1. LECTURE 5 ENERGY BALANCE Ch 61

2. 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

3. 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

4. Paul Ashall, 2008 Energy balance system mass in H in mass out H out W Q

5. 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

6. STEADY STATE/NON-STEADY STATE  Non steady state - accumulation/depletion of energy in system

7. 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

8. 1) No mass transfer (closed or batch) ∆E = Q + W 2) No accumulation of energy, no mass transfer (m 1 = m 2 = m) ∆E = 0 Q = -W

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

10. 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?

11. SAMPLE PROBLEM Water 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.

12. ENTHALPHY BALANCE  p.e., k.e., W terms = 0  Q = H 2 – H 1 or Q = Δ H, 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

13. continued  Q –ve (H1>H2), heat removed from system  Q +ve (H2>H1), heat supplied to system.

14. 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

15. continued Ignore k.e. and p.e. terms relative to enthalpy changes for processes involving phase changes, chemical reactions, large temperature changes etc Q = Δ H (enthalpy balance) Basis for calculation 1 min. Steady state Q = Hout – Hin Q = [295 x 2793.4] – [(120 x 125.7) + (175 x 272)] Q = + 7.67 x 10 5 kJ/min

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

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

18. INVOLVING CHEMICAL REACTIONS Heat of Reaction: ∆H rxn = ∑∆H f o (products) - ∑∆H f o (reactants)

19. SAMPLE PROBLEM: Calculate the ∆H rxn for the following reaction: 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g) Given the following ∆H f o /mole at 25 o C. NH 3 (g): -46.191 kJ/mol NO(g):+90.374 kJ/mol H 2 O(g): -214.826 kJ/mol

20. ∆H = -904.696 kJ/mol

21. ENERGY BALANCE THAT ACCOUNT FOR CHEMICAL REACTIONS : Most common: 1) What is the temperature of the incoming or exit streams 2) How much material must be introduced into the entering stream to provide for a specific amount of heat transfer.

22. aA + bB → cC + dD T4T4 T2T2 T3T3 T1T1 reactants products A B C D

23. SAMPLE PROBLEM: An iron pyrite ore containing 85.0% FeS 2 and 15.0% inert materials (G) is roasted with an amount of air equal to 200% excess air according to the reaction 4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8 SO 2 in order to produce SO 2. All the inert materials plus the Fe 2 O 3 end up in the solid waste product, whick analyzes at 4.0% FeS 2. Determine the heat transfer per kg of ore to keep the product stream at 25 o C if the entering streams are at 25 o C.

24. Products ComponentsAmount kmole∆Hf kJ/molnx∆H FeS 2 -177.9 Fe 2 O 3 -822.156 N2N2 0 O2O2 0 SO 2 -296.90

25. Products ComponentsAmount kmole∆Hf kJ/molnx∆H FeS 2 -177.9 Fe 2 O 3 -822.156 N2N2 0 O2O2 0 SO 2 -296.90

26. Material Balance First!!!!

27. ∆E = 0, W = 0, ∆E = 0, ∆PE = 0, ∆KE = 0 Therefore Q = ∆H

28. Products ComponentsAmount kmole∆Hf kJ/molnx∆H FeS 2 0.0242-177.9-4.305 Fe 2 O 3 0.342-822.156-281.177 N2N2 21.99800 O2O2 3.93800 SO 2 1.368-296.90-406.159

29. Reactants ComponentsAmount kmole∆Hf kJ/molnx∆H FeS 2 0.7803-177.9-126.007 Fe 2 O 3 0-822.156 N2N2 21.9830 O2O2 5.84370 SO 2 0-296.90

30. SAMPLE PROBLEM Q =

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

32. Mechanical energy balance  Consider mechanical energy terms only  Application to flow of liquids Δ P + Δ v 2 + g Δ h +F = W ρ 2 where W is work done on system by a pump and F is frictional energy loss in system (J/kg) Δ P = P 2 – P 1 ; Δ v 2 = v 2 2 –v 1 2 ; Δ h = h 2 –h 1  Bernoulli equation (F=0, W=0)

33. Paul Ashall, 2008 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.

34. continued Δ P/ ρ + Δ v 2 /2 + g Δ h +F = W Δ P = P 2 – P 1 (Pa) Δ v 2 = v 2 2 – v 1 2 Δ h = h 2 - h 1 (m) F= frictional energy loss (mechanical energy loss to system) (J/kg) W = work done on system by pump (J/kg) ρ = 1000 kg/m 3

35. continued Volumetric flow is 20/(1000.60) m 3 /s = 0.000333 m 3 /s v 1 = 0.000333/( π(0.0025) 2 ) = 16.97 m/s v 2 = 0.000333/ ( π(0.005) 2 ) = 4.24 m/s (101325 - P 1 )/1000 + [(4.24) 2 – (16.97) 2 ]/2 + 9.81.50 = 0 P 1 = 456825 Pa (4.6 bar)

36. 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 Δ H = integral Cp(T)dT between limits T 2 and T 1  Use of mean heat capacities/specific heats over a temperature range  Use of simple empirical equations to describe the variation of Cp with T

37. continued e.g. Cp = a + bT + cT 2 + dT 3,where a, b, c, d are coefficients Δ H = integralCpdT between limits T 2, T 1 Δ H = [aT + bT 2 + cT 3 + dT 4 ] 2 3 4 Calculate values for T = T 2, T 1 and subtract Note: T may be in deg cent or K - check units for Cp!

38. 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 -2 T + 0.572 x 10 -5 T 2 – 2.87 x 10 -9 T 3, where T is in deg cent)

39. 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, 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, pp321-324  ‘PhysProps’

40. Example – change of phase A 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.]

41. Paul Ashall, 2008 Energy balances on systems involving chemical reaction  Standard heat of formation ( Δ H 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 ) Δ H o fA, Δ H o fB, Δ H o fC, Δ H o fD (heats of formation) Δ H o R = c Δ H o fC + d Δ H o fD - a Δ H o fA - b Δ H o fB

42. Paul Ashall, 2008 Heat (enthalpy) of reaction  Δ H o R –ve (exothermic reaction)  Δ H o R +ve (endothermic reaction)

43. Paul Ashall, 2008 ENTHALPHY BALANCE - REACTOR Qp = H products – H reactants + Qr Qp – heat transferred to or from process Qr – reaction heat ( ζ Δ H o R ), where ζ is extent of reaction and is equal to [moles component,i, out – moles component i, in]/ ν i

44. system Qr H reactants H products Qp +ve -ve Note: enthalpy values must be calculated with reference to a temperature of 25 deg cent

45. Paul Ashall, 2008 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

46. Paul Ashall, 2008 ENERGY BALANCE TECHNIQUES  Adiabatic temperature: Qp = 0

47. EXAMPLES  Reactor  Crystalliser  Drier  Distillation

48. Paul Ashall, 2008 References  The Properties of Gases and Liquids, R. Reid  Elementary Principles of Chemical Processes, R.M.Felder and R.W.Rousseau

49. SAMPLE PROBLEM: Limestone (CaCO 3 ) is converted into CaO in a continuous vertical kiln. Heat is supplied by combustion of natural gas (CH 4 ) in direct contact with limestone using 50% excess air. Determine the kg of CaCO 3 that can be processed per kg of natural gas. Assume that the following heat capacities apply. C p of CaCO 3 = 234 J/mole – o C C p of CaO = 111 j/mole - o C