LECTURE 5ENERGY BALANCE

Ch 61

ENERGY BALANCE Concerned with energy changes and

energy flow in a chemical process. Conservation of energy – first law of

thermodynamics i.e. accumulation of energy in a system = energy input – energy output

Forms of energy Potential energy (mgh) Kinetic energy (1/2 mv2) Thermal energy – heat (Q) supplied to or removed from a

process Work energy – e.g. work done by a pump (W) to transport

fluids Internal energy (U) of molecules

m – mass (kg)g – gravitational constant, 9.81 ms-2

v – velocity, ms-1

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Energy balance

systemmass in

Hin

mass out

Hout

W

Q

IUPAC convention

- heat transferred to a system is +ve and heat transferred from a system is –ve

- work done on a system is+ve and work done by a system is -ve

STEADY STATE/NON-STEADY STATE

Non steady state - accumulation/depletion of energy in system

Uses Heat required for a process Rate of heat removal from a process Heat transfer/design of heat exchangers Process design to determine energy requirements of a

process Pump power requirements (mechanical energy balance) Pattern of energy usage in operation Process control Process design & development etc

1) No mass transfer (closed or batch)∆E = Q + W2) No accumulation of energy, no mass transfer (m1 = m2= m)

∆E = 0 Q = -W

3. No accumulation of energy but with mass flowQ + W = ∆ [(H + K + P)m]

Air is being compressed from 100 kPa at 255 K (H = 489 kJ/kg) to 1000kPa and 278 K (H = 509 kJ/kg) The exit velocity of the air from the compressor is 60 m/s. What is the power required (in kW) for the compressor if the load is 100 kg/hr of air?

SAMPLE PROBLEMWater is pumped from the bottom of a well 4.6 m deep at a rate of 760 L/hour into a vented storage tank to maintain a level of water in a tank 50 m above the ground. To prevent freezing in the winter a small heater puts 31,600 kJ/hr into the water during its transfer from the well to the storage tank. Heat is lost from the whole system at a constant rate of 26,400 kJ/hr. What is the temperature of the water as it enters the storage tank assuming that the well water is at 1.6oC? A 2-hp pump is used. About 55% of the rated horse power goes into the work of pumping and the rest is dissipated as heat to atmosphere.

ENTHALPHY BALANCE p.e., k.e., W terms = 0 Q = H2 – H1 or Q = ΔH

, where H2 is the total enthalpy of output streams and H1is the total enthalpy of input streams, Q is the difference in total enthalpy i.e. the enthalpy (heat) transferred to or from the system

continued Q –ve (H1>H2), heat removed from

system Q +ve (H2>H1), heat supplied to

system.

Example – steam boiler

Two input streams: stream 1- 120 kg/min. water, 30 deg cent., H = 125.7 kJ/kg; stream 2 – 175 kg/min, 65 deg cent, H= 272 kJ/kg

One output stream: 295 kg/min. saturated steam(17 atm., 204 deg cent.), H = 2793.4 kJ/kg

continuedIgnore k.e. and p.e. terms relative to enthalpy

changes for processes involving phase changes, chemical reactions, large temperature changes etc

Q = ΔH (enthalpy balance)Basis for calculation 1 min.Steady stateQ = Hout – HinQ = [295 x 2793.4] – [(120 x 125.7) + (175 x 272)] Q = + 7.67 x 105 kJ/min

STEAM TABLES Enthalpy values (H kJ/kg) at various P, T

Enthalpy changes Change of T at constant P Change of P at constant T Change of phase Solution Mixing Chemical reaction crystallisation

INVOLVING CHEMICAL REACTIONS

Heat of Reaction:∆Hrxn= ∑∆Hf

o(products) - ∑∆Hf

o(reactants)

SAMPLE PROBLEM:

Calculate the ∆Hrxn for the following reaction:4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Given the following ∆Hfo/mole at 25oC.

NH3(g): -46.191 kJ/mol

NO(g): +90.374 kJ/molH2O(g): -214.826 kJ/mol

∆H = -904.696 kJ/mol

ENERGY BALANCE THAT ACCOUNT FOR CHEMICAL REACTIONS:

Most common:1) What is the temperature of the

incoming or exit streams2) How much material must be introduced

into the entering stream to provide for a specific amount of heat transfer.

aA + bB → cC + dD

T4

T2

T3T1

reactants productsA

B

C

D

SAMPLE PROBLEM:An iron pyrite ore containing 85.0% FeS2 and 15.0% inert materials (G) is roasted with an amount of air equal to 200% excess air according to the reaction

4FeS2 + 11O2 → 2Fe2O3 + 8 SO2

in order to produce SO2. All the inert materials plus the Fe2O3 end up in the solid waste product, whick analyzes at 4.0% FeS2. Determine the heat transfer per kg of ore to keep the product stream at 25oC if the entering streams are at 25oC.

ProductsComponents Amount

kmole∆Hf kJ/mol nx∆H

FeS2 -177.9

Fe2O3 -822.156

N2 0

O2 0

SO2 -296.90

ProductsComponents Amount

kmole∆Hf kJ/mol nx∆H

FeS2 -177.9

Fe2O3 -822.156

N2 0

O2 0

SO2 -296.90

Material Balance First!!!!

∆E = 0, W = 0, ∆E = 0, ∆PE = 0, ∆KE = 0

Therefore Q = ∆H

ProductsComponents Amount

kmole∆Hf kJ/mol nx∆H

FeS2 0.0242 -177.9 -4.305

Fe2O3 0.342 -822.156 -281.177

N2 21.998 0 0

O2 3.938 0 0

SO2 1.368 -296.90 -406.159

ReactantsComponents Amount

kmole∆Hf kJ/mol nx∆H

FeS2 0.7803 -177.9 -126.007

Fe2O3 0 -822.156

N2 21.983 0

O2 5.8437 0

SO2 0 -296.90

SAMPLE PROBLEM

Q =

Latent heats (phase changes) Vapourisation (L to V) Melting (S to L) Sublimation (S to V)

Mechanical energy balance Consider mechanical energy terms only Application to flow of liquidsΔP + Δ v2 + g Δh +F = Wρ 2where W is work done on system by a pump and F

is frictional energy loss in system (J/kg)ΔP = P2 – P1; Δ v2 = v2

2 –v12; Δh = h2 –h1

Bernoulli equation (F=0, W=0)

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Example - Bernoulli eqtn.

Water flows between two points 1,2. The volumetric flow rate is 20 litres/min. Point 2 is 50 m higher than point 1. The pipe internal diameters are 0.5 cm at point 1 and 1 cm at point 2. The pressure at point 2 is 1 atm..

Calculate the pressure at point 2.

continuedΔP/ρ + Δv2/2 + gΔh +F = W

ΔP = P2 – P1 (Pa)Δv2 = v2

2 – v12

Δh = h2 - h1 (m)F= frictional energy loss (mechanical energy loss to

system) (J/kg)W = work done on system by pump (J/kg)ρ = 1000 kg/m3

continuedVolumetric flow is 20/(1000.60) m3/s= 0.000333 m3/sv1 = 0.000333/(π(0.0025)2) = 16.97 m/s

v2 = 0.000333/ (π(0.005)2) = 4.24 m/s

(101325 - P1)/1000 + [(4.24)2 – (16.97)2]/2 + 9.81.50 = 0

P1 = 456825 Pa (4.6 bar)

Sensible heat/enthalpy calculations ‘Sensible’ heat – heat/enthalpy that must be transferred to

raise or lower the temperature of a substance or mixture of substances.

Heat capacities/specific heats (solids, liquids, gases,vapours) Heat capacity/specific heat at constant P, Cp(T) = dH/dT or

ΔH = integral Cp(T)dT between limits T2 and T1

Use of mean heat capacities/specific heats over a temperature range

Use of simple empirical equations to describe the variation of Cp with T

continuede.g. Cp = a + bT + cT2 + dT3

,where a, b, c, d are coefficients

ΔH = integralCpdT between limits T2, T1

ΔH = [aT + bT2 + cT3 + dT4] 2 3 4

Calculate values for T = T2, T1 and subtract

Note: T may be in deg cent or K - check units for Cp!

Example

Calculate the enthalpy required to heat a stream of nitrogen gas flowing at 100 mole/min., through a gas heater from 20 to 100 deg. cent.

(use mean Cp value 29.1J mol-1 K-1 or Cp = 29 + 0.22 x 10-2T + 0.572 x 10-5T2 – 2.87 x 10-9 T3, where T is in deg cent)

Heat capacity/specific heat data Felder & Rousseau pp372/373 and Table B10 Perry’s Chemical Engineers Handbook The properties of gases and liquids, R. Reid et al, 4th

edition, McGraw Hill, 1987 Estimating thermochemical properties of liquids part 7-

heat capacity, P. Gold & G.Ogle, Chem. Eng., 1969, p130 Coulson & Richardson Chem. Eng., Vol. 6, 3rd edition, ch.

8, pp321-324 ‘PhysProps’

Example – change of phaseA feed stream to a distillation unit contains an

equimolar mixture of benzene and toluene at 10 deg cent.The vapour stream from the top of the column contains 68.4 mol % benzene at 50 deg cent. and the liquid stream from the bottom of the column contains 40 mol% benzene at 50 deg cent.

[Need Cp (benzene, liquid), Cp (toluene, liquid), Cp (benzene, vapour), Cp (toluene, vapour), latent heat of vapourisation benzene, latent heat of vapourisation toluene.]

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Energy balances on systems involving chemical reaction

Standard heat of formation (ΔHof) – heat of reaction

when product is formed from its elements in their standard states at 298 K, 1 atm. (kJ/mol)

aA + bB cC + dD-a -b +c +d (stoichiometric

coefficients, νi)

ΔHofA, ΔHo

fB, ΔHofC, ΔHo

fD (heats of formation)

ΔHoR = c ΔHo

fC + d ΔHofD - a ΔHo

fA - bΔHofB

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Heat (enthalpy) of reaction ΔHo

R –ve (exothermic reaction) ΔHo

R +ve (endothermic reaction)

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ENTHALPHY BALANCE - REACTOR

Qp = Hproducts – Hreactants + Qr

Qp – heat transferred to or from processQr – reaction heat (ζ ΔHo

R), where ζ is extent of reaction and is equal to [moles component,i, out – moles component i, in]/ νi

system

Qr

HreactantsHproducts

Qp

+ve

-ve

Note: enthalpy values must be calculated with reference to a temperature of 25 deg cent

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ENERGY BALANCE TECHNIQUES Complete mass balance/molar balance Calculate all enthalpy changes between

process conditions and standard/reference conditions for all components at start (input) and finish (output).

Consider any additional enthalpy changes Solve enthalpy balance equation

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ENERGY BALANCE TECHNIQUES Adiabatic temperature: Qp = 0

EXAMPLES Reactor Crystalliser Drier Distillation

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References The Properties of Gases and Liquids, R.

Reid Elementary Principles of Chemical

Processes, R.M.Felder and R.W.Rousseau

SAMPLE PROBLEM:Limestone (CaCO3) is converted into CaO in a continuous vertical kiln. Heat is supplied by combustion of natural gas (CH4) in direct contact with limestone using 50% excess air. Determine the kg of CaCO3 that can be processed per kg of natural gas. Assume that the following heat capacities apply.Cp of CaCO3 = 234 J/mole –oC

Cp of CaO = 111 j/mole - oC