reaksi kimia (brief)

8/10/2019 Reaksi Kimia (Brief)

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6. Bab 15 : Reaksi Kimia

15-1 Bahan Bakar & Pembakaran

15-2 Proses Pembakaran Teoritik & Aktual

15-3 Entalpi Pembentukan & Entalpi Pembakaran


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Bahan Bakar & Pembakaran

FIGURE 15 1Kebanyakan bahan bakar hidrokarbon cair dihasilkan daridestilasi minyak bumi


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Pembakaran adalah reaksi kimia dimana bahan bakardioksidasi & sebagian besar energi dihasilkan .

Pembakaran (Combustion)

FIGURE 15 3Tiap kmol O 2 di udara terdapat3,76 kmol N 2.


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FIGURE 15 4Pada proses pembakaran tunak, komponen-komponen yang masuk ruang

bakar disebut sebagai REAKTAN & komponen yg keluar disebut

sebagai PRODUK .


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Rasio udara bahan bakar (air fuel ratio, AF) biasanya dinyatakan berdasarkan basis massa & didefinisikan

sebagai :

FIGURE 15 6Rasio udara - bahan bakar (air fuel ratio, AF) menunjukkan jumlahmassa udara yg digunakan per satuan massa bahan bakar selama

proses pembakaran.


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EXAM PL E 15 1Balancing the Combustion Equation

1 kmol oktana (C 8H 18) dibakar dengan udara yg mengandung 20kmol O 2. Asumsikan produk hanya mengandung CO 2, H 2O, O 2,dan N2, tentukan :

a. Jumlah mol setiap gas dalam produk dan b. Rasio AF pada proses pembakaran ini.


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Solution : Jumlah bahan bakar & oksigen di udara diberikan.Jumlah produk dan rasio AF ditanyakan.

Asumsi : Produk pembakaran hanya mengandung CO 2, H 2O, O 2, dan N 2 .

Proper ties : Massa molar udara M air = 28.97 kg/kmol = 29.0 kg/kmol(Table A 1).

Analysis :

Persamaan kimia untuk proses pembakaran ini dapat ditulis sbb :

Massa Molar dari Udara dapat ditentukan dari :


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Jumlah massa atau jumlah mol setiap komponen dari reaktan &produk harus sama , sehingga :

Dengan mensubstitusikan koefisien di atas, persamaan pembakaranmenjadi :


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Rasio Udara Bahan Bakar (AF) :

Kesimpulan : 24,2 kg udara digunakan untuk membakar 1 kgbahan bakar selama proses pembakaran ini.


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Pembakaran Teoritik & Aktual

Proses pembakaran dikatakan lengkap / sempurna jika semua komponen yg dapat terbakar dari bahanbakar terbakar sempurna (tak ada O 2 pada produk)

PembakaranAktual

Pembakaran

Teoritik


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Proses pembakaran ideal dimana bahan bakar terbakar secarasempurna dengan udara teoritik disebut pembakaran teoritik /stoikiometri.

Sebagai contoh, pembakaran teoritik metana sbb :

FIGURE 15 9Proses pembakaran lengkap tanpa

oksigen berlebih pada produkpembakaran disebut :pembakaran teoritik .


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EXAM PL E 15 2Pembakaran Dengan Udara Berlebih

Etana (C 2H 6) dibakar dengan 20% udara berlebih selama proses pembakaran. Asumsikan pembakaran sempurna / lengkap yang berlangsung pada tekanan 100 kPa. Tentukan :

a. Rasio AF


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Persamaan Reaksi Pembakaran :

Koefisien a th ditentukan dari kesetimbangan O 2 antara reaktan & produk sbb :


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Sehingga Persamaan Reaksi Pembakaran menjadi :

(a ) Rasio AF dapat dihitung sbb :


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Entalpi Pembentukan (Enthalpy of Formation) danEntalpi Pembakaran (Enthalpy of Combustion)

FIGURE 15 14

Bentuk energi mikroskopik darisuatu zat terdiri dari :

1. Energi Sensibel2. Energi Laten

3. Energi Kimia4. Energi Nuklir


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Sifat ini disebut Entalpi Reaksi (Enthalpy of Reaction, h R ) yangdidefinisikan sebagai perbedaan antara entalpi Produk pada suatukondisi tertentu dengan entalpi Reaktan untuk suatu rekasi sempurna .

FIGURE 15

15

Saat ikatan kimia yg ada rusak& terbentuk ikatan baru yg lainselama proses pembakaran

sejumlah besar energisensibel diserap ataudilepaskan.


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Untuk Proses Pembakaran Entalpi Reaksi biasanya dikenal sebagai

Entalpi Pembakaran (Enthalpy of Combustion, h C ) yang menunjukkan

sejumlah kalor yang dilepaskan selama proses pembakaran tunaksaat 1 kmol (atau 1 kg) bahan bakar secara sempurna terbakar padasuhu & tekanan tertentu.

Entalpi Reaksi (h R ) atau Entalpi Pembakaran (h C) dapat dituliskan sbb :


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FIGURE 15 17

Entalpi Pembakaran menunjukkansejumlah energi yang dilepaskan saatbahan bakar terbakar selama proses

pembakaran tunak pada kondisi tertentu.


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Sifat ini disebut juga sebagai Entalpi Pembentukan (Enthalpy ofFormation, h f ) saat dipandang sebagai entalpi dari suatu zat padakondisi tertentu karena komposisi kimia pembentuknya .

FIGURE 15 18Entalpi Pembentukan dari suatu senyawaatau molekul menunjukkan energiyang diserap atau dilepaskan saatkomponen terbentuk dari elemen-elemennya selama proses pembakaran

tunak pada kondisi tertentu.( Tabel A-26 )


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Nilai Kalor (Heating value) suatu bahan bakar didefinisikan sebagai jumlah kalor yang dilepaskan saat bahan bakar terbakar sempurna .

Dengan kata lain, Nilai Kalor bahan bakar sama dengan HargaMutlak Entalpi Pembakaran dari bahan bakar , & dapat dituliskansbb :

Nilai Kalor tergantung pada fasa H 2O pada Produk .

Nilai Kalor disebut HHV (Higher Heating Value) jika H 2O padaproduk berada pada fasa cair.

Nilai Kalor disebut LHV (Lower Heating Value) jika H 2O padaproduk berada pada fasa uap.


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FIGURE 15 19

HHV suatu bahan bakar

sama dengan jumlah dari LHVdan Kalor Laten PenguapanH 2O pada produk.

Hubungan antara HHV dan LHV sbb :


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EXAM PL E 15 5Evaluation of the Enthalpy of Combustion

Tentukan Entalpi Pembakaran dari Oktana cair (C 8H 18) pada kondisi25C dan 1 atm, menggunakan data Entalpi Pembentukan (Tabel A-26) .Asumsikan H 2O pada produk berfasa cair .


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Tabel A-26 Entalpi Pembentukan pada 25C dan 1 atm sbb :- CO 2 (g) = - 393,520 kJ/kmol- H 2O (l) = - 285,830 kJ/kmol- C 8H18 (l) = - 249,950 kJ/kmol

Reaktan & Produk pada kondisi referensi standar (25C dan 1 atm).

N2 dan O 2 adalah elemen yang stabil sehingga entalpi pembentukannya sama dengan NOL .

Persamaan stoikiometri pembakaran C 8H 18 sbb :


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Sehingga :

Sehingga Entalpi Pembakaran C 8H 18 menjadi :


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SELESAI


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FIRST-LAW ANALYSISOF REACTING SYSTEMSC8H18()


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Steady-Flow Systems

the enthalpy of a component on a unit mole basis as (Fig. 15 21)

FIGURE 15 21The enthalpy of a chemical componentat a specified state is the sum of the

enthalpy of the component at 25C, 1atm ( hf ), and the sensible enthalpy ofthe component relative to 25C, 1 atm.


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The steady-flow energy balance relation can be expressed for a chemically reacting steady-flow system more explicitly as

Rate of net energy transfer in by heat,work, and mass

Rate of net energy transfer out by heat,work, and mass

In combustion analysis, it is more convenient to work with quantities expressed permole of fuel . Such a relation is obtained by dividing each term of the equation above

by the molal flow rate of the fuel, yielding

Energy transfer in per mole of fuel by

heat, work, and mass

Energy transfer out per mole of fuel by

heat, work, and mass


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Taking heat transfer to the system and work done by the system to be positivequantities, the energy balance relation just discussed can be expressed more compactlyas

or as

where

If the enthalpy of combustion h C for a particular reaction is available, the

steady-flow energy equation per mole of fuel can be expressed as


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A combustion chamber normally involves heat output but no heat input. Then theenergy balance for a typical steady-flow combustion process becomes

Energy in by mass permole of fuel

Energy out by mass permole of fuel


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Closed Systems

The general closed-system energy balance relation can be expressed for a stationarychemically reacting closed system as

To avoid using another property the internal energy of formation u f we utilizethe definition of enthalpy ( u h Pv or u f u u h Pv ) and express theabove equation as (Fig. 15 22)


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FIGURE 15 22An expression for the internal energyof a chemical component in terms ofthe enthalpy.


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EXAM PL E 15 6First-Law Analysis of Steady-Flow Combustion

Liquid propane (C3H8) enters a combustion chamber at 25C at a rate of0.05 kg/min where it is mixed and burned with 50 percent excess air thatenters the combustion chamber at 7C, as shown in Fig. 15 23. An analysisof the combustion gases reveals that all the hydrogen in the fuel burnsto H2O but only 90 percent of the carbon burns to CO2, with the remaining10 percent forming CO. If the exit temperature of the combustion gases is 1500K, determine ( a ) the mass flow rate of air and ( b) the rate of heattransfer from the combustion chamber.


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Solution Liquid propane is burned steadily with excess air. The mass flowrate of air and the rate of heat transfer are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air and the combustiongases are ideal gases. 3 Kinetic and potential energies are negligible.

Analysis We note that all the hydrogen in the fuel burns to H2O but 10 percent of the carbon burns incompletely and forms CO. Also, the fuel is burned with excess air and thus there is some free O2 in the product gases.

The theoretical amount of air is determined from the stoichiometric reaction to be

O2 balance:


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Then the balanced equation for the actual combustion process with 50 percentexcess air and some CO in the products becomes

(a) The air fuel ratio for this combustion process is

Thus,


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(b) The heat transfer for this steady-flow combustion process is determined fromthe steady-flow energy balance E out E in applied on the combustion chamber perunit mole of the fuel,

or

Assuming the air and the combustion products to be ideal gases, we have h h (T), and we form the following minitable using data from the property tables:


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The h f of liquid propane is obtained by subtracting the h fg of propane at 25Cfrom the h f of gas propane. Substituting gives

Thus 363,880 kJ of heat is transferred from the combustion chamber for each kmol(44 kg) of propane. This corresponds to 363,880/44 8270 kJ of heat loss perkilogram of propane. Then the rate of heat transfer for a mass flow rate of 0.05kg/min for the propane becomes


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EXAM PL E 15 7First-Law Analysis of Combustion in a Bomb

The constant-volume tank shown in Fig. 15 24 contains 1 lbmol of methane (CH4)gas and 3 lbmol of O2 at 77F and 1 atm. The contents of the tank are ignited, andthe methane gas burns completely. If the final temperature is 1800 R, determine ( a )the final pressure in the tank and ( b) the heat transfer during this process.


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Solution Methane is burned in a rigid tank. The final pressure in the tank and theheat transfer are to be determined.

Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO2 and all the hydrogen to H2O. 2 The fuel, the air, and the combustiongases are ideal gases. 3 Kinetic and potential energies are negligible. 4 There are nowork interactions involved.

Analysis

The balanced combustion equation is


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(a ) At 1800 R, water exists in the gas phase. Using the ideal-gas relation for both thereactants and the products, the final pressure in the tank is determined to be

Substituting, we get


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(b) Noting that the process involves no work interactions, the heat transfer duringthis constant-volume combustion process can be determined from the energy

balance E in E out E system applied to the tank,

Since both the reactants and the products are assumed to be ideal gases, all theinternal energy and enthalpies depend on temperature only, and the Pv terms in

this equation can be replaced by RuT . It yields

since the reactants are at the standard reference temperature of 537 R.


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From h f and ideal-gas tables in the Appendix,

Substituting, we have


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Discussion On a mass basis, the heat transfer from the tank would be 308,730/1619,300 Btu/lbm of methane.


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ADIABATIC FLAME TEMPERATURE

In the absence of any work interactions and any changes in kinetic or potentialenergies, the chemical energy released during a combustion process

either is lost as heat to the surroundings or is used internally to raise thetemperature of the combustion products. The smaller the heat loss, thelarger the temperature rise. In the limiting case of no heat loss to the surroundings(Q 0), the temperature of the products reaches a maximum,which is called the adiabatic flame or adiabatic combustion temperatureof the reaction (Fig. 15 25).

FIGURE 15 25The temperature of a combustionchamber becomes maximum whencombustion is complete and no heatis lost to the surroundings ( Q 0).


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The adiabatic flame temperature of a steady-flow combustion process is determinedfrom Eq. 15 11 by setting Q 0 and W 0. It yields

FIGURE 15 26The maximum temperature

encountered in a combustion chamberis lower than the theoretical adiabaticflame temperature.


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EXAM PL E 15 8Adiabatic Flame Temperature in Steady Combustion

Liquid octane (C8H18) enters the combustion chamber of a gas turbine steadily at1 atm and 25C, and it is burned with air that enters the combustion chamber atthe same state, as shown in Fig. 15 27. Determine the adiabatic flame temperaturefor ( a) complete combustion with 100 percent theoretical air, ( b) completecombustion with 400 percent theoretical air, and ( c) incomplete combustion (someCO in the products) with 90 percent theoretical air.


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Solution Liquid octane is burned steadily. The adiabatic flame temperature is to bedetermined for different cases.

Assumptions 1 This is a steady-flow combustion process. 2 The combustionchamber is adiabatic. 3 There are no work interactions. 4 Air and the combustiongases are ideal gases. 5 Changes in kinetic and potential energiesare negligible.

Analysis

(a ) The balanced equation for the combustion process with the theoretical amount ofair is


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The adiabatic flame temperature relation H prod H react in this case reduces to

since all the reactants are at the standard reference state and h f 0 for O2 and N2.The h f and h values of various components at 298 K are


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Substituting, we have

which yields

It appears that we have one equation with three unknowns. Actually we haveonly one unknown the temperature of the products T prod since h h (T )for ideal gases. Therefore, we have to use an equation solver such as EES ora trial-and-error approach to determine the temperature of the products.


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A first guess is obtained by dividing the right-hand side of the equation by the totalnumber of moles, which yields 5,646,081/(8 + 9 + 47) 88,220 kJ/kmol. Thisenthalpy value corresponds to about 2650 K for N2, 2100 K for H2O, and 1800 K

for CO2. Noting that the majority of the moles are N2, we see that T prod should beclose to 2650 K, but somewhat under it. Therefore, a good first guess is 2400 K. Atthis temperature,

This value is higher than 5,646,081 kJ. Therefore, the actual temperature isslightly under 2400 K. Next we choose 2350 K. It yields


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which is lower than 5,646,081 kJ. Therefore, the actual temperature of the productsis between 2350 and 2400 K. By interpolation, it is found to be

(b) The balanced equation for the complete combustion process with 400 percent theoretical air is

By following the procedure used in ( a), the adiabatic flame temperature in this caseis determined to be

Notice that the temperature of the products decreases significantly as a result ofusing excess air.


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(c) The balanced equation for the incomplete combustion process with 90 percenttheoretical air is

Following the procedure used in ( a), we find the adiabatic flame temperature inthis case to be

Discussion Notice that the adiabatic flame temperature decreases as aresult of incomplete combustion or using excess air. Also, the maximum

adiabatic flame temperature is achieved when complete combustion occurswith the theoretical amount of air .

reaksi kimia (brief)

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