JENIS PANAS, PROSES REVERSIBLE JENIS PANAS, PROSES REVERSIBLE DAN IRREVERSIBLE, NERACA PANAS DAN IRREVERSIBLE, NERACA PANAS

Latifa Hanum Lalasari, ST., MTLatifa Hanum Lalasari, ST., MT

Program Starta – 1, Teknik Kimia - Program Starta – 1, Teknik Kimia - ITIITI

2012 2012

Azas Teknik Kimia II

ISOLATED SYSTEM, CLOSED SYSTEM AND OPEN SYSTEM

• Isolated system:  No mass transfer and no energy transfer across the system boundary.

• Closed system:  No mass transfer, but energy may transfer across the system boundary.

• Open system:  Both mass and energy allowed to transfer across the system boundary.

THE GENERAL ENERGY BALANCE

Energy balance without chemical reaction :

ISOLATED SYSTEM

No mass transfer and no energy transfer across the system boundary, but only transfer between bodies within the system.

solated (Insulated) System:   U = constantSince there is no mass change,  Q = 0

Finding Equilibrium Temperature: T  From Q = 0Qw + Qb = 0mwcw(T-Tw) = - mbcb(T-Tb)T =  (mwcwTw + mbcbTb) / (mwcw + mbcb)

Task 1 :

A 1.8 Kg steel block at 475 oC is placed in an insulated container of 15 L of water at 20 oC. Find the equilibrium temperatur ?

CLOSED SYSTEM

A closed system has no mass transfer, but allows energy transfer across the system boundary. Closed systems are also called non-flow systems, control-mass systems or constant-mass systems.

A closed system has no mass transfer, but allows energy transfer across the system boundary. Closed systems are also called non-flow systems, control-mass systems or constant-mass systems.

A closed system has no mass transfer, but allows energy transfer across the system boundary. Closed systems are also called non-flow systems, control-mass systems or constant-mass systems.

A closed system has no mass transfer, but allows energy transfer across the system boundary. Closed systems are also called non-flow systems, control-mass systems or constant-mass systems.

A solid is a closed system (no mass transfer across boundary)

Expanding or compressing a gass can also be a closed system if there is no flow or major leakages.

Non-flow Energy Equation (1st Law of Thermodynamics);      U2-U1 = Q - WWhereU = internal energy (J)Q = heat transfer (J) (+ = heat IN)W = work transfer (J) (+ = work OUT)  

Non-flow, No phase change, Constant c;      U2-U1 = mct t

WhereU = internal energy (J)m = mass (kg)c = specific heat capacity (J/kgK) t= (t2-t1) = (tfinal-tintitial) = temperature change (K or oC)

Non-flow, With phase change, Constant c;      U2-U1 = mc1(t1-t0) + mcphaset1 + mc2(t2-t1)t0 = initial solid temperaturet1 = fusion temperature (melting/freezing)t2 = final liquid temperature(For example, melting ice.... Energy change (or heat added) = (heat ice from ? to 0oC) + (latent heat of fusion) + (Heat water to ?oC)

A tank holds 2.3 kg of water at 24 oC and absorbs 178 KJ of heat. The final temperatur is 47 oC. (a) Find internal energy change. How much work must have been done?

Taks 2 :

HEAT EFFECTS:

1. Sensible heat : associated with simple temperature changes. No phase changes (liquid to vapor or liquid to solid) and no chemical reactions are considered.

2. Latent heat is the heat effect associated with a phase change (heat of vaporization or heat of fusion)

3. Heats of reaction involve energy produced or consumed by chemical reactions. This heat is negative for exothermic and positive for endothermic reactions

Sensible HeatSensible Heat

Untuk gas Ideal :

Dimana :

Cp – Cv = R

Task :

Ethane gas is to heated from 400 oF to 600 oF in a heat exchanger. The flow rate of the gas is 40.000 lbm/hr. Assuming constant-pressure operation and the methane is an ideal gas. How much heat must be transferred into the gas to raise its temperatur ?

Latent HeatLatent HeatPersamaan Clausius-Clapyron :

Untuk Penguapan : V = Vv - VL

Heat of Reaction :Heat of Reaction :

HR = H products – H Reactans

HF = H compounds – ∑H elements forming the compound

Dimana:Panas pembentukan dari elemen murni seperti H2, O2, N2, C adalah nol

ENTHALPY DATA:ENTHALPY DATA:

Enthalpies of vapor and liquid for a pure components

H = Cv (T – To) + HvEnthapy of saturated vapor :

Enthapy of saturated liquid : h= CL (T – To)

Multicomponent systems : H = ∑ yiHi

h = ∑ xiHi

Contoh Aplikasi NE untuk Tugas Akhir Contoh Aplikasi NE untuk Tugas Akhir ::

Neraca Energi pada HE :Neraca Energi pada HE :

Counter current Cocurrent

Panas keluar hot stream Cold Stream

Jika Cp = Konstan, maka Q = Fc. Cp (Tc out – Tc in)

Neraca Energi untuk proses reversible Neraca Energi untuk proses reversible dan irreversible:dan irreversible:

Proses reversible :prosese dimana pengaruh luar diabaikan, ditunjukkan tidak ada friksi,viscositas, dan lain-lain. Atau pengaruh luar tidak begitu besar.

Proses irreversible : pengaruh luar diperhatikan karena menunjukkan perubahan yang signifikan. Akibatnya menghasilkan kerja yang tidak maksimum

Inputs:Temperature of the entering reactant stream = 11 °CTemperature of the entering glycol stream = 252 °CSensible heat of the reactant stream =m.CP.ΔT= 9.3182×3.8353×(11-0) =362 kWSensible heat of the glycol stream =m.CP.ΔT= 9.3182×4.0124×(252-0) =9413 kWTotal Sensible heat entering the heat exchanger = 9775 kW

Outputs:Temperature of the leaving reactant stream = 200 °CTemperature of the leaving glycol stream = 70 °CSensible heat of the reactant stream = m.CP.ΔT = 9.3182×3.8353×(200-0)=7148 kWSensible heat of the glycol stream = m.CP.ΔT = 9.3182×4.0124×(70-0)= 2627 kWTotal heat leaving the heat exchanger = 9775 kW

Energy balance around the Heat Exchanger

LMTD MethodLMTD Method

A Methodology for Heat ExchangerDesign Calculations

- The Log Mean Temperature Difference (LMTD) Method -• A form of Newton’s Law of Cooling may be applied to heat exchangers by using a log-mean value of the temperature difference between the two fluids:

1mq U A T

1 2

11 21n /m

T TT

T T

Evaluation of depends on the heat exchanger type.1 2 and T T

• Counter-Flow Heat Exchanger:

1 ,1 ,1

, ,

h c

h i c o

T T TT T

2 ,2 ,2

, ,

h c

h o c i

T T TT T

LMTD Method (cont.)LMTD Method (cont.)

• Parallel-Flow Heat Exchanger:

1 ,1 ,1

, ,

h c

h i c i

T T TT T

2 ,2 ,2

, ,

h c

h o c o

T T TT T

Note that Tc,o can not exceed Th,o for a PF HX, but can do so for a CF HX.

For equivalent values of UA and inlet temperatures,

1 , 1 ,m CF m PFT T

• Shell-and-Tube and Cross-Flow Heat Exchangers:

1 1 ,m m CFT F T

Energy BalanceEnergy Balance

Overall Energy Balance

• Assume negligible heat transfer between the exchanger and its surroundings and negligible potential and kinetic energy changes for each fluid.

, ,h i h ohq m i i

, ,c c o c iq m i i

fluid enthalpyi

• Assuming no l/v phase change and constant specific heats,

, , ,p h h i h ohq m c T T , ,h h i h oC T T

, , ,c p c c o c iq m c T T

, ,c c o c iC T T

, Heat capacity r s ateh cC C

• Application to the hot (h) and cold (c) fluids:

Tugas :

Text book : David M. Himmelblau, edisi 6

Soal : Hal. 422, soal No : 1,3,4,5,6,8 Hal. 438, soal No : 1,3,5,6

Dikumpulkan : Hari Minggu, Tgl 12 Februari 2012 Pada saat Presentasi group Membawa ringkasan 1 lembar A4 (persiapan UTS)