Neraca energi

Basic Principles

Mass and energy balance

Types of Energy to write an energy balance, we need to know what kinds of energy can enter or leave a system 1. A system could gain or lose kinetic energy, if we're analyzing a moving system. 2. Again, if the system is moving, there could be potential energy changes. 3. Heat could enter the system via conduction, convection, or radiation. 4.Work (either expansion work or shaft work) could be done on, or by, the system. 5.energi listrik

expressions for the different types of energy: • From physics, recall that . If the system itself is not moving,

this is zero.

• The gravitational potential energy of a system is GPE = mgh where g is the gravitational constant and h is the height of the center of mass of the system. If the system does not change height, there is no change in GPE.

• The heat entering the system is denoted by Q, regardless of the mechanism by which it enters (the means of calculating this will be discussed in a course on transport phenomenon). According to this book's conventions, heat entering a system is positive and heat leaving a system is negative, because the system in effect gains energy when heat enters.

• The work done by or on the system is denoted by W. Work done BY a system is negative because the system has to "give up" energy to do work on its surroundings. For example, if a system expands, it loses energy to account for that expansion. Conversely, work done ON a system is positve.

Law of Conservation of Energy Ein − Eout = Eaccumulated Accumulation of anything is 0 at steady state, and energy is no exception Total energy carried in the flow itself is:

When a mass stream flows into a system it expands or contracts and therefore performs work on the system. An expression for work due to this expansion is: Since this work is done on the system, it enters the energy balance as a positive quantity

Definition of enthalpy H = U + PVIn stream i, if only KE, GPE, internal energy, and expansion work are considered, the energy carried by mass flow is:

Overall steady - state energy balance Steady State Energy Balance on an Open System

Some important points:1. If the system is closed AND at steady state that means the total heat flow must

equal the total work done in magnitude, and be opposite in sign. However, according to another law of thermodynamics, the second law, it is impossible to change ALL of the heat flow into work, even in the most ideal case.

2. In an adiabatic system with no work done, the total amount of energy carried by mass flows is equal between those flowing in and those flowing out. However, that DOES NOT imply that the temperature remains the same, as we will see in a later section. Some substances have a greater capacity to hold heat than others, hence the term heat capacity.

3. If the conditions inside the system change over time, then we CANNOT use this form of the energy balance. The next section has information on what to do in the case that the energetics of the system change.

NP total pada sistem alir ( flow system) pada keadaan steady state :

Persamaan di atas sering dipakai untuk kasus transportasi fluida, Yaitu persamaan Bernoulli.

Neraca Energi untuk proses kimia ( non flow system ).

Sistem non alir dianggap terjadi di dalam alat-alat proses, misal alat penukar panas (HE =heat exchanger), reaktor, dan alat-alat transfer massa lainnya.Pada sistem ini, biasanya EP dan EK <<< Q dan W, sehingga EP dan EK dapat diabaikan dan NP menjadi :

Untuk beberapa proses, biasanya nilai W sangat kecil. Sehingga : H2 – H1 = Q = ΔHDengan, H1 = entalpi arus masuk (titik satu), H2 = entalpi arus keluar (titik dua).

Neraca Panas di Reaktor

jika di reaktor reaksi tidak dijalankan pada kondisi standar, maka dipikirkan:

1. suhu umpan (TF) diturunkan atau dinaikkan sampai suhu standar,kemudian2. direaksikan pada kondisi standar (TR), lalu3. suhu produk dinaikkan suhunya sampai suhu keluar reaktor (TP).

Tampak bahwa untuk menyelesaikan neraca panas pada proses kimia perludiselesaikan terlebih dahulu neraca massanya.

NERACA PANAS DENGAN REAKSI KIMIA1. Suatu konverter digunakan untuk mengoksidasi SO2 menjadi SO3. Oksigen disuplai dari udara. Dianggap berisi 21% mol O2 dan 79% mol N2.

Jika digunakan udara berlebihan 25% dan diinginkan SO3 yang terbentuk adalah 180 mol/jam. Tentukan kebutuhan SO2 dan udara umpan jika konversi hanya 80%. Jika suhu SO2 umpan adalah 40oC, suhu udara umpan 30oC dan suhu gas keluar konverter adalah 60oC, berapa panas

yang dihasilkan konverter itu.

2. Gas metan dibakar dengan oksigen. Seratus lima puluh kgmol/jam umpan terdiri atas 20% metan, 60% O2 dan 20% CO2 diumpankan ke reaktor. Konversi limiting reactant = 90%. Jika suhu gas umpan 50oC dan suhu gas keluar dari ruang pembakaran 190oC, tentukan panas yang dibutuhkan/

dihasilkan dari ruang pembakaran itu.

3. Gas metan dibakar dengan oksigen. Seratus lima puluh kgmol/jam umpan terdiri atas 20% metan, 60% O2 dan 20% CO2 diumpankan ke furnace. Hasil analisis gas hasil furnace menunjukkan gas hasil berisi gas metan 1,5 kgmol/jam. Jika suhu gas umpan 27oC dan suhu gas keluar dari ruang pembakaran 327oC, tentukan panas yang dibutuhkan/dihasilkan dari ruang pembakaran itu.

4. Reaktor digunakan untuk mengoksidasi SO2 menjadi SO3. Umpan terdiri atas 12% SO2, 8% O2, dan 80% N2 dengan suhu umpan 427oC. Jika konversi SO2 adalah 50%, dan gas hasil keluar reaktor pada suhu 527oC serta 100 mol/jam gas diumpankan. Tentukan :

a. komposisi gas hasil reaktor. b. Panas reaksi yang dihasilkan/dibutuhkan reaktor itu.

5. Reaksi amonia dijalankan pada reaktor fase gas, reaksi : 4NH3 + 5O2 -------- 4NO + 6H2O Oksigen disuplai dari udara yang diumpankan ke reaktor dengan 25% berlebihan. Jika

diumpankan 100 gmol/jam NH3 dengan suhu 30oC dan udara pada suhu 40oC. Gas hasil keluar reaktor pada suhu 50oC.

a . Berapa udara umpan? b . Jika konversi hanya 80%, tentukan komposisi gas hasil! c .Tentukan panas reaksi reaktor itu! d . Eksotermis atau endotermiskah reaktor itu?

6. Suatu ketel digunakan untuk membuat uap air. Panas yang digunakan adalah panas pembakaran gas metan. Gas metan bertekanan 1 atm, suhu 473oC dan berkecepatan 38.786 L/j diumpankan ke furnace, sedangkan suhu udara yang diumpankan 373oC. Agar terjadi pembakaran sempurna, udara yang diumpankan 50% berlebihan. Gas hasil pembakaran keluar furnace pada suhu 1473oK. Tentukan :

• a. Kecepatan arus udara umpan.• b. Kecepatan dan komposisi gas hasil furnace.• c. Panas yang dihasilkan.

Latent heat • the amount of energy released or

absorbed by a chemical substance during a change of state that occurs without changing its temperature, meaning a phase transition such as the melting of ice or the boiling of water

• Two of the more common forms of latent heat (or enthalpies or energies) encountered are latent heat of fusion (melting) and latent heat of vaporization (boiling).

Specific latent heat • specific latent heat of fusion the amount

of energy required to convert 1 kg (or 1 lb) of a substance from solid to liquid (or vice-versa) without a change in the temperature of the surroundings

• specific latent heat of vaporization the amount of energy required to convert 1 kg (or 1 lb) of a substance from liquid to gas (or vice-versa) without a change in the external temperature

Enthalpy of Moist and Humid Air The enthalpy of moist and humid air consist of sensible heat and latent heat - enthalpy is used to calculate cooling and heating processes Moist air is a mixture of dry air and water vapor. In atmospheric air, water vapor content varies from 0 to 3% by mass. The enthalpy of moist and humid air includes the :•enthalpy of the dry air - the sensible heat - and •the enthalpy of the evaporated water - the latent heat

Specific enthalpy - h - (J/kg, Btu/lb) of moist air is defined as the total enthalpy (J, Btu) of the dry air and the water vapor mixture - per unit mass (kg, lb) of moist air

Specific Enthalpy of Moist Air

• H = ha + x hw         (1)where• H = specific enthalpy of moist air (kJ/kg, Btu/lb)• ha = specific enthalpy of dry air (kJ/kg, Btu/lb)• x = humidity ratio (kg/kg, lb/lb)• hw = specific enthalpy of water vapor (kJ/kg,

Btu/lb)

Specific Enthalpy of Dry Air - Sensible Heat Assuming constant pressure conditions the

specific enthalpy of dry air can be expressed as:• ha = cpa t         (2)where• cpa = specific heat capacity of air at constant

pressure (kJ/kgoC, kWs/kgK, Btu/lboF)• t = air temperature (oC, oF)• For air temperature between -100oC (-150oF)

and 100oC (212oF) the specific heat capacity can be set to

• cpa = 1.006 (kJ/kgoC)    = 0.240 (Btu/lboF)

Specific Enthalpy of Water Vapor - Latent Heat Assuming constant pressure conditions the

specific enthalpy of water vapor can be expressed as:

• hw = cpw t + hwe         (3)where• cpw = specific heat capacity of water vapor

at constant pressure (kJ/kgoC, kWs/kgK)• t = water vapor temperature (oC)• hwe = evaporation heat of water at 0oC

(kJ/kg)

For water vapor the specific heat capacity can be set to

• cpw = 1.84 (kJ/kgoC)•     = 0.444 (Btu/lboF) The evaporation heat (water at 0oC) can be set to• hwe = 2501 kJ/kg) = 970 (Btu/lb) • Using (2) and (3), (1) can be modified to

• H = cpa t + x [cpw t + hwe]         (1b)(1b) in metric units

• H = (1.006 kJ/kgoC) t + x [(1.84 kJ/kgoC) t + (2501 kJ/kg)]         (1c)

where • H = enthalpy (kJ/kg)• x = mass of water vapor (kg/kg)• t = temperature (oC) (1b) in Imperial units

• h = (0.240 Btu/lboF) t + x [(0.444 Btu/lboF) t + (970 Btu/lb)]         (1d)

where • h = enthalpy (Btu/lb)• x = mass of water vapor (lb/lb)• t = temperature (oF)

Example - Enthalpy in Moist Air • The enthalpy of humid air at 25oC with specific

moisture content x = 0.0203 kg/kg (saturation), can be calculated as:

• h = (1.006 kJ/kgoC) (25oC) + (0.0203 kg/kg) [(1.84 kJ/kgoC) (25oC) + (2501 kJ/kg)]

= (25.15 kJ/kg) + (0.93 kJ/kg) + (51.70 kJ/kg) = 77.8 (kJ/kg)• Note! The latent heat due to evaporation of

water is the major part of the enthalpy. The sensible heat due to heating evaporated water vapor can be almost neglected.