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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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**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) • 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**Energy balance**W system mass out mass in Hin Hout Q Paul Ashall, 2008**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 + W 2) No accumulation of energy, no mass transfer (m1 = m2= m) ∆E = 0 Q = -W**3. No accumulation of energy but with mass flow**Q + 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 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.**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**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**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= ∑∆Hfo(products) - ∑∆Hfo(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/mol H2O(g): -214.826 kJ/mol**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.**aA + bB→ cC + dD**T3 T1 C A products reactants T2 D B T4**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.**∆E = 0, W = 0, ∆E = 0, ∆PE = 0, ∆KE = 0**Therefore Q = ∆H**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 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)**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**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**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)**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**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!**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)**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 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.]**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**Heat (enthalpy) of reaction**• ΔHoR –ve (exothermic reaction) • ΔHoR +ve (endothermic reaction) Paul Ashall, 2008**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**system**Hproducts Hreactants Qr +ve -ve Qp Note: enthalpy values must be calculated with reference to a temperature of 25 deg cent**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**ENERGY BALANCE TECHNIQUES**• Adiabatic temperature: Qp = 0 Paul Ashall, 2008**EXAMPLES**• Reactor • Crystalliser • Drier • Distillation**References**• The Properties of Gases and Liquids, R. Reid • Elementary Principles of Chemical Processes, R.M.Felder and R.W.Rousseau Paul Ashall, 2008**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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