Thermodynamics revisited

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Justin Burke
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1 Thermodynamics revisited How can I do an energy balance for a reactor system? 1 st law of thermodynamics (differential form): de de = = dq dq--dw dw Energy: de = du + de kin + de pot + de other du = Work: dw = d(pv) + dw s flow work other work Definition of Enthalpy: dh = du + d(pv) Energy balance: internal energy dh dh = = flow work du = L13-1 The The energy energy balance balance for for a a reactor reactor typically typically involves involves the the enthalpy enthalpyand and heat heat flows flows in in and and out out of of the the reactor reactor as as well well as as any any other other work work done done to to the the system. system.

2 Energy Balances Energy balances completely equivalent to mass balances! input New! heat exchange accumulation & production (or consumption) New! shaft work output Input Output + Production - Exchange = Accumulation Input: Output: Production: Q Q in = i out = Q R = j i Exchange: W + Q = W + U A ( T T ) S ex s ext ( Steady state: = 0!! ) L13-2

3 Energy Balance, cont d L13-3 Energy balance so far: Σ F io C p,im T in - Σ F i C p,im T out + Σ ΔΗ R r V - U A (T T ext ) +W s all terms (need to) have dimensions [energy/time] heat transfer coefficient U: [U] = [energy/area/temperature/time] = dh/dt = 0 (steady state) Let s simplify the first two terms (i.e. the heat flow in & out of the system): 1. Let s rewrite in terms of mass based rather than molar quantities: (with m i denoting a mass flow [mass/time]!) 2. Let s further assume that the heat capacity of the reaction mixture is only very weakly dependent on T and on composition (i.e. C p,i C p ) 3. However, mass is conserved in a chemical reaction (unlike moles!!), hence:. m total,in = m total,out =,... Q in Q out = therefore:

4 L13-4 Energy Balance, simplified So, the general but simplified! form for the energy balance becomes: or: (assumption: C p independent of T and composition!)

5 Energy Balance: CSTR Q Accumulation = Input - Output + Production - Steady state, W s = 0: T T 0 = Need to replace r = f(c A,T) )! Combine with mass balance: C A0 C A = Exchange dh dt = ρcv = V ρc ( T T) + ( ΔH ) Vr UAT ( T ) W dt dt p p 0 R ext S (C p, V = const, single reaction) Two coupled algebraic equations, which need to be solved simultaneously! L13-5

6 Example You want to produce propylene glycol through the hydrolysis of propylene oxide: CH 2 -CH-CH 3 + H 2 O -> CH 2 -CH-CH 3 \ / O OH OH You have a 300 gal. CSTR available, your reactor feed is 2500 lb/h PO in a solvent stream of 2300 lb/h methanol, mixed with lb/h of water. The feed streams are at 58F, but feed temperature rises to 75F before entering the reactor due to the heat of mixing. The reaction kinetics has been measured and found to be first order in PO (zero order in water due to the strong excess of water), with k 0 = h -1 and E A /R = o R. You have to maintain one constraint on your reaction system: if the operating temperature exceeds 125F, you will loose to much PO (b.p. at 1 atm: 94F) by vaporisation through the reactor vent system. What will the conversion in your CSTR be - and should you even use this reactor at all? L13-6

7 Example, cont d L13-7 I) CSTR design equation: II) rate law: III) stoichiometry: IV) Solving for X: V) Energy Balance: divide by V 0 : Can be solved by math program (Matlab( or similar), by simple iteration (not always successful) or graphically! V = -r A = C A = X = V 0 ρ C p (T 0 -T) ΔH R V r = 0 ρ C p (T 0 -T) ΔH R τ r = ρ C p (T 0 -T) ΔH R C A0 X = 0 X =

8 Example, cont d L13-8 Values: τ = V/V 0 = 40 ft 3 /( ft 3 /h) = 0.12 h ρ = m tot /V 0 = ( lb/h)/( ft 3 /h) = 59 lb/ft 3 C p (Btu/lb F): CH 3 OH: 0.61, PO: 0.79, H 2 O: 1.0 C p = Σ m i C pi / m tot = ( )/19300 Btu/lb F = MB: X = τ k / (1 + τ k) EB: = 0.93 Btu/lb F (compare this to 1.0 Btu/lb F for water!) ΔH R = H f,pg (T in ) H f,po (T in ) H f,w (T in ) = Btu/(lb mol) C A0 = F A0 /V = (lb mol)/ft 3 X = exp{ /(T/ o R)} / [ exp{ /(T/ o R)}] = ρ C ( T T) p R 0 ΔH C A0 = ( Btu / ft / F)(( ) R T ) ( Btu / ft ) 54.87(1/ F)(535 R T) = = (535 T / R) 4805

9 L Example, cont d X X PO + H2O -> PG T / R X

10 Example, still cont d Solve by iteration: i T(i) T(i+1) delt/t [%] L

11 Example, yet one more L Solve the the previous example with with Matlab! + quick, easy, convenient + always works! -- You You need to to know Matlab For more complex systems, the last solution is often the only one e that works. (Graphical solution also always works, but can be painfully complex and doesn t give high accuracy solution. Simple iteration very often fails s to converge!)

12 Energy Balance: BR Q Accumulation = Input - Output + Production - dh dt With W s = 0: Combine with mass balance: dx A /dt= Exchange dt = ρ C V = ( ΔH ) V r U A ( T T ) W dt p R ext S (C p, V = const, single reaction) Two coupled differential equations, to be solved simultaneously! L

13 Energy Balance: PFR z dv Q z + dz differentiation: const. C p, const. density -> V = V 0 with dv = A dz and V 0 /A = u : energy balance, PFR (Steady State, no external work): 0 = Input - Output + Production - Exchange 0 V ρ cp T V ρ cp T + ( ΔH R) r dv daextu ( T Text) = z z+ dz ( ρ p ) d V C T dv da 0 = + r ( ΔHR ) U( T Text ) dz dz dz where a V is the specific surface area: a V = da ext /dv = A ext /V What s a V for a tubular plug-flow reactor? ext L13-13 mass balance:

14 Isothermal Heat Load A main consideration in reactor design concerns the question, how to control the reactor temperature, i.e. how much heat to withdraw from an exothermal reaction or to supply to an endothermal reaction to keep the reactor temperature within a pre-set limit. In an ideal case, this would mean keeping the reaction (and hence the reactor) perfectly isothermal this is called the isothermal heat load. Let s do an example: Q = U A ( T T ) = Q ext reaction You need to run a first order liquid phase reaction (k(300k) = 0.05 min -1 ) isothermally at 300K. The heat of reaction is -20 kj/mol, the reactor has a volume of 10l and is fed with 2 mol/l of the reactant. At what rate do you need to withdraw heat, if your reactor is (a) a BR, (b) a CSTR, (c) a PFR and your target conversion is 90%? (a) heat production in reactor: dq/dt = mass balance, BR 1 st order reaction: C A = hence heat load: dq/dt = L do do the the CSTR CSTR & PFR PFR yourself! yourself!

15 Isothermal Heat Load Example (b) CSTR: dq/dτ = (b) PFR: analogous to BR, hence: dq/dt = Note that the rate of heat removal from the CSTR is constant! This makes the CSTR once again the much simpler reactor to design and operate! L

CHE 404 Chemical Reaction Engineering. Chapter 8 Steady-State Nonisothermal Reactor Design

Textbook: Elements of Chemical Reaction Engineering, 4 th Edition 1 CHE 404 Chemical Reaction Engineering Chapter 8 Steady-State Nonisothermal Reactor Design Contents 2 PART 1. Steady-State Energy Balance