ENERGY BALANCE

Nur Istianah,ST.,MT.,M.Eng THP UB 2017

HEAT

• Heat is a kind of energy that is related to the temperature or phase change. It involved in thermodynamic transformations

• Sensible heat is defined as the energy transferred between two bodies at different temperatures

• Latent heat is the energy associated withphase transitions

EXOTERM

System Energy

Environment

Q<0

ENDOTERM

System Energy

Environment

Q>0

Sensibel heat

SENSIBEL HEAT

-40

-20

0

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40

60

80

100

120

140

Tem

pe

ratu

r

Entalpi

Sensibel heat

∆H = H2 – H1 = Q = m 𝐶𝑝𝑑𝑇 = 𝑚𝐶𝑝

(𝑇2− 𝑇1)

𝑇2

𝑇1

PANAS LATEN (PERUBAHAN/TRANSISI FASA)

Q= mhf atau Q = mhv

Laten heat

Laten heat

-40

-20

0

20

40

60

80

100

120

140

Tem

pe

ratu

r

Entalpi

Heat of Fusion (+)

Heat of Vaporization (+)

Heat of Solidification (-)

Heat of Condensation (-)

Relationship between sensible Heat and latent Heat

ICE at -50 C ICE at 0 C

water at 0 C water at

100 C

vapor at

100 C

vapor at

150 C

Q1 = sensible heat

Q3 = sensible heat

Q5 = sensible heat

Q2 = Latent heat

Of Fusion

Q 4 = Latent heat

Of vaporization

Heat term

• Units of heat: joule (J), Calorie (Cal), or British thermal unit (Btu)

• 1 Calorie: amount of heat needed to raise the temperature of 1 gram of water by 1 C0 (from 14.50C to 15.50C)

• 1 Btu: amount of heat needed to raise the temperature of 1 lb of water by 1 F0 (from 630F to 640F)

• 1 cal = 10-3 kcal = 3.969 x 10-3 Btu = 4.186 J

ENTHALPY

• Enthalpy is the absolute value of which cannot be measured directly.

• Heat can be defined as enthalpy changes of materials

• Specific heat capacity (cp) : amount of energy needed to raise the temperature of 1 kg of a substance by 1K ( kJ/kgC or kJ/kg.K)

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Enthalpy dan specific heat

Enthalpy: sensible

ΔH = CpΔT = Cp (T-Tref)

Laten: Presented in experimental data

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Specific heat

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Enthalpy dan specific heat

Enthalpy:

ΔH = CpΔT = Cp (T-Tref)

Siebel`s Eq.:

Cavg = 0,4 F + 0,2 SNF + M BTU/(lb0F) or

Cavg = 1674,72 F + 837,36 SNF + 4186,8 M J/(KgK)

Where: fat(F), solids non-fat (SNF), dan moisture (M)

1

– Heldman and Singh (1981) proposed the following expression based on the components of a food product [Introduction to Food Engineering, 4th ed., page 258,

equation 4.4]:

– where X is the mass fraction; the subscripts on the right-hand side are h (carbohydrate); p (protein); f (fat); a (ash); and w (moisture)

Cp (J.kg-1.K-1) =1424Xh +1549Xp + 1675Xf + 837Xa + 4187Xw

Enthalpy dan specific heat

2

Example

– A apple with 0,15 kg of weight was left at 0oC until the temperature raise up to 20oC. Calculate the heat transfered if the apple has composistion; water 84,4%, Protein 0,2%, carbohydrate 14,5%, fat 0,6%, dan Ash 0,3% ?

– Defined: m = 0,15 kg

T1 = 0oC

T2 = 20oC

Xw = 0,844, Xp = 0,002, Xh = 0,145, Xf = 0,006, Xa = 0,003

– Q = ?

– Solusi: ∆H = Q = m CpdT𝑇2

𝑇1

• For unknown Cp, Cp can be calculated from Siebel Eq.

•

• ∆H = Q = m CpdT𝑇2

𝑇1

= 0,15 kg x 3,76 kJ/kg.oC x (20-0) oC

= 11,28 kJ

Cp = 1,424Xh +1,549Xp + 1,675Xf + 0,837Xa + 4,187Xw = 1,424 (0,145) +1,549 (0,002) + 1,675 (0,006) + 0,837 (0,003) + 4,187 (0,844) = 3,76 kJ/kg.K

– A formulated food has the following composition:

water, 80%; protein, 2%; carbohydrate,17%; fat, 0.1%; and ash, 0.9%. Estimate the heat capacity of the food!

ENTALPHY of FOOD – For processes where temperature changes, we must

use predictive models of specific heat that include temperature dependence. Choi and Okos (1986) presented a comprehensive model to predict specific heat based on composition and temperature.

[Introduction to Food Engineering, 4th ed., page 258,

equation 4.5]. Their model is as follows

Cp = Σ(XiCpi) = X1Cp1 + X2Cp2 + X3Cp3 + X4Cp4 + XnCpn

3

Choi Okos Eq.

Exercise

– Calculate the specific of milk at 60oC

2% lemak (Cpf=1,675 kJ/ kgoC),

89,2% air (Cpw=4,187 kJ/kgoC),

3,3% protein (Cpp=1,549 kJ/kgoC),

4,8% karbohidrat (Cph=1,424 kJ/kgoC) dan

0,7% abu (Cpa=0,837 kJ/kgoC).

Solution

– By using Eq. of Choi dan Okos (1986), this is the specific heat of milk:

Cp milk = Σ(XiCpi) = XwCpw + XfCpf + XpCpp + XhCph + XaCpa

Cp milk = (0,892)(4,187) + (0,02)(1,675) + (0,033)(1,549) + (0,048)(1,424) + (0,007)(0,837)

Cp milk = 3,894 kJ/kgK

Problem

– Calculate the specific of milk at 60oC:

2% lemak (Cpf= ?),

89,2% air (Cpw=?),

3,3% protein (Cpp=?),

4,8% karbohidrat (Cph=?) dan

0,7% abu (Cpa=?).

There is information of specific heat of components

Choi Okos Eq.

Solution

THANKS FOR YOUR ATTENTION

The best person is one give something useful always

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