Energy Balance 1

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 2

Forms of energy Potential energy (mgh) Kinetic energy (1/2 mv 2 ) 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 3

Energy balance system mass in Hin mass out Hout W Q 4

IUPAC convention - heat transferred to a system is positive and heat transferred from a system is negative. - work done on a system is positive and work done by a system is negative 5

Steady state/non-steady state Non steady state: accumulation/depletion of energy in system 6

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 7

Enthalpy balance p.e., k.e., W terms = 0 Q = H 2 – H 1 or Q = ΔH where H 2 is the total enthalpy of output streams and H 1 is the total enthalpy of input streams Q is the difference in total enthalpy i.e. the enthalpy (heat) transferred to or from the system 8

Q negative (H1>H2), heat removed from system Q positive (H2>H1), heat supplied to system. 9

Example – steam boiler Two input streams: stream 1: 120 kg/min. water, 30 deg cent., H = 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 = kJ/kg 10

Ignore 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 – no moving part) Basis for calculation 1 min. Steady state Q = H out – H in = [295 x ] – [(120 x 125.7) + (175 x 272)] = x 10 5 kJ/min 11

Steam tables Enthalpy values (kJ/kg) H at various P, T 12

Enthalpy changes Change of T at constant P Change of P at constant T Change of phase Solution Mixing Chemical reaction Crystallisation 13

Latent heats (phase changes) Vaporisation (L to V) Melting (S to L) Sublimation (S to V) 14

Mechanical energy balance 15

Example - Bernoulli Equation. 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 1. 16

ΔP/ρ + Δv 2 /2 + gΔh + F = W ΔP = P 2 – P 1 (Pa) Δv 2 = v 2 2 – v 1 2 Δh = h 2 - h 1 (m) F= frictional energy loss (mechanical energy loss to system) (J/kg) = 0 W = work done on system by pump (J/kg) = 0 ρ = 1000 kg/m 3 17

Volumetric flow is 20/(1000x60) m 3 /s = m 3 /s v 1 = /( π(0.0025) 2 ) = m/s v 2 = / ( π(0.005) 2 ) = 4.24 m/s ( P 1 )/ [(4.24) 2 – (16.97) 2 ]/ = 0 P 1 = Pa (4.6 bar) 18

Sensible heat - Enthalpy calculations 19

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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 °C. Use mean Cp value 29.1J mol -1 K -1 or Cp = x T x T 2 – 2.87 x T 3, where T is in °C 21

Heat capacity/specific heat data Felder & Rousseau pp. 372/373 and Table B10 Perry’s Chemical Engineers Handbook The properties of gases and liquids, R. Reid et al, 4 th 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, 3 rd edition, ch. 8, pp

Example – change of phase A feed stream to a distillation unit contains an equimolar mixture of benzene and toluene at 10 °C.The vapour stream from the top of the column contains 68.4 mole % benzene at 50 °C and the liquid stream from the bottom of the column contains 40 mole % benzene at 50 °C. Need Cp (benzene, liquid), Cp (toluene, liquid), Cp (benzene, vapour), Cp (toluene, vapour), latent heat of vaporisation benzene, latent heat of vaporisation toluene. 23

Energy balances on systems involving chemical reaction Standard heat of formation (ΔH o f ) – 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, ν i ) ΔH o fA, ΔH o fB, ΔH o fC, ΔH o fD (heats of formation) ΔH o R = c ΔH o fC + d ΔH o fD - a ΔH o fA - bΔH o fB 24

Heat (enthalpy) of reaction ΔH o R negative (exothermic reaction) ΔH o R positive (endothermic reaction) 25

Enthalpy balance equation - reactor Qp = H products – H reactants + Qr Qp – heat transferred to or from process Qr – reaction heat (ζ ΔH o R ), where ζ is extent of reaction. ζ=[moles component_i_out – moles component_i_in]/ ν i 26

System Qr H reactants H products Qp positive negative Note: enthalpy values must be calculated with reference to a temperature of 25 °C 27

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 (reactions) Solve enthalpy balance equation 28

Energy balance techniques Adiabatic flame temperature: Qp = 0 29

Examples Reactor Crystalliser Drier Distillation 30

References The Properties of Gases and Liquids, R. Reid et al. Elementary Principles of Chemical Processes, R.M.Felder and R.W.Rousseau 31