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Lecture 5 Energy Balance

Lecture 5 Energy 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 )

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Lecture 5 Energy Balance

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  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 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

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

  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 (m1 = m2= 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 = 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

  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 = ΔH (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 105 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: ∆Hrxn= ∑∆Hfo(products) - ∑∆Hfo(reactants)

  19. 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/mol H2O(g): -214.826 kJ/mol

  20. ∆H = -904.696 kJ/mol

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

  22. aA + bB→ cC + dD T3 T1 C A products reactants T2 D B T4

  23. 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.

  24. Products

  25. Products

  26. Material Balance First!!!!

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

  28. Products

  29. Reactants


  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 ΔP + Δ v2+g Δh +F = W ρ 2 where W is work done on system by a pump and F is frictional energy loss in system (J/kg) ΔP = P2 – P1; Δ v2 = v22 –v12; Δh = h2 –h1 • Bernoulli equation (F=0, W=0)

  33. 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. Paul Ashall, 2008

  34. continued ΔP/ρ + Δv2/2 + gΔh +F = W ΔP = P2 – P1 (Pa) Δv2 = v22 – 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

  35. continued Volumetric flow is 20/(1000.60) m3/s = 0.000333 m3/s v1 = 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)

  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 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

  37. continued e.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!

  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-1or Cp = 29 + 0.22 x 10-2T + 0.572 x 10-5T2 – 2.87 x 10-9 T3, 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, 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’

  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. 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, ΔHofB, ΔHofC, ΔHofD (heats of formation) ΔHoR = c ΔHofC + d ΔHofD - a ΔHofA - bΔHofB Paul Ashall, 2008

  42. Heat (enthalpy) of reaction • ΔHoR –ve (exothermic reaction) • ΔHoR +ve (endothermic reaction) Paul Ashall, 2008

  43. ENTHALPHY BALANCE - REACTOR Qp = Hproducts – Hreactants + Qr Qp – heat transferred to or from process Qr – reaction heat (ζ ΔHoR), where ζ is extent of reaction and is equal to [moles component,i, out – moles component i, in]/ νi Paul Ashall, 2008

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

  45. 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 Paul Ashall, 2008

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

  47. EXAMPLES • Reactor • Crystalliser • Drier • Distillation

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

  49. 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

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