Theoretical Flame Temperature

air or oxygen without loss or gain of heat is called the Theoretical flame temperature.

WORKED EXAMPLES

9.1 Calculate the enthalpy of sublimation of Iodine from the following reactions and data

(a) H₂ (g) + I₂ (s) Æ 2HI(g) DH = 57.9 kJ (b) H₂ (g) + I₂ (g) Æ 2HI(g) DH = –9.2 kJ The desired reaction is I₂(s) Æ I₂ (g)

Solution: (a) – (b) = DH = 67.1 kJ

9.2 Find the enthalpy of formation of liquid ethanol from the following

data: _{–DH, Heats of reaction, kJ (1) C}

2H5OH (l) + 3O2(g) Æ 2CO2 (g) + 3H2O(l) – 1367.8 (2) C

(graphite) + O₂(g) Æ CO₂(g) – 393.5 (3) H₂(g) + ½O₂(g) Æ H₂O(l) – 285.8 Solution: [2 ¥ (2) – (1) + (3 ¥ (3))] = 2C + 3H₂ + ½O₂Æ C₂H₅OH. The enthalpy of formation of ethanol = –276.6 kJ

9.3 200 kg of Cadmium at 27 °C is to be melted. (The melting point is 320.9 °C). The heat supply is

from a system, which supplies 210 kcal/ kg, at steady state. Find the quantity of heat to be supplied by the system.

Atomic weight of Cadmium = 112.4, C_p = (6 + 0.005T) kcal/kmole °C and T in °C. Latent heat of fusion = 2050 kcal/kmole

Basis: 200 kg of Cadmium º 1.78 katoms

©¸ ¦µ

2

Sensible heat = 1.78t tª¹ = 3885.5 kcal ª¹

Ǽ

Latent heat of fusion = 1.78 ´ 2050 = 3649.0 kcal Total heat to be supplied = 7534.5 kcal

Quantity of steam to be supplied = 7534.5= 35.88 kg₂₁₀

9.4 An evaporator is to be fed with 10,000 kg/h of a solution having 1% solids. The feed is at 38 °C.

It is to be concentrated to 2% solids. Steam at 108 °C is used. Find the weight of vapour formed and the weight of steam used. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kcal/kg, steam is 540 kcal/kg and that of the vapour is 644 kcal/kg.

Basis: One hour.

Feed = 10,000 kg/h

Vapour formed is 5000 kg/h Thick liquor is 5000 kg/h

Enthalpy of feed = 10,000 ´ 38.1 = 38.1 ´ 104 kcal

Enthalpy of the thick liquor = 100.8 ´ 5,000 = 5,04,000 kcal. Enthalpy of the vapour = 644 ´ 5,000 = 32,20,000 kcal. Heat supplied by steam = M_sl_s = M_s ´ 540 kcal.

Heat balance:

Heat input by steam + heat in, by feed

= Heat out, in vapour + Heat out, thick liquor or, [M_s.(540) + 38.1 ´ 104] = (32,20,000 + 5,04,000) \ M_s (540) = 33,43,000.

Thus the weight of steam required, M_s = 6,190.75 kg/hr.

9.5 Calculate the standard heat of reaction:

CaC₂ + 2H₂O ® Ca(OH)₂ + C₂H₂

DH_f cal/mole –15,000 –68,317.4 –2,35,800 54,194

Solution:

29,971.2 cal/mole

9.6 How much heat must be added to raise the temperature of 1 kg of a 20% caustic solution from 7

°C to 87 °C? Take datum temperature as 0 °C. Data:

Specific heat at 7 °C = 3.56 and at 87 °C = 3.76 kJ/kg K

Solution: Q = (m.C_p.t)₁ – (mC_pt)₂ = 1 [(3.76 ´ 87) – (3.56 ´ 7)] = 302.2 kJ

9.7 How many Joules are needed to heat 60 kg of sulphur trioxide from 273.16 K to 373.16 K?

C_pSO3 = 34.33 + 42.86 ´ 10–3T – 13.21 ´ 10–6T2 J/mole K Solution: Number of moles of the trioxide = 60 = 0.75 kmole₈₀

At 373.16 K,

Q = nòC_pSO3 dt

At 273.16 K

Q = [{34.33 ´ (373.16 – 273.16)} + {(42.86 ´ 10–3/2)(373.162 – 273.162)} + {(–13.21 ´ 10–6/3)(373.163 – 273.163)}] Q = 3,509.25 kJ/kmole.

9.8 Using the following data of heats of combustion in cal/g mole, calculate the following:

(a) Heats of combustion of benzene to water (b) Heat of vaporization of benzene – cal/g mole (i) C₆H₆ (l) to CO₂ (g) and H₂O (l) = 7,80,980 (ii) C₆H₆ (g) to CO₂ (g) and H₂O (g) = 7,59,520 (iii) H₂ (g) to H₂O (l) = 68,317 (iv) H₂ (g) to H₂O (g) = 59,798 (v) Graphite to CO₂ (g) = 94,052 Desired reactions:

(a) C₆H₆ (l) + 7.5O₂ ® 6CO₂ (g) + 3H₂O (l) (b) C₆H₆ (l) ® C₆H₆ (g)

(a) Equation (i) itself gives value, DH_c = – 7,80,980 cal/g mole. (ii) C₆H₆ (g) + 7.5O₂ ® 6CO₂ (g) + 3H₂O (g)

(iii) H₂ (g) + ½O₂ ® H₂O (l) (iv) H₂ (g) + ½O₂ ® H₂O (g) (v) C + O₂ ® CO₂ (g)

We can obtain the reaction (b) from the reaction (i) to (v) using

suitable multiplication factor for each step and adding or subtracting the equations as shown below: i.e. Steps for equation. (b) = (i) + 3(iv) – (ii) – 3(iii) l = 8,097 cal/g mole

9.9 Find the heat of formation of ZnSO₄ from its elements and from these

data: _kcal/mole

(i) ZnS ® Zn + S 44

(ii) 2ZnS + 3O₂ ® 2ZnO + 2SO₂ –221.88 (iii) 2SO₂+ O₂ ® 2SO₃ –46.88

(iv) ZnSO₄ ® ZnO + SO₃ 55.1

Desired equation: Zn + S + 2O₂ ® ZnSO₄ kcal/mole Steps: ½ [(ii) + (iii) – 2(i) – 2(iv)] = –233.48 kcal/mole.

9.10 Steam that is used to heat a batch reaction vessel enters the steam chest, which is segregated

from the reactants, at 250 °C, is saturated and completely condensed. The reaction absorbs 1000 Btu/lb of charge in the reactor. Heat loss from the steam chest to the surroundings is 5000 Btu/h. The reactants are placed in the vessel at 70 °F. At the end of the reaction, the materials are at 212 °F. If the charge contains 325 lb of material and the products and reactants have an average C_p of 0.78 Btu/1b °F, how many lb of steam are needed per lb of charge. The charge remains for an hour in the vessel.

Basis: One hour: Datum 70 °F Btu

Reaction absorbs heat = 1000 ´ 325 = 3,25,000 Heat loss to surroundings = 5,000

Heat in products: 325 ´ 0.78 (212 – 70) = 36,000 \ Q = total heat = 3,66,000

From steam tables at 482 °F(250 °C) l_s = 734.9 Btu/lb we have, Q = m_sl_s = (m_s) (734.9) = 3,66,000 Btu

\ m_s = 498.2 lb/h

lb of steam/lb of charge = 498.2= 1.533₃₂₅

9.11 Pure ethylene is heated from 30 °C to 250 °C at a constant pressure. Calculate the heat added per

kmole C_p = 2.83 + 28.601 ´ 10–3T – 87.26 ´ 10–7T2 where C_p is in kcal/kmole K and T in K T2

DH = n CdTp T1 = 303 K, T2 = 523 K∫

T1 n = 1 kmole Or, DH = [2.83 {T₂ – T₁} + (28.601 ´ 10–3/2) {T2 – T2}_{2 1} – {87.26 ´ 10

–7_/3}{T3 _{– T}3

2 1}] Heat added = 2,886.11 kcal

9.12 Calculate the amount of heat given off when 1 m3 of air at standard conditions cools from 500 °C to –100 °C at constant pressure. C_p air = 6.386 + 1.762 ´ 10–3 T – 0.2656 ´ 10–6 T2, where C_p is in kcal/kmole K and T in K. 1 m3 ₌1 _{= 0.0446 kmole} 22.414 173 Q = 0.0446 ∫C dT_{= 0.0446 [6.386 ´ 600 + (1.762 ´ 10}–3_/2)(1732 _{– 773}2₎ p 773 – (0.2656 –6_{/3) (173}3 _{– 773}3_{)]´ 10} Q = –191.345 kcal,

Hence, heat is given off

9.13 Air being compressed from 2 atm and 460 °C (enthalpy 210.5 Btu/lb) to 10 atm and 500 °R

(enthalpy 219 Btu/lb). The exit velocity of air is 200 ft/s. What is the horse power required for the compressor if the load is 200 lb of air/hour?

Basis: One hour.

(W_s = shaft work, Btu/lb v: velocity, ft/s)

W_s = (219 – 210.5) = 8.5 Btu/lb.

22'v¦µ (200) = 0.8 Btu/lb¨· §¶ tt 778)

Horse power needed = (8.5 + 0.8) ´200 _{= 0.73 HP} 2545

[1 Btu = 778 ft. lbf; 1 HP = 2545 Btu/h]

9.14 Find the heat of reaction at 1200 K. C₂H₆ ® C₂H₄ + H₂ 'H_{f,C H26} 84,720 kJ/kmole

'H_{f,C H24} 52,280 kJ/kmole

DH°_rxn, _{298 K}: 52,280 – (–84,720) = 1,37,000 kJ

DH_rxn = DH°_rxn – nC_pReactants(1200 – 298) + nC_pProducts (1200 – 298)

= (1,37,000) – [1 ´ 100 ´ (1200 – 298)] + [{(1 ´ 78.7) + (1 ´ 29.7)}(1200 – 298)] = 1,37,000 – 90,200 + 97,776.8

= 1,44,576.8 kJ/kmole (Heat to be supplied)

9.15 Calculate the heat input to raise the temperature of 132 kg of CO₂ from 100°C to 1000°C. Perform the calculation in the following ways. (a) by integrating the expression for C_p and (b) by using mean heat capacity value C_p in kcal/kmole K.

C_p = 6.85 + 8.533 ´ 10–3 T – 2.475 ´ 10–6 T2, kcal/kmole K Basis: 132 kg of CO₂ º 132/44 = 3 kmoles

T2

T1

– (2.475 –6 /3)(12733 – 3733)]´ 10 DH = 3[10,826.29] = 32,478.87 kcal

C_p values at 1273 K and 373 K are:

C_p at 1273 K = 13.702 kcal/kmole K and C_p at 373 K = 9.6884 kcal/kmole K C_pav = 11.6952 kcal/kmole K DH = òm C_pav dT = 3 ´ 11.6952 ´ (1273 – 373) = 31,577.04 kcal

9.16 SO₂ gas is oxidized in 100% excess air with 70% conversion to SO₃. The gases enter the converter at 400 °C and leave at 450 °C. How many kcals are absorbed in the heat exchanger of the converter per kmole of SO₂ sent?

Basis: 1 kmole SO₂.

SO₂ + ½O₂ ® SO₃ O₂ sent = 0.5 ´ 2 = 1. SO₃ formed = 0.7 kmole N₂ in air = 1 ´ 79/21 = 3.76 kmole. SO₂ remaining = 0.3 kmole

SO₃ SO₂ O₂ N₂ Total gases leaving kmoles 0.7 0.3 0.65 3.76 5.41 C_p mean, cal/g mole °C 15.5 11.0 7.5 7.1 —

H_rxn = –23, 490 cal/g mole (given)D DH_rxn = –23,490 ´ 0.7 = –16,443 kcal Datum: 0°C

Heat in Heat out

SO2 = 1 ´ 11 ´ 400 = 4,400 kcal SO2 = 0.3 ´ 11 ´ 450 = 1,485.00 kcal O2 = 1 ´ 7.5 ´ 400= 3,000 kcal O2 = 0.65 ´ 7.5 ´ 450 = 2,193.75 kcal N2 = 3.76 ´ 7.1 ´ 400 = 10,676 kcal N2 = 3.76 ´ 7.1 ´ 450 = 12,013.20 kcal

SO3 = 0.7 ´ 15.5 ´ 450 = 4,882.50 kcal Total = –18,076 kcal Total = + 20,574.45 kcal DHrxn –16,443 kcal \ Heat in –34,519.00 kcal

Hence, heat absorbed in heat exchanger = –13,944.5 kcal

9.17 From the following data compute the enthalpy change of formation for NH₃ at 480 °C. D H_f at 25°C for NH₃ = –10.96 kcal/kmole

C_p N₂ = 6.76 + (6.06 ´ 10–4T) + (13 ´ 10–8T2)

C_p H₂ = 6.85 + (2.8 ´ 10–5T) + (22 ´ 10–8T2)

C_p NH₃ = 6.703 + (0.0063T) where T is in K, Reaction: N₂ +3H₂ ® 2NH₃

D H_f/kmole 0 0 –10.96

Basis: 1 mole of N₂ (Feed at 273 K) DH_rxn 298 K: (2 ´ –10.96) = –21.92 kcal. Find DC_p = 2NH₃ – (N₂ + 3H₂) Da = (2 ´ 6.703) – [7.76 + (3 ´ 6.85)] = –13.9 Db = (2 ´ 0.0063) – [6.06 ´ 10–4 + (3 ´ 2.8 ´ 10–5)] = 0.0119 Dg = (2 ´ 0) – (13 ´ 10–8 + 3 ´ 22 ´ 10–8) = –7.9 ´ 10–7 D H o = D H rxn – Da T – '¦µ _{T2 –}'¦µ _{T3§¶ §¶} ¨· ¨· 0.0119 t 298 27.9 10 73 D H o = –21.92 – (–13.9 ´ 298) – 2 – t t₂₉₈ 3 DH_o= 3,598.87 kcal 480 °C = 753 K H_{rxn, 480°C} = DH_o + DaT + '¦µ T2 + '¦µ T3D§¶ _{§¶¨· ¨·} 0.0119 = 3,598.87 + (–13.9 ´ 753) + ¦µ_{´ (753)2§¶} ¨· 7.9 107¦µ_{´ (753)}3_{+ t§¶}

¨·

\DH_{rxn 480°C} = –3,606.56 kcal/kmole

9.18 Calculate the calorific value of a blast furnace gas analyzing 25% CO, 12.5% CO₂ and 62.57% N₂.

(a) C + O₂ ® CO₂; DH_rxn: –94 kcal (b) C + ½O₂ ® CO; DH_rxn: –26 kcal

Also, calculate the theoretical flame temperature for the combustion of this gas assuming theoretical amount of air is used, the combustion reaction is complete and reactants enter at 25 °C.

C_p = a + bT + cT2, cal/kmole K

where a, b and c are all dimensional constants and available in literature 3c ´ 105Gas a b ´ 10

CO₂ 10.55 2.16 –2.04 N₂ 6.66 1.02 —

C CO CO

2

CO II Combustion2I CombustionO2 N2Air N2

Basis: 100 g moles of inlet gas

CO entering 25 g moles, CO₂ exit = 25 + 12.5 = 37.5 g moles

¦µ79 _{= 109.52 g moles}

2 needed 12.5 g moles, N2 = t§¶O¨·

CO + ½O₂ ® CO₂

Reaction (a) – (b) gives

DH_rxn = –94 + 26 = –68 kcal/kmole

Calorific value: Heat given out = 68 ´ 25 = 1,700 kcal Exit gases carry this heat away.

This gas temperature is called Theoretical flame temperature which is calculated as follows: –17 ´ 105 cal = 37.5 [{10.55 (T – 298)} + 2.16 ´ 103(T2 – 2982)/2

– 2.04 ´ 10–5(T3 – 2983)/3] + 109.52[6.66 (T – 298) + 1.02 ´ 10–3(T2 – 2982)/2] Solving the above equation we have T = 2721.085K 2721 K = 2448 °C.

9.19 An inventor thinks he has developed a new catalyst which can make the gas phase reaction CO₂

+ 4H₂ ® CH₄ + 2H₂O proceed to 100% conversion. Estimate the heat that must be provided or

removed if the reactants enter and products leave at 500 °C (in effect, we have to calculate the heat of reaction at 500 °C).

DH_f CO₂ CH₄ H₂O kcal/kmole –94,052 –17,889 –57,798 at 298 K \D H_rxn = [–17,889 – (2 ´ 57,798)] – [–94,052]

C_p for CO₂ = 6.339 + 10.14 ´ 10–3T – 3.415 ´ 10–6T2 H₂ = 6.424 + 1.039 ´ 10–3T – 0.078 ´ 10–6T2 H₂O = 6.97 + 3.464 ´ 10–3T – 0.483 ´ 10–6T2 CH₄ = 3.204 + 18.41 ´ 10–3T – 4.48 ´ 10–6T2 Da = [3.204 + (2 ´ 6.97) – 6.339 – (4 ´ 6.424)] = –14.891 Db = [18.41 + (2 ´ 3.464) – 10.14 – (4 ´ 1.039) = 11.047 ´ 10–3 = [–4.48 – (2 ´ 0.483) – {–3.415 – (4 ´ 0.078)}] = –1.719 ´ 10–6Dg DC_p = –14.891 + 11.047 ´ 10–3T – 1.719 ´ 10–6 T2

We find DH_o using data at 298 K

'¦µ _T2 _– '¦µ _T3_DH o = DHrxn – DaT – §¶ §¶ ¨· ¨· = –39,433 – (–14.891 ´ 298) – 11.047 10 3¦µ2 tt§¶2¨· ¦µ 3 – 1.716 10 6 tt§¶ 3_{¨· DH}

o= –39,433 + 4,430 – 491 + 15.16 = –35,479 kcal Next we find DHrxn at 500 °C (or) 773

K: '¦µ _T3_DH 773 = DHo + DaT + '¦µ T2 + §¶§¶ ¨·¨· = –35,479 + (–14.891 ´ 773) + 11.047 10 32 tt 773 2 1.719 10 ₆₃ + t t₇₇₃ 3 = – 43,943 kcal/kmole

\ 43,943 kcal of heat must be removed.

9.20 CO at 50 °F is completely burnt at 2 atm pressures with 50% excess air, which is at 1000 °F.

the combustion chamber in terms of Btu/lb of CO entering.

Basis: 1 lb mole of CO = 28 lb, O₂ needed = 0.5 lb mole CO + ½O₂ ® CO₂ O₂ supplied = 0.5 ´ 1.5 = 0.75 lb mole (50% excess)

Air supplied = 3.57 lb mole and N₂ = 2.82 lb moles DH_rxn = –1,21,745 Btu/lb mole

Q = DH_rxn + DH_products – DH_reactants O₂ remaining = 0.25 lb mole

CO₂: 1 lb mole, N₂: 2.82 lb moles Datum: 32°F

Data: DH (Btu/lb mole)

Temperature , °F CO Air O₂ N₂ CO₂ 50 125.2 — — — — 77 313.3 312.7 315.1 312.2 392.2 800 — — 5,690 5,443 8,026 1000 — 6,984 — — — DHproducts = DH800°F – DH77°F = 1(8,026 – 392.2) + 2.82(5,443 – 312.2) + 0.25(5,690 – 315.1) = 23,446 Btu/lb mole DH reactants: (DH_1000°F – DH_77°F)_air + (DH_50°F – DH_77°F)_CO = 3.57 (6,984 – 312.7) + 1(125.2 – 313.3) = 23,612 Btu/lb mole

Q = –1,21,745 + 23,446 – 23,628 = –1,21,927 Btu/lb mole Heat evolved by combustion =

1,21,927/28

= 4,354.5 Btu/lb of CO 9.21 Pure CO is mixed with 100% excess air and completely burnt at constant pressure. The reactants are originally at 200 °F. Determine the heat added or removed, if the product temperatures are 200 °F, 500 °F, 1000 °F, 1500 °F, 2000 °F and 3000 °F.

Basis: 1 lb mole of CO

CO + ½O₂ ® CO₂

O₂ supplied = 1 lb mole, N₂ = 3.76 lb moles

Exit: CO₂ : 1 lb mole, O₂ : 0.5 lb mole, N₂ : 3.76 lb moles

Assuming a base temperature of 25 oC, (77 oF) and using mean heat capacities, D H = Hp – H_R; Q = DH

DH = SnC_ppr(77 – 200) + DH_{rxn77 °F} + SnC_pR (t – 77)

Reactants: DH_rxn = –1,21,745 Btu/lb mole

Gas n C_p nC_p

CO 1 6.95 6.95 O₂ 1 7.10 7.10 N₂ 3.76 6.95 26.13 Total 40.18 \SnC_ppr(77 – 200) = –(40.18 ´ 123) = – 4,942 Btu DH = – 4,942 –1,21,745 + SnC_pR (t – 77°) 200 °F 500 °F 1000 °F 1500 °F 2000 °F 3000 °F n —— — —— — — — — — — — Cp nCp Cp nCp Cp nCp Cp nCp Cp nCp Cp nCp CO2 1.0 9.15 9.15 9.9 9.9 10.85 10.85 11.5 11.5 12.05 12.05 12.75 12.75 O2 0.5 7.10 3.55 7.25 3.63 7.15 3.88 7.8 3.9 8.0 4.0 8.3 4.15 N2 3.76 6.95 26.13 7.0 26.32 7.15 26.88 7.35 27.64 7.55 28.39 7.88 29.42 SnCpR — — 38.83 — 39.85 — 41.51 — 43.04 — 44.44 — 46.32 SnCp (t–77) 4,776 16,857 38,314 61,246 83,758 1,35,810 Q = DH –21,911 –1,09,830 –88,373 –65,441 –41,229 +9,123

9.22 Coal is burnt to a gas of the following composition: CO₂ : 9.2, CO : 1.5, O₂: 7.3, N₂: 82%. What is the enthalpy difference for this gas between the bottom and the top of the stack if the temperature at the bottom is 550 °F and at the top is 200 °F?

C_p of N₂ = 6.895 + 0.7624 ´ 10–3 T – 0.7 ´ 10–7 T2C_p of O₂ = 7.104 + 0.7851 ´ 10–3 T – 0.5528 ´ 10–7 T2C_p of CO₂ = 8.448 + 5.757 ´ 10–3 T – 21.59 ´ 10–7 T2 + 3 ´ 10–10 T3C_p of CO = 6.865 + 0.8024 ´ 10–3 T – 0.736 ´ 10–7 T2Basis: 1 lb mole of CO₂:

Multiplying these equations by the respective mole fractions of each component and adding them together, we have for N₂ = 0.82 ´ C_pN2 for O₂ = 0.073 ´ C_pO2 for CO₂ = 0.092 ´ C_pCO2 for CO = 0.015 ´ C_pCO C_pnet = 7.049 + 1.2243 ´ 10–3 T – 2.6164 ´ 10–7 T2 + 0.2815 ´ 10–10 T3 200 ∫Cp netdT\DH = 550 = 7.049 (200 – 550) + 1.2243 103 t§¶¦µ₍₂₀₀2 _{– 550}2_{) ¨·}2 – 2.6164 107¦µ₍₂₀₀3 _{– 550}3_{) + 0.2815}1010 t§¶ 3

t§¶ ¦µ 4

(2004 _{– 550}4_{) ¨· ¨·}

or, DH = –2,465 – 160.6 + 13.8 – 0.633 = –2,612 Btu 9.23 Calculate the theoretical flame temperature for CO burnt at constant

pressure with 100% excess air? The reactants enter at 200 °F. CO + ½O₂ ® CO₂

Basis: 1 g mole CO

Temperature of reactants: 200 °F = 93.3 °C

Gases entering: CO–1, O₂–1, N₂–3.76 (all in moles) Gases leaving: CO₂–1, O₂–0.5, N₂–3.76 (all in moles) \DH_rxn 25 °C = – 67,636 cal.

Gas mole DTC_pm DH = nC_pmDT CO 1.0 (93.3 – 25) 6.981 476 Air 4.76 (93.3 – 25) 6.993 2,270

Total 2,746 cal

Next we have DH_product = – (DH_reactants – DH_rxn)

= – (2,746 + 67,636) = –70,382 cal Let us assume exit temperature as 1800 °C, then DT = (1800 – 25) = 1775 °C Gas mole DTC_pm DH

CO₂ 1.0 1775 12.94 23,000 O₂ 0.5 1775 8.35 7,400 N₂ 3.76 1775 7.92 52,900

This total of –83,300 cal is not matching with –70,382 cal, the value calculated.

Let the Theoretical flame temperature be 1500 °C, then DT = 1475 °C DH = (1 ´ 12.7 ´ 1475) + (0.5 ´ 8.31 ´ 1475) + (3.76 ´ 7.88 ´ 1475)

= 68,460 cal

Making linear interpolation for the theoretical flame temperature, we have, Theoretical flame temperature

= 1500 + 70,382 68,460 ËÛ_{´ (1800 – 1500)} 83,300ÌÜ ÍÝ = 1500 + 39 = 1539 °C º 2798 °F

9.24 Calculate the theoretical flame temperature of a gas having 20% CO and 80% N₂ when burnt with 150% excess air. Both air and gas being at 25 °C.

Data: Heat of formation of CO₂ = – 94,052 cal/g mole, CO = –26,412 cal/g mole at 25 °C.

C_pm: CO₂: 12.1, O₂: 7.9, N₂: 7.55 cal/g mole K (from literature) Basis: 1 g mole CO, CO + 0.5O₂ ® CO₂

O₂ supplied = (0.5 ´ 2.5) = 1.25 g moles (150% excess) 80 N 2 in feed = 1 ÈØ ÉÙ = 4 g molesÊÚ N 2 in air = ÈØ79 ​ÉÙ = 4.7 g moles_ÊÚ21

Exit gas: CO₂: 1 g mole, O₂: 0.75 g mole, N₂: 8.7 g moles Q = SH_products + SH_rxn – SH_reactants. (Datum 298 K)

SH_reactants is zero, since air and gas are at 25 °C.

DH_rxn = DH_CO2 – DH_CO

= –94,052 – (–26,412) = –67,640 cal/g mole. Let the “Theoretical Flame Temperature” be T, K

67,640 = [1 ´ 12.1 ´ (T – 298)] + [8.7 ´ 7.55 ´ (T – 298)] + [0.75 ´ 7.9 ´ (T – 298)], 67,640 = 83.71 T – 24,945.6

\ T = 1106.03 K ∫ 833.03 °C 1106 K ∫ 833 °C.

9.25 Find the theoretical flame temperature of a gas containing 30% CO and 70% N₂ when burnt with 100% excess air. The reactants enter at 298 K.

DH_f CO₂ = –3,93,700 kJ/kmole; DH_f CO = –1,10,600 kJ/kmole Mean molar specific heat, kJ/kmole K at different temperatures is given below:

Temperature K CO₂ O₂ N₂ 800 45.4 31.6 30.3 1000 47.6 32.3 30.6 1200 49.4 33.0 31.2 1400 50.8 33.6 31.8 1600 52.0 34.0 32.3 1800 53.2 34.4 32.7

N₂ in feed = 70/30 = 2.34 kmoles O₂ supplied = 0.5 ¥ 2 = 1 kmole N₂ from air = 3.76 kmoles

Exit gas consists of:

CO₂: 1, O₂: 0.5, N₂: (3.76 + 2.34) = 6.1 kmoles Let us consider the equation

Q = SHproducts + DHrxn –SHreactants

where SH_reactants = Zero at 298 K (_Q Datum is 298 K)

\DH_rxn = (–3,93,700) – (–1,10,600) = –2,83,100 kJ/mole By iteration method: Let the theoretical flame temperature be 1400 K:

DT = (1400 – 298) = 1102 K

DH_pr = (1 ¥ 50.8 ¥ 1102) + (0.5 ¥ 33.6 ¥ 1102) + (6.1 ¥ 31.8 ¥ 1102) = 2,88,261 kJ/kmole π 2,83,100 kJ/kmole

Let the theoretical flame temperature be 1200 K \DT (1200 – 298) = 902 K

DH_pr = (1 ¥ 49.4 ¥ 902) + (0.5 ¥ 33 ¥ 902) + (6.1 ¥ 31.2 ¥ 902) = 2,31,110 kJ/kmole π 2,83,100 kJ/kmole

So theoretical flame temperature lies in between these two values (by interpolation ) \ Theoretical flame temperature

2,83,100 2,31,110 = 1,200 + ÌÜ

2,88,261 2,31,110 ´ (1400 – 1200)ÍÝ

So, the temperature of the exit gases is 1382 K = 1109 °C.

9.26 The analysis of 15,000 lit of a gas mixture at standard condition is as follows: SO₂: 10%, O₂: 12% and N₂: 78%. How much heat must be added to this gas to change its temperature from 30 °C to 425 °C?

The C_pm values are in cal/g mole °C

Gas SO₂ O₂ N₂C_{pm 30 °C} 10 6.96 6.80 C_{pm 425 °C} 11 7.32 7.12 The amount of gas mixture = 15,000 litres

º

15,000 22.414

= 669.2 g moles

we can then write the amount of each component SO₂ : 669.2 ´ 0.1 = 66.92 g moles O₂ : 669.2 ´ 0.12 = 80.30 g moles N₂ : 669.2 ´ 0.78 = 521.98 g moles 669.20 g moles Reference temperature: 0 °C \ Q = 66.92 [(11 ´ 425) – (10 ´ 30)] + 80.3 [(7.32 ´ 425) – (6.96 ´ 30)]

+ 521.98 [(7.12 ´ 425) – (6.8 ´ 30)] = 19,98,849.22 cal 9.27 Estimate the theoretical flame

temperature of a gas containing 20% CO and 80% N₂ when burnt with 100% excess air. Both air and gas are initially at 25 °C.

C_p CO₂ = 6.339 + 10.14 ´ 10–3 T – 3.415 ´ 10–6 T2 C_p O₂ = 6.117 + 3.167 ´ 10–3 T – 1.005 ´ 10–6 T2 C_p N₂ = 6.457 + 1.389 ´ 10–3 T – 0.069 ´ 10–6 T2

The values of C_p are in kcal/kmole K and temperature is in K DH_rxn 25 °C = –67,636 kcal

Basis: 1 kmole of CO; N₂ = 4 kmoles

Air supplied: O₂: 1 kmole, N_{2 from air}: 3.76 kmoles Exit gas: CO₂: 1, O₂: 0.5, N₂: 7.76 kmoles

Datum 25 °C = 298 K

DH_rxn = –DH products + 67,636 = [1 ´ òC_pCO2 ´ (T – 298)] + [0.5 ´ òC_pO2 ´ (T – 298)] + [7.76 ´ òC_pN2 ´ (T – 298)] 22 67,636 = 6.339( T – 298) + 10.14 ´ 10 –3 T 298 2 33 T 298_{– 3.415 ´ 10}–6 T 298 3+ 6.117 ´2 22T33 + 3.167 ´ 10 –3 T 298_{– 1.005 ´ 10}–6 298 _{4 6 T} 22 + 7.76 ´ 6.457

´ ( T – 298) + 7.76 ´ 1.389 ´ 10 –3 298 2 T 33 – 7.76 ´ 0.069 ´ 10 –6 298 3

Solving for theoretical flame temperature = T = 1216 K = 943 °C

9.28 Dry methane and dry air at 298 K and 1 bar pressure are burnt with 100% excess air. The

standard heat of reaction is –802 kJ/g mole of methane. Determine the final temperature attained by the gaseous products if combustion is adiabatic and 20% of heat produced is lost to the surroundings. Data: C_pm values (J/g mole K) for the components are: O₂: 31.9, N₂: 32.15, H₂O : 40.19, CO₂ :

51.79.

Basis: 1 g mole of methane.

Datum: 298 K

CH₄ + 2O₂ ® CO₂ + 2H₂O

\ Oxygen supplied = 2 ´ 2 = 4 g moles N₂ entering = 4

79 _{= 15.05 g moles´} 21

Gases leaving are: CO₂: 1, H₂O : 2, O₂: 2, and N₂: 15.05 g moles. Heat given out = 802 kJ Heat loss = (802 ´ 0.2) = 160.4 kJ

\ Q = Heat in exit gases = (802 – 160.4) = 641.6 kJ

Q = [1 ´ 51.79 ´ (T – 298)] + [2 ´ 40.19 ´ (T – 298)] + [2 ´ 31.9 ´ (T – 298)] + [15.05 ´ 32.15 ´ (T –

298)] = [679.8275 ´ (T – 298)] = 641.6 ´ 103 J. \ T = 1242 K = 969 °C.

9.29 An iron pyrite ore contains 85% FeS₂ and 15% gangue. It is roasted with 200% excess air to get SO₂. The reaction is given. All the gangue plus Fe₂O₃ end up in the solid waste produced which analyzes 4% FeS₂. Determine the standard heat of reaction in kJ/kg of ore roasted and the analysis of

the solid waste. Heat of formation data is in kJ/g mole.

4FeS₂ + 11O₂ ® 2Fe₂O₃+8SO₂ Weights 479.4 352 319.4 512 Heats of formation –177.9 0 –88.2 –296.9 (kJ/g mole)

Basis: 1 kg of ore containing FeS₂ = 0.85 kg and gangue = 0.15 kg Let x kg of FeS₂ be in the solid waste, 319.4 ËÛ_{then, FeS} 2 reacted: (0.85 – x) kg; Fe2O3 formed (0.85 x)​ÌÜ ÍÝ = 0.666(0.85 – x) x = 0.04{(0.15) + 0.666(0.85 – x) + x};

Solving the above, we find x = 0.029 kg

Solid waste = (0.15) + (0.029) + [0.666 ´ (0.85 – 0.029)] = 0.725786 kg A summary of the composition of the solid waste is given:

Solid waste kg Weight %

Gangue 0.150 20.66 FeS₂ 0.029 3.99 Fe₂O₃ 0.547 75.35 Total 0.726 100.00

FeS₂ reacted = (0.85 – 0.029) = 0.821 kg = 6.85 ´ 10–3 kmole\ Fe₂O₃ formed = 0.547 kg = 3.424 ´ 10–3 kmole SO 2 formed = ​ÌÜ ËÛ512_{= 1.068 kg = 0.016688 kmole ÍÝ}479.4 Heat of reaction = –[(0.016688 ´ 296.9) + (3.424 ´ 10–3 _{´ 822.3)} –(6.85 ´ 10–3 ´ 177.9)] = –6.55 kJ

9.30 For the following reaction, estimate the heat of reaction at 298 K. A + B ® C + D

Compound DH°_f(kcal/g mole) A –269.8

B –195.2 C –337.3 D –29.05

= [–337.3 – 29.05] – [–269.8 – 195.2] = 98.65 kcal

9.31 Estimate the standard heat of reaction DH°₂₉₈ for the reaction. A + B ® C Standard heats of combustion are:

DH_{c, 298} for A = –328000 cal/g DH_{c, 298} for B = –212000 cal/g DH_{c, 298} for C = –542000 cal/g DH_{c, 298} = SDH°_c, reactants – SDH°_c, products = [–328000 – 212000] – [–542000] = 2000 cal

9.32 Calculate the heat of formation of CHCl₃ from the following data: CHCl₃ + 1 _O 2 + H2O ® CO2 + 3HCl, DH = –509.93 kJ (1)2 H₂ + 1 O₂ ® H₂O; DH = –296 kJ (2)₂ C + O₂ ® CO₂; DH = –393.78 kJ (3) 1 H₂ + 1 Cl₂ ® HCl; DH = –167.5 kJ (4)_{2 2} CO₂ + 3HCl 1 O₂ + H₂O; DH = –509.93 kJ® CHCl₃ + ₂ H₂ + 1 O₂ ® H₂O; DH = –296 kJ₂ C + O₂CO₂; DH = –393.78 kJ Eq. (4) × 3 + Eq. (3)/2, gives

3 _H

2 + Cl2 + 1 C + 1 O2 ® 3HCl + 1 CO22 2 2 2

9.33 Standard heat of reaction accompanying any chemical change is equal to the algebraic sum of the

standard heat of formation of the products minus the algebraic sum. Calculate the standard heat of reaction, DH°_{f, 298} for the reaction

2FeS₂ + 1.5O₂ ® Fe₂O₃ + 4SO₂ The standard heats of formation are: FeS₂ = –42,520 cal/g mole

Fe₂O₃ = –1,96,500 cal/g mole SO₂ = –70,960 cal/g mole From the reaction,

DH°_{r, 298} = DH_Fe2O3 + 4 [DH_SO2] – 2[DH_FeS2]

= –19,6500 + 4[–70,960] – 2[42,520] = –39,5300 cal

9.34 The heat of reaction at 300 K and 1 atm pressure for the reaction A + 3B ® C is 30,000 cal/mole

In document Process Calculations (Page 183-200)