

= (2175.823 +1449.1) (313.15 – 273.15) + 6083.214 ´ 10^–3 (313.15² – 273.15²)/2 Let x be the amount of uncondensed CO₂ (kmol/h)

(94.7 – x) 10 166 = 203 999 94.7 – x = 20.067

x = 74.633 kmol/h CO₂ condensed = 20.067 kmol/h

Inerts = 0.8 + 3.0 + 1.5 = 5.3 kmol/h

Uncondense CO2

inerts = 74.633

5.3 = 14.083 kmol/kmol

At T = 273.15 K, saturation pressure of CO₂, p_S1 = 3485.9 kPa (Ref. 24) total pressure = 3000 kPa

Applying Raoult's law, Moles of CO

Moles of inerts² = 3485.9

6.781 (4000 – 3485.9)= Thus calculated ratios do not match.

Trial 2: Assume liquefaction temperature (T₂) of 275.15 K.

Enthalpy of gas mixture from 313.15 K to 275.15 K

= 193 955 kJ/h º 53.876 kW

l_v at 275.15 K = 225.05 kJ/kg = 9904.3 kJ/kmol (94.7 – x) 9904.3 = 193 955

94.7 – x = 19.532

x = 74.1168 kmol/h CO2

Inerts = 74.168 14.183 5.3 = CO₂ condensed = 20.555 kmol/h

At T = 275.15 K p_S2 = 3673.3 kPa Heat given up by gas from 313.15 K

to 276.15 K = 188 926 kJ/h º 52.479 kW (94.7 – x) 9749.1 = 188 926

x = 75.322 kmol/h Uncondensed CO2

Inerts = 75.322 14.211 5.3 =

Saturation pressure of CO₂ at 276.15 K, p₅₃ = 3768.5 kPa Uncondensed CO2

Inerts = 3768.5 16.279

(4000 3768.5)=

-Since ratio is reversed, the liquefaction temperature is between 275.15 K and 276.15 K.

Trial 4: Assume T₄= (275.15 + 276.15)/2

= 275.65 K or t₄ = 2.5°C l_v= (225.05 + 221.53)/2

= 223.29 kJ/kg º 9826.9 kJ/kmol

Inerts = 3720.9 13.332

(4000 3720.9)=

-Further trials are unwarranted as both ratios are close enough.

CO₂ condensed = 94.7 – 75.219 = 19.481 kmol/h Hence liquefaction temperature = 269.85 K (–3.3 °C)

CO₂ liquefied = (94.7 – 75.219) 100/94.7

= 20.57 % Ans. (b)

Enthalpy of saturated liquid CO₂ at 37.21 bar a and 2.5°C (275.65)

= 206.13 kJ/kg (Ref. 24)

Liquefied gas is precooled in the revert gas exchanger before dry gas production.

Note that liquid CO₂ temperature after the revert gas exchanger but before the snow tower will be at higher then – 11°C.

Refer Example 5.19,

100 ´ 206.13 = (100 – y) 422.61 + (–148.39 y) 571.0 y = 21 648

y = 37.91 kg dry ice/100 kg liquid CO₂ For 100 kmol raw gas input, dry ice production

= 19.481 ´ 0.3791

= 7.385 kmol/h º 325.2 kg/h For 100 kg dry ice production, raw gas requirement

= 100 ´ 100 ´ 42.803/325.02

= 1316.93 kg/h Ans. (c)

º 30.767 kmol/h

Duty of water cooler = (30.767 ´ 191.441)/100

= 58 901 kJ/h

º 16.361 kW Ans. (a)

Note: Application of Raoult's law may not be fully justified as pressure of liquid CO₂ is high (40 bar a). Use of modified Raoult's law or pure component pressures (Sec. 2.6.3) is more appropriate.

EXERCISE 5.22

Basis: 32 000 kg of total ammonia (liquid + vapour) in each tank wagon Temperature of ammonia in wagon = –29°C (244.15 K)

Let x = quantity of ammonia vapours, contained in the wagon (in kg) Liquid ammonia = 32 000 – x kg

Total volume of each wagon = 60.663 m³ x

1 0844. + ( – ) . 32 000

676 25 x = 60.663

x = 14.49 kg vapours of ammonia Liquid NH₃ = 320 00 – 14.49 = 31 985.51 kg Total enthalpy, H₁ = 14.49 ´ 1425.06 + 31 985.51 ´ 68.32

= 2205 899.1 kJ

Similar values of liquid ammonia and vapour ammonia at different temperatures can be calculated.

Temperature, NH₃ vapours, liquid NH₃, total enthalpy,

°C(K) kg kg kJ

–29 (244.15) 14.49 31 985.51 2205 899.1

–27 (246.15) 15.62 31 984.38 2493 097.8

–25 (248.15) 16.85 31 983.15 2781 056.0

–23 (250.15) 18.12 31 981.88 3069 691.4

–21 (252.15) 19.44 31 980.56 3359 336.6

–19 (254.15) 20.84 31 978.16 3649 276.3

–17 (256.15) 22.27 31 377.73 3940 422.8

Temperature change from 244.15 K to 246.15 K

Average temperature, T = 246.15 K Ambient temperature T_a= 298.15 K Heat gain, f₁ = 1.3 (298.15 – 245.15) + 3.76 = 72.66 kW

º 261 576 kJ/h Enthalpy gain between two temperatures,

H₁¢ = 2493 097.8 – 2205 899.1

= 287 198.7 kJ

Time, required for increase in temp. = 287 198.7/261 576 = 1.098 h Similar calculations are listed below:

Rise in temperature Enthalpy change, Heat transfer Time, h

K kJ rate, f, kJ/h

From to

244.15 246.15 287 198.7 261 576 1.098

246.15 248.15 287 958.2 252 216 1.142

248.15 250.15 288 635.4 242 856 1.189

250.15 252.15 289 645.2 233 496 1.240

252.15 254.15 289 939.7 224 136 1.294

254.15 256.15 291 146.5 214 776 1.356

Total 1734 523.7 7.319

Ans. (a) Average T = 244.15 + 256.15

2 = 250 K

f = 1.3 [298.15 – 250.15] + 3.76

= 66.16 kW º 238 176 kJ/h

Time q = 1734 523.7/238 176 = 7.283 h Ans. (b)

EXERCISE 5.23

Basis: 1650 Nm³/h air flow rate

Molar flow rate = 1650/22.414 = 73.615 kmol/h Mass flow rate = 73.615 ´ 28.84 = 2123.06 kg/h Temperature drop from 463.15 K to 453.15 K:

Heat loss rate at 463.15 K (190°C) = 575.8 W/m [Ref. Fig. (5.35)]

Heat loss rate at 459.15 K (180°C) = 532.3 W/m [Ref. Fig. (5.35)]

Average heat loss rate = (575.8 + 532.3)/2457.5

= 554.05 W/m

Enthalpy to be removed from air = 2123.06 ´ 1.006 ´ 10

= 21 358 kJ/h º 5.9328 kW Length of pipe, required for cooling, L₁ = 5.9328 ´ 1000/554.05 L₁ = 10.71 m

Similar calculations are listed below, keeping enthalpy removal quantity as same (i.e. 5.9328 kW) for 10 K drop.

Temp, Heat loss rate, Avg. heat loss Length of pipe

°C(K) ref. Fig. (5.35), W/m rate, W/m m

190 (463.15) 575.8 — —

180 (453.15) 532.3 554.05 10.71

170 (443.15) 481.1 506.7 11.71

160 (433.15) 434.4 457.75 12.96

150 (423.15) 400.0 417.2 14.22

140 (413.15) 364.8 382.4 15.51

130 (403.15) 324.6 344.7 17.21

120 (393.15) 292.6 308.6 19.22

110 (383.15) 266.4 279.5 21.23

100 (373.15) 244.3 255.35 23.23

Total — — 146.00

Balance length of pipe = 150 – 146 = 4 m Heat loss rate at 363.15 K (90°C) = 216.4 W/m Avg. heat loss rate between 373.15 K (100°C) and 363 K(90°C)

= (244.3 + 216.4)/2 = 230.35 W/m Length of pipe, required to cool to 363 K (90 °C)

= 5932.8/230.35

= 25.76 m However, actual length to be cooled = 4 m

Drop in temperature = (4/25.76) 10

= 1.55 K Temperature of air at the end of 150 m pipe = 373.15 – 1.55

= 371.6 K (98.45°C) Ans.

EXERCISE 5.24

Basis: 1 kmol mixture of benzene and toluene containing 0.6 kmol benzene.

Antoine equation:

for benzene: log p = 4.018 14 – 1203.835 ( – 53.226)T for toluene: log p = 4.078 27 – 1343.943

( – 53.773)T At p = 101.325 kPa, = 1.013 25 bar

T_s1= 353.25 K, T_s2 = 383.77 K t_s1= 80.1°C t_s2 = 110.62 °C Initial guess, T₁= 353.60 ´ 0.6 + 383.77 ´ 0.4

= 365.67 K say 366 K

Bubble point: component 1 = benzene and component 2 = toluene. All pressures are kPa.

Temperature, K

x₁ T₁ = 366 T₂ = 360 T₃ = 364 T₅ = 362.75

p_si y_i p_si y_i p_si y_i p_si y_i

0.6 147.653 0.8743 124.155 0.7352 139.468 0.8259 134.533 0.7966 0.4 59.416 0.2346 48.928 0.1932 55.737 0.2200 53.132 0.2121

1.0 1.1089 0.9284 1.0459 1.0087

At Sy_i = 1.00. Bubble point, T_BP = 362.75 K or 89.6°C Ans. (a–i) Dew point:

Temperature, K

y₁ T₁ = 366 T₂ = 368 T₃ = 370 T₄ = 369

p_si x_i p_si x_i p_si x_i p_si x_i

0.6 147.653 0.4117 156.206 0.3892 165.137 0.3681 160.62 0.3785

0.4 59.416 0.6821 63.283 0.6405 67.350 0.6018 65.29 0.6208

1.0 1.0938 1.0297 0.9699 0.9993

At Sx_i = 1.00 dew point, T_DP= 369 K or 95.85 °C Ans. (a – ii) Enthalpy (sensible heat) of liquid mixture at T:

Reference temperature: T₀ = 273.15 K(0 °C) H₁ = ^BP

= 12 699.6 kJ/kmol benzene H₂ = ^BP

= 14 152.0 kJ/kmol toluene H_sol= S (H_i × x_i)

= 0.6 ´ 12 699.6 + 0.4 ´ 14 152.0

= 7619.8 + 5660.8

= 13 280.6 kJ/kmol liquid mixture From Fig. 6.2, H_sol= 13 733 kJ/kmol liquid mixture Enthalpy of vapour mixture at T_DP :

l_v1=

562.05 – 369.0 0.38

562.05 – 353.3

591.75 – 369.0 0.38

591.75 – 383.8

+ 5056.235 ´ 10^–3 (369.0² – 273.15²)/2

– 14 292.204 ´ 10^–6 (369.0³ – 273.15³)/3 + 14 419.754 ´ 10^–9 (369.0⁴ – 273.15⁴)/4

= – 46 547.5 + 155 999.5 – 142 626.2 + 46 883.5

= + 13 709.3 kJ/kmol benzene H₂¢ = ^DP

273.15^Tò C_l2 ×dT

= – 56.3627 (369.0 – 273.15)

+ 1768.423 ´ 10^–3 (369.0² – 273.15²)/2

– 5192.623 ´ 10^–6 (369.0³ – 273.15³)/3 + 5497.39 ´ 10^–9 (369.0⁴ – 273.15⁴)/4

= – 5416.5 + 54 561.0 – 51 818.7 + 17 873.9

= + 15 199.7 kJ/kmol toluene Enthalpy of benzene vapour at T_DP= 13 709.3 + 29 820.7

= 43 530 kJ/kmol benzene vapour Enthalpy of toluene vapour at T_DP= 15 199.7 + 34 058.3

= 49 258 kJ/kmol toluene vapour Enthalpy of vapour mixture = S ¢ ×

b

H yi i

g

= 0.6 ´ 43 530 + 0.4 ´ 49 258

= 45 821 kJ/kmol vapour mixture Ans. (b) From Fig. 6.2, enthalpy of vapour mixture

= 45 666.7 kJ/kmol vapour mixture

EXERCISE 5.25

Basis: 1 kmol natural gas

Temperature n_i K_i Values of L_i for trial value of L

L = 0.5 0.1 0.01 0.041

CH₄ 0.8957 2.70 0.2421 0.0354 0.0033 0.0140

CO₂ 0.0112 0.90 0.0059 0.0012 0.0001 0.0005

C₂H₆ 0.0526 0.38 0.0381 0.0019 0.0014 0.0053 C₃H₃ 0.0197 0.098 0.0179 0.0105 0.0019 0.0060 i-C₄H₁₀ 0.0068 0.038 0.0065 0.0051 0.0014 0.0036 n-C₄H₁₀ 0.0047 0.024 0.0046 0.0039 0.0014 0.0030 C₅H₁₂ 0.0038 0.0075 0.0038 0.0035 0.0022 0.0032 C₆H₁₄ 0.0031 0.0019 0.0031 0.0030 0.0026 0.0030 C₇H₁₈ 0.0024 0.0007 0.0024 0.0024 0.0022 0.0024

Total 1.0000 0.3244 0.0769 0.0165 0.041

V = 1 – 0.041 = 0.959 kmol and L = 0.041 kmol Ans.

EXERCISE 5.26

Basis: 1 kmol crude gas mixture from demethanizing unit Cryogenic temp. in cold box = 111 K (– 162.15 °C)

Component y_i K*_i L₁ = L₂ = L₃ = L₄ = V 0.5 0.25 0.22 0.215

H₂ 0.750 49.73 0.0148 0.0050 0.0042 0.0041 0.7459 CH₄ 0.200 0.057 0.1892 0.1708 0.1664 0.1655 0.0345 C₂H₆ 0.045 9.45 ´ 10^–5 0.0450 0.0450 0.0450 0.0450 Nil N₂ 0.005 1.158 0.0022 0.0011 0.0010 0.0010 0.0040 1.000 — 0.2512 0.2219 0.2166 0.2156 0.7844

* K_i values are calculated from Table 5.15 of the text at T = 111 K.

New basis: crude gas stream flow = 18 720 Nm³/h

= 835.19 kmol/h Ingoing gas stream:

components n_i Molar (m )_i H_i , kJ/kg m_i × H_i

kmol/h mass kg/h at 2.75 MPa a kW

and 300.15 K

H₂ 626.4 2 1252.8 3972.1 1382.29

CH₄ 167.03 16 2672.5 888.43 659.54

C₂H₆ 37.58 30 1127.4 607.82 190.35

N₂ 4.18 28 117.0 305.67 9.93

Total 835.19 5169.8 2242.11

Product gas stream:

component kmol/h Molar kg/h H_i, kJ/kg m_i× H_i ni mass (m )_i at 2.75 MPa a, kW

and 297.65 K

H₂ 622.98 2 1246.0 3935.7 1362.19

CH₄ 28.81 16 461.0 883.44 113.13

C₂H₈ Nil 30 Nil 604.90 Nil

N₂ 3.34 28 93.5 303.17 7.83

Total 655.13 1800.5 1483.19

Tail gas stream:

component ni Molar (m )i H_i, kJ/kg m_i× H_i

kmol/h mass kg/h at 140 KPa a, kW

and 297.15 K

H₂ 3.42 2 6.8 3917.6 7.40

CH₄ 138.22 16 2211.5 907.34 557.38

C₂H₆ 37.58 30 1127.4 665.26 208.34

N₂ 0.84 28 23.5 308.14 2.01

Total 180.06 3369.3 775.13

Net heat change, f₁ = 775.13 + 1483.19 – 2242.11

= 16.21 kW (endothermic) Heat infiltration, f₂ = 20.9 kW

Refrigeration requirement, f₃ = 20.9 – 16.21 = 4.69 kW º 1.334 TR

Recovery of hydrogen = (622.98 ´ 100)/626.4 = 99.45%

Purity of product hydrogen stream = 622.98 ´ 100/655.13

= 95.1% Ans.

EXERCISE 5.27

(a) C₇H₁₆(g) + 11 O₂(g) ¾¾¾® 7 CO₂(g)+ 8 H₂O(g)

Heat of combustion of n-heptane (g), DH°_c = – 4501.46 kJ/kmol 7 C(s) + 7 O₂(g) ¾¾¾® 7 CO₂(g)

Heat of combustion of carbon (s), DH°c= – 393.51 kJ/kmol 8 H₂(g)+ 4 O₂(g) ¾¾¾® 8 H₂O(g)

Heat of combustion of hydrogen (g), DH°c (g) = – 241.82 kJ/kmol Heat of formation of n-heptane (g) = Heat of combustion of carbon +

Heat of combustion of hydrogen – Heat of combustion of n-heptane DH°_f = – 7 ´ 393.51 – 8 ´ 241.82 + 4501.46

= – 187.67 kJ/kmol Ans.

From Table AV.2 DH°_f = – 187.78 kJ/kmol (b) C₂H₅OH (g)¾¾¾® 2 CO₂(g)+ 3 H₂O(g)

Heat of combustion of ethyl alcohol (g), DH°c = – 1277.53 kJ/kmol 2 C(s) + 2 O₂(g) ¾¾¾® 2 CO₂(g)

3 H₂(g) + 1.5 O₂(g) ¾¾¾® 3 H₂O(g)

D H°_f of C₂H₅OH (g) = – 2 ´ 393.51 – 3 ´ 241.82 + 1277.53

= – 234.95 kJ/kmol Ans.

From Table AV.2, DH°_f = – 234.95 kJ/kmol (c) C₄H₆(l) + 5.5 O₂ (g) ¾¾¾® 4 CO₂(g) + 3 H₂O(g)

Heat of combustion of butadiene (1,3), DH°c = – 2386.69 kJ/kmol 4 C(s) + 4 O₂(g) ¾¾¾® 4 CO₂(g)

3 H₂(g) + 1.5 O₂(g) ¾¾¾® 3 H₂O(g)

D H°_f of C₄H₆(l) = – 4 ´ 393.51 – 3 ´ 241.82 + 2386.69

= 87.19 kJ/kmol Ans.

From Table AV.2, DH°_f = 87.19 kJ/kmol

EXERCISE 5.28

(a) DH°_c(g) for hydrogen = – 241.82 kJ/kmol l_v at 298.15 for water = 2442.5 kJ/kg

º 44.002 kJ/mol D H°_f(l) = – 241.82 – 44.002

= – 285.82 kJ/mol Ans.

(b) D H°_f(g) for methanol = – 200.94 kJ/mol Watson equation:

l_v at 298.15 K for methanol = 35 210 512.5 – 298.15 ^0.38 512.5 – 337.7

é ù

ê ú

ë û

= 38 048 kJ/kmol

D H°_f(l)= – 200.94 – 38.05 = – 238.99 kJ/mol Ans.

NIST equation:

For methand, p_c= 80.84 bar, T_c = 512.5 K At T = 298.15 K T_r = 0.5818 l_v at 298.15 K= 45.3 ´ e–(–0.31)(0.5818) ´ (1 – 0.5818)^0.4241

= 37.485 kJ/mol DH°_f (l) = –200.94 – 37.485

= –238.43 kJ/mol Ans.

(c) D H°_f(g) for CS₂= 117.36 kJ/mol Watson equation:

l_v= 26 740

552 – 298.15 0.38

552 – 319.04

é ù

ê ú

ë û

= 27 627 kJ/kmol D H°_f(l) = 117.36 – 27.64

= 89.72 kJ/mol Ans.

NIST equation:

For CS₂, p_c= 73.00 bar T_c= 552.0 K

At T = 298.15 K T_r = 298.15/552 = 0.5786 l_v at 298.15 K = 37.07 ´ e–(0.2264 ´ 0.5786) ´ (1 – 0.5786)^0.2264

= 26.739 kJ/mol

DH°_f(l) = +117.36 – 26.739 = 90.62 kJ/mol Ans.

EXERCISE 5.29

Basis: 1 kg NaHCO₃

ro

H

D = (– 1130.68 – 393.51 – 241.82) – 2(– 950.81)

= + 135.61 kJ/2 mol NaHCO₃

º + 67.805 kJ/mol NaHCO₃ (endothermic) Molar mass of NaHCO₃= 84.0066

Moles of NaHCO₃= 1000/84.0066 = 11.904 mol/kg Heat to be supplied for dissociation = 67.805 ´ 11.904

= 807.15 kJ/kg NaHCO₃ Ans.

EXERCISE 5.30

ro

H

D = DH_f^o DH

products fo

reactants

d i

^–

d i

= (– 1412.2 – 393.51) – (– 1130.68 – 824.2)

= 149.17 kJ/mol Na₂CO₃ or mol Fe₂O₃ Ans.

EXERCISE 5.31

Basis: 100 kg CH₄ reformed Input heat balance:

Gas kg Molar kmol Tempera- (H^° – DH°_o + DH°_f)_i n_i × (H° – DH°_o mass n_i ture K(°C) kJ/kmol + DH°_f), kJ

CH₄ 100 16 6.250 698 (425) – 38 416 – 240 100

O₂ 100 32 3.125 698 (425) + 2 102 + 65 691

H₂O 100 18 5.55 1253 (980) – 192 447 – 1068 081

Total 300 — 14.925 — — – 1242 490

Outlet Heat balance:

Gas kg Molar kmol Tempera- (H^° – DH°_o – DH°f)_i n_i × (H° – DH°_o mass n_i ture K(°C) kJ/kmol +DH°f)_i, kJ H₂ 27.3 2 13.65 1198 (925) + 35 134 + 479 579 H₂O 79.2 18 4.40 1198 (925) – 194 882 – 857 481 CO 149.8 28 5.10 1198 (925) – 76 942 – 392 404 CO₂ 50.7 44 1.15 1198 (925) – 339 706 – 390 662

Total 300.0 — 24.30 — — 1160 968

Heat of reaction, D ° = 1160 968 – (1242 490)Hr

= 81 522 kJ (endothermic) Ans.

EXERCISE 5.32

Basis: 0.1 kg coke and N₂ flow of 0.025 L × s^–1. These are same bases as those of Exercises 3.24 and 4.1.

Reaction: CH_0.6 + 1.6 O₂ = CO₂ + 0.6 H₂

– 101.1 1.6 ´ 0 – 393.51 0.6 ´ (– 241.82) DH°_f Hr

D ° = – 393.51 – 0.6 ´ 241.82 + 101.1

= – 393.51 – 145.09 + 101.1

= – 437.5 kJ/1.6 mol oxygen

(exothermic) Actual oxygen consumed = 0.128 mol

Heat generated = 437.5 ´ 0.128/1.6

= 35 kJ Ans.

EXERCISE 5.33

Many data are taken from Example 4.20.

Heat capacity Data of RII Exit Gas stream

Component n._i Heat capacity (C°_mpr2) equation constants kmol/h a_i× n._i b_i× n._i ´ 10³ c_i× n._i ´ 10⁶ d_i× n._i ´ 10⁹

CH₃OH 1.415 35.2 72.0 83.0 – 63.9

HCHO 218.267 10 520.5 — — —

CO₂ 16.481 352.1 1059.5 – 676.6 161.5

CO 1.857 53.9 – 5.2 21.6 – 8.7

H₂ 3.714 106.3 3.8 – 0.5 2.9

CH₄ 1.16 22.3 60.5 13.9 – 13.1

(CH₃)₂O 1.858 122.3 — — —

O₂ 159.270 4145.1 1872.2 – 373.1 – 89.6

N₂ 1082.930 32 044.9 – 5567.3 14 276.2 – 5380.0

H₂O 268.04 8709.2 21.3 3541.0 – 1218.9

Total 1754.852 56 111.8 – 2483.2 16 885.5 – 6609.8 In the gas to gas heat exchanger, air is preheated from 308.15 K to 523.15 K. Air flow rate is 1395.1 kmol/h against 1363.1 kmol/h in Example 4.10.

From Example 5.37 (Table 5.31),

Heat duty of heat exchanger f₆ = (1395.1/1363.1) 308 15

z

C°^mpa

523 15 . .

= 8935 667 kJ/h º 2482.13 kW

In the static mixer (I), 1395.1 kmol/h of air at 523.15 K is mixed with 93 kmol/

h methanol at 351.47 K (i.e. at the outlet of evaporator).

From Table 5.31, Enthalpy of reactor input stream over 298.15 K,

f₂ = (93/125) 298 15

z

C°^mpme Heat of reactions at 298.15 K in RI,

HrI

D ° = 20320 839 ´ 92.07/123.75

= 15118 704 kJ/h º 4199.64 kW Heat Capacity Data for RI Exit Gas stream

Component n._i, kmol/h Heat capacity (C_pmr1° ) equation constants a_i × n._i b_i × n._i ´ 10³ c_i × n._i ´ 10⁶ d_i × n._i ´ 10⁹

Total 1543.714 46 861.7 – 2282.4 15 119.6 – 6048.8 Enthalpy of RI exit stream, at 613.15 K over 298.15 K,

¢

f₃ = ^{298 15613 15}

z

^..C_mpr1° ^dT

= 15 260 251 kJ/h º 4238.96 kW Heat transfer in RI = 9546 895 + 15 118 704 – 15 260 251

= 9405 348 kJ/h º 2612.60 kW

Enthalpy of additional methanol vapours added to RI exit stream

f¢¢₃ = 140 548 125

. ^{298 15351 47}

z

^..C_mpa° ^dT

= 344 999 kJ/h º 95.83 kW Total enthalpy of gas mixture, entering RII

= 15 260 251 + 344 999

= 15 605 250 kJ/h Let T₂ be the temperature of stream, entering RII.

By Mathcad, T₂ = 582.3 K or 309.15 °C Enthalpy of RII exit gas stream,

f₄ = ^{298 15613 15}

z

^..C°^{mpr2 dT}

= 18 246 584 kJ/h º 5068.50 kW Heat of reaction in RII = 140 063

123 75 .

. ´ 20 320 639

= 22 999 351 kJ/h º 6388.71 kW

Heat transfer in RII = 15 260 251 + 22 999 351 – 18 246 584

= 20 013 018 kJ/h º 5559.17 kW

RII exit gases leave the gas heat exchanger at 383.15 K (110°C).

Heat transfer in BFW heater and heat exchanger

= ^{383 15613 15}

z

^..C°^mpr2 = 13 404 122 kJ/h

Heat transfer in BFW heater,

f₅ = 13 404 122 – 8935 667

= 4468 455 kJ/h º 1241.24 kW Total heat transfer taken place for steam generation

= 9405 348 + 20 013 018 + 4468 455

= 33 886 821 kJ/h º 9413.01 kW Steam generated = 33 886 821

2629 3.

= 12 888.2 kg/h Total methanol evaporated = 93 + 140.548

= 233.548 kmol/h Heat duty of evaporator f₁ = 233 548 4551 263

125

. ´

= 8503 506 kJ/h º 2362.01 kW Steam consumption in evaporator, m_s = 8503 506

2119 7.

= 4011.7 kg/h Ans.

EXERCISE 5.34

(a) Basis: 1 kmol CH₄

Reaction: CH₄ + C₂H₄ = C₃H₈ (gaseous)

Gas DH°_f Constants of C°_mp equation

kJ/kmol a b ´ 10³ c ´ 10⁶ d ´ 10⁹

CH₄ – 74 520 19.2494 52.1135 11.973 – 11.3173

C₂H₄ + 52 550 4.1261 155.0213 – 81.5455 16.9755 C₃H₈ – 104 680 – 4.2227 306.264 – 158.6316 32.1455

H r

D °_r= – 104 680 – (– 74 520 + 52 550) = – 82 710 kJ/kmol Da = – 4.2227 – 4.1261 – 19.2494 = – 27.5982

Db ´ 10³= 306.264 – (155.0213 + 52.1135) = 99.1292 Dc ´ 10⁶= – 158.6316 + 81.5455 – 11.973 = – 89.0591

Dd ´10⁹= 32.1455 – 16.9755 + 11.3173 = 26.4873

DH°_o = DH°_RT – DaT – (Db/2)T² – (Dc/3)T³ – (Dd/4)T⁴ For T = 298.15 K, D °H_rT = HD °_r = – 82 710 kJ/kmol

H0

D ° = – 82 710 + 27.5982 ´ 298.15 – (99.1297 ´ 10^–3/2) (298.15)² + (89.0591 ´ 10^–6/3)(298.15)³ –

(26.4873 ´ 10^–9/4)(298.15)⁴

= – 82 710 + 8828 – 4406 + 787 –52

= – 77 553 kJ/kmol – heat of reaction at 0 K HrT

D ° = – 78 154 – 27.598 T + 49.565 ´ 10^–3 T² – 29.686 ´ 10^–6 T³ + 6.622 ´ 10^–9 T⁴Ans.

(b) CO(g) + H₂O(g) = CO₂(g) + H₂(g)

D H°_f – 110.53 – 241.82 – 393.51 0 all in kJ/mol D °H_r = – 393.51 – (– 110.53 – 241.82)

= – 41.16 kJ/mol

Gas Constants of C°_mp equation

a b ´ 10³ c ´ 10³ d ´ 10³

CO₂ 21.3655 64.2841 – 41.0506 9.7999

H₂ 28.6105 1.0194 – 0.1476 0.769

CO 29.0277 – 2.8165 11.6437 – 4.7063

H₂O 32.4921 0.0796 13.2107 – 4.5474

Da = 21.3655 + 28.6105 – 29.0277 – 32.4921 = – 11.5438 Db ´ 10³= 64.2841 + 1.0194 – (– 2.8165 + 0.0796) = 68.0404 Dc ´ 10³= – 41.0506 – 0.1476 – (11.6437 + 13.2107) = – 66.0526 Dd ´ 10³= 9.7999 + 0.769 – (– 4.7063 – 4.5474) = 19.8226

H0

D ° = – 41 160 + 11.5438 (298.15) + 68.0404 ´ 10^–3 (298.15)²/2 + 66.0526 ´ 10^–6 (298.15)³/3 – 19.8226 ´ 10^–9 (298.15)⁴/4

= –41 160 + 3442 – 3024 + 583 – 39

= –40 378 kJ/kmol at 0 K H

D °_r = –40 378 – 11.544 T + 34.02 ´ 10^–3 T² – 22.018 ´ 10^–6 T³

+ 4.956 ´ 10^–9 T⁴ Ans.

(c) Basis = 1 kmol CO consumed or CH₃OH produced CO(g) + 2 H₂(g) = CH₃OH(g)

Gas DHº_f Constants of Cº_mp equation

kJ/mol a b ´ 10³ c ´ 10³ d ´ 10³

CH₃OH – 200.94 24.8692 50.8755 58.6278 – 45.1266 CO – 110^.53 29.0277 – 2.8165 11.6437 – 4.7063

Let H¢ be enthalpy in kJ/kg, T be temperature in K andp be pressure in bar a

H = H¢

´

1.055 056 2.204 623

= H¢

2 326.

t = 1.8T – 459.67

and p¢ = p

14 503 77. or p = 14.503 77 p¢

Substitute these values in equation.

H

EXERCISE 5.36

Basis: 100 kmol outgoing gas mixture

Gas kmol n_i Constants of Cº_mp equation

incoming outgoing a b ´ 10³ c ´ 10⁶ d ´ 10⁹ CO₂ 58.9 57.10 21.3655 64.2841 – 41.0506 9.7999

H₂ 47.94 41.10 28.6105 1.0194 – 0.1476 0.769

CH₄ — 1.68 19.2494 52.1135 11.973 – 11.3173

CO — 0.12 29.0277 – 2.8165 11.6437 – 4.7063

Total 107.84 100.00

= – 164.65 kJ/mol CO₂ reacted (exothermic) Actual CO₂ consumed as per reaction (1) = 1.68 kmol

Heat evolved = 164 650 ´ 1.68 = 276 612 kJ

CO₂(g) + H₂(g) = CO(g) + H₂O(g) (2)

DHº_r2= – 241.82 – 110.53 + 393.51

= 41.16 kJ/mol CO₂ (endothermic) Actual CO₂ consumed as per reaction (2) = 0.12 kmol Heat absorbed = 41 160 ´ 0.12 = 4939 kJ

Total enthalpy of product gas mixture

= 1129 704 + 276 612 – 4939 = 1401 377 kJ Let T₃ be the final temperature of product gas mixture.

H₁= ³ _i

(T²₃ – 298.15²)/2 + (– 41.0506 ´ 57.1 – 0.1476 ´ 41.1 + 11.973 ´ 1.68 + 11.973 ´ 1.68 + 11.6437 ´ 0.12)

´ 10^–6 (T₃² – 298.15²)/3

+ (9.7999 ´ 57.1 + 0.769 ´ 41.1 – 11.3173 ´ 1.68

– 4.7063 ´ 0.12) ´ 10^–9 (T₃– 298.15)/4

H₂= 2431.68 (T₃ – 298.15) + 3799.7 ´ 10^–3 (T₃² – 298².15)/2

– 2328.54 ´ 10^–6 (T³₃ – 298³.15)/3 + 571.6 ´ 10^–9 (T⁴₃ – 298⁴.15)/4

= 2431.68 T₃ + 1899.87 ´ 10^–9 T⁴₃ – 776.23 ´ 10^–6 T⁹₃

– 142.9 ´ 10^–9 T⁴₃ – 724 641 – 168 716 + 20 542 – 1127

= 1401 377 Using Mathcad

T₃= 669.61 K or 396.46°C Ans.

EXERCISE 5.37

Basis: 100 kmol dry gas mixture, entering the shift converter.

Refer Exercise 5.32 (b).

T₁= 618.15 K

DH º_rT1= – 41 160 – 7134 + 12 993 – 5197 + 723

= – 39 775 kJ/kmol CO T₂= 783.15 K

DH º_rT2= – 41 160 – 9039 + 20 857 – 10 570 + 1863

= – 38 049 kJ/kmol CO

Average heat of reaction, DHº_rT = [– 39 775 + (– 38 049)]/2

= – 38 912 kJ/kmol CO

Gas (Hº – Hº + DHº_f), kJ/kmol Average Cº_mp between

618 K 783 K 618 K and 783 K,

CO – 95 804 – 90 644 31.273

H₂ 17 723 22 585 29.467

CO₂ – 370 400 – 362 262 49.321

N₂ 18 029 23 126 30.891

Ar — — 39.946 (constant)

H₂O – 218 133 – 211 933 37.576

Partial oxidation of NG:

Sn_i× Cº_mpi= (0.37 ´ 31.272 + 0.6 ´ 29.467 + 0.2 ´ 49.321 + 0.01 ´ 39.948) 100

= 3063.7 kJ/(100 kmol dry gas × K)

Partial oxidation of FO: For steam, accompanying the gas mixture,

n_i× Cº_mpt= 37.576 ´ 100 ´ x

= 3757.6 x kJ/(100 kmol × K) Rise in temperature in shift converter,

DT = 38912

Basis: Flowrate of offgases from uranium oxide dissolver

= 38.3 Nm³/h Total heat generated = 455 567 ´ 1.025

= 466 956 kJ/d º 5.405 kW Theoretical NH₃ required = 4 ´ 1.025/3 = 1.367 kmol/d

Actual NH₃ fed = 2 Nm³/h º 0.089 kmol/h º 2.141 kmol/d

Excess NH₃ = [(2.141 – 1.367)/1.367]/100

= 56.6% Ans. (a)

Reactants:

Total 41.099 1336.31 94.86 464.49 – 178.3

Enthalpy of reactants at 393.15 K (120ºC) over 298.15 K (25ºC):

f₁= 1336.31 (393.15 – 298.15) + 94.86 ´ 10^–3 (393.15² – 298.15²)/2

+ 464.49 ´ 10^–6 (393.15³ – 298.15³)/3 – 178.3 ´ 10^–9 (393.15⁴ – 298.15⁴)/4

= 126 949 + 3114 + 5301 – 712

= 134 652 kJ/d º 1.558 kW Total enthalpy of product gas mixture

= f₁ + DHº_r

= 134 652 + 466 956

= 601 608 kJ/d º 6.963 kW Products:

Constants of Cº_mp equation Component n.

Total 44.005 1375.07 – 11.83 539.14 – 196.05 Enthalpy of products at T K over 298.15 K,

f₂= 1375.07 (T – 298.15) – 11.83 ´ 10^–3 (T² – 298.15²)/2

+ 539.14 ´ 10^–8 (T³ – 298.15³)/3 – 196.05 ´ 10^–3 (T⁴ – 298.15⁴)/4

= 1375.07 – 5.92 ´ 10^–3 T² + 179.71 ´ 10^–6 T³ – 49.01 ´ 10^–9 T⁴

– 409 771 + 525 – 4756 + 387

= 601 608 kJ/d º 6.963 kW

By trial and error or by Mathcad, T = 703.78 K or 430.63ºC Ans. (b)

EXERCISE 5.39

Basis: 38.3 Nm³/h offgases

Flow rate = 41.01 kmol/d (Ref. Exercise 5.38) NO₂ entering the reactor = 41.01 ´ 0.035

= 1.435 kmol/d

Theoretical NH₃ required = 1.435 ´ 4/3 = 1.913 kmol/d Excess NH₃ = [(2.141 –1.913)/1.913]100 = 11.92%

Total heat generated = 1.435 ´ 455 567 = 653 739 kJ/d º 7.566 kW Reactants:

Constants of Cº_mp Components n.

Total 43.151 1329.84 118.95 446.17 – 173.58 Enthalpy of reactants at 393.15 K (120ºC) over 298.15 K (25ºC),

f₁= 1329.84 (393.15 – 298.15) + 118.95 ´ 10^–3 (393.15² – 298.15²)/2 + 446.17 ´ 10^–6 (393.15³ – 298.15³)/3 – 173.58 ´ 10^–9 (393.15⁴ – 298.15⁴)/4

= 126 335 + 3904 + 5092 – 693

= 134 638 kJ/d º 1.558 kW

This enthalpy does not include enthalpy of nitrogen (x kmol/d) at 303.15 K (30ºC).

Total enthalpy of product gas mixture (excluding that of N₂ stream), f¢₁= 134 638 + 653 739

= 788 377 kJ/d º 9.125 kW NH₃ consumed= 1.913 kmol/d

N₂ produced = 1.913 ´ 7/8 = 1.674 kmol/d Total N₂ in product stream = 10.868 + 1.674 = 12.542 kmol/d

H₂O produced = 1.913 ´ 12/8 = 2.87 kmol/d Total H₂O in product stream = 26.656 + 2.87 = 29.526 kmol/d Products (excluding extra nitrogen supply):

Constants of Cº_mp equation Components n.

Total 44.347 1389.72 – 30.39 550.68 – 198.43 Enthalpy of product gas stream (excluding extra N₂ supply) at 753.15 K over 298 K, f¢₂ = 1389.72 (753.15 – 298.15)

– 30.39 ´ 10^–3 (753.15² – 298.15²)/2

+ 550.68 ´ 10^–6 (753.15³ – 298.15³)/3

– 198.43 ´ 10^–9 (753.15⁴ – 298.15⁴)/4

= 632 323 – 7266 + 73 515 – 15 558

= 683 014 kJ/d º 7.905 kW Enthalpy carried by extra nitrogen in feed stream

f¢₁ – f¢₂ = 788 377 – 683 014

= 105 363 kJ/d º 1.219 kW

Enthalpy of extra nitrogen at 753.15 K over 303.15 K

= 29.5909 (753.15 – 303.15) – 5.141 ´ 10^–3 (753.15² – 303.15²)/2 + 13.1829 ´ 10^–6 (753.15³ – 303.15³)/3 – 4.968 ´ 10^–9 (753.15⁴ – 303.15⁴)/4

= 133 16 – 1222 + 1754 – 389

= 13 459 kJ/kmol N₂

Flow of nitrogen required to control the temperature

= 105 363/13 459

= 7.828 kmol/d

º 7.31 Nm³/h Ans.

EXERCISE 5.40

Basis: 1 kmol toluene

O₂ requirement for benzaldehyde production = 1 kmol Excess O₂ = 100%

Actual O₂ supply = 2 kmol

N₂ supply = 2 ´ 79/21 = 7.524 kmol Air supply = 2 + 7.524 = 9.524 kmol

DH^o_r = – 36.7 – 241.82 – (50.17) = – 328.69 kJ/mol Side reaction is the standard combustion reaction.

DH^o_c = –3772.0 kJ/mol (NHV)

Toluene burnt to CO₂ and H₂O = 0_.005 kmol Total toluene converted = 0.13 kmol

Total oxygen consumed = 0.125 + 0.005 ´ 9

= 0.17 kmol

Toluene reacted to benzaldehyde = 0.13 – 0.005 = 0.125 kmol Total heat produced at 298.15 K (25ºC)

= 0.125 ´ 328 690 + 0.005 ´ 377 200

= 42 972 kJ Inlet gas mixture is at 448.15 K (175ºC):

Constants of Cº_mp equation

Components kmol mole % n_i×a_i n_i×b_i ´ 10³ n_i×c_i ´ 10⁶ n_i×d_i ´ 10⁹ n_i

C₇H₈ 1.0 9.5 – 35.19 563.18 – 349.80 82.59

O₂ 2.0 19.0 52.05 23.51 – 4.69 – 1.12

N₂ 7.524 71.5 222.64 – 38.68 99.19 – 37.38

Total 10.524 100.0 239.500 548.01 – 255.30 44.09 Enthalpy of ingoing gas mixture at 448.15 K over 298.15 K,

H₁= 239.5 (448.15 – 298.15) + 548.01 ´ 10^–3 (448.15² – 298.15²)/2

– 255.3 ´ 10^–6 (448.15³ – 298.15³)/3 + 44.09 ´ 10^–9 (448.15⁴ – 298.15⁴)/4

= 35 925 + 30 673 – 5404 + 357

= 61 551 kJ

Outlet gas mixture at 468.15 K (195ºC):

Constants of Cº_mp equation Components kmol mole % n_i× a_i n_i× b_i ´ 10³ n_i× c_i ´ 10⁶ n_i× d_i ´ 10⁹

n_i

C₇H₈ 0.87 8.36 – 30.62 489.97 – 304.32 71.85

O₂ 1.83 17.59 47.63 21.51 – 4.29 – 1.03

N₂ 7.524 72.32 222.64 – 38.68 99.19 – 37.38

C₇H₈O 0.125 1.20 16.25 — — —

CO₂ 0.035 0.34 0.75 2.25 1.44 0.34

H₂O 0.020 0.19 0.65 — 0.26 – 0.09

Total 10.404 100.00 257.30 475.05 – 210.60 33.69 Enthalpy of outcoming gas mixture at 468.15 K over 298.15 K,

H₂= 257.30 (468.15 – 298.15) + 475.05 ´ 10^–3 (468.15² – 298.15²)/2

– 210.6 ´ 10^–6 (468.15³ – 298.15³)/3 + 33.69 ´ 10^–9 (468.15⁴ – 298.15⁴)/4

= 43 741 + 30 943 – 5342 + 338 = 69 680 kJ Heat liberated = 61 551 + 42 972 – 69 680

= 34 843 kJ Ans.

EXERCISE 5.41

Basis: 1 kmol S₂(g)

Stoichiometric CH₄ requirement = 0.5 kmol Actual supply of natural gas = 2 kmol

Actual CH₄ supply = 2 ´ 0.6 = 1.2 kmol Excess CH₄ = 1.2 – 0.5 = 0.7 kmol

% Excess = 0.7 ´ 100/05 = 140 Ans. (a)

Conversion = 80%

S₂ reacted = 0.8 kmol

CS₂ produced = 0.8/2 = 0.4 kmol Ratio = CS produced

S fed Antoine constants for CS₂

A = 6.0677, B = + 1169.1 and C = – 31.55 at T = 298.15 K

Basis: 3 kmol H₂S supply, including the bypass

Out of 3 kmol, 1 kmol H₂S enters the furnace and 2 kmol bypass the furnace.

Out of 3 kmol, 1 kmol H₂S enters the furnace and 2 kmol bypass the furnace.

In document 1.DISPOSICIONES GENERALES (página 32-35)