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Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Heated Tank 5.0 context and direction From Lesson 3 to Lesson 4, we increased the dynamic order of the process, introduced the Laplace transform and block diagram tools, took more account of equipment, and discovered how control can produce instability. Now we change the process: our system models have previously depended on material
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  Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Heated Tank 5.0   context and direction From Lesson 3 to Lesson 4, we increased the dynamic order of the  process, introduced the Laplace transform and block diagram tools, took more account of equipment, and discovered how control can produce instability. Now we change the process: our system models have  previously depended on material balances, but now we will write the energy balance. We will also introduce the integral mode of control in the algorithm. DYNAMIC SYSTEM BEHAVIOR 5.1   a heated tank We consider a tank that blends and heats two inlet streams. The heating medium is a condensing vapor at temperature T c  in a heat exchanger of surface area A. F 1  T 1 F 2  T 2 F T o T c V AF 1  T 1 F 2  T 2 F T o T c V A  For the present, we continue to assume constant mass in an overflow tank. Writing the material balance, FFF 21  ρ=ρ+ρ  (5.1-1) This is not yet the time for complications: we will approximate the  physical properties of the liquid (density, heat capacity, etc.) as constants. We will also simplify the problem by assuming that the flow rates remain constant in time. The energy balance is ( ) )TT(FC)TT(UA)TT(CF)TT(CF)TT(VC dtd ref o pocref 2 p2ref 1 p1ref o p  −ρ−−+−ρ+−ρ=−ρ (5.1-2) revised 2006 Mar 31 1  Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Heated Tank where the overall heat transfer coefficient is U and the thermodynamic  reference is T ref  . We identify a steady-state operating  reference condition with all variables at their desired values. ( ) )TT(FC)TT(UA)TT(CF)TT(CF0)TT(VC dtd ref or  por cr ref r 2 p2ref r 1 p1ref or  p  −ρ−−+−ρ+−ρ==−ρ  (5.1-3) We subtract (5.1-3) from (5.1-2), define deviation variables, and rearrange to standard form. 'c p'2 p p2'1 p p1'o'o p p TUAFCUATUAFCCFTUAFCCFTdtdTUAFCVC +ρ++ρρ++ρρ=++ρρ  (5.1-4) To make some sense of the equation coefficients, define the tank residence time FV R   =τ  (5.1-5) and a ratio of the capability for heat transfer to the capability for enthalpy removal by flow.  p FCUA ρ=β  (5.1-6) β  thus indicates the importance of heat transfer in the mixing of the fluids. We now use (5.1-5) and (5.1-6) to define the dynamic parameters: time constant and gains. β+τ=τ 1 R   (5.1-7) Thus the dynamic response of the tank temperature to disturbances is faster as heat transfer capability ( β ) becomes more significant. For no heat transfer ( β  = 0) the time constant is equal to the residence time. β+= 1FFK  11  (5.1-8) β+= 1FFK  22  (5.1-9) revised 2006 Mar 31 2  Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Heated Tank Gains K  1  and K  2  show the effects of inlet temperatures T 1  and T 2  on the outlet temperature. For example, a change in T 1  will have a small effect on T o  if the inlet flow rate F 1  is small compared to overall flow F. β+β= 1K  3  (5.1-10) Gain K  3  shows the effect of changes in the temperature T c  of the condensing vapor. For high heat transfer capability, β  is large, and gain K  3  approaches unity. Our system model of the heated tank is finally written 'c3'22'11'o'o TK TK TK T dtdT ++=+τ  (5.1-11) which shows a first-order system with three inputs. As is our custom, we will take the initial condition on T ′ o  as zero. 5.2   solving by laplace transform Solution by Laplace transform is straightforward: { }{ } { } { } { }( ) )s(T 1sK )s(T 1sK )s(T 1sK )s(T )s(TK )s(TK )s(TK )s(T)0(T)s(sT TLK TLK TLK TL dtdTLTK TK TK LT dtdTL 'c3'22'11'o'c3'22'11'o'o'o'c3'22'11'o'o'c3'22'11'o'o +τ++τ++τ=++=+−τ ++=+ ⎭⎬⎫⎩⎨⎧τ++=⎭⎬⎫⎩⎨⎧+τ  (5.2-1) Output T ′ o  is related to three inputs, each through a first-order transfer function. 5.3   response of system to step disturbance Suppose the tank is disturbed by a step change Δ T 1  in temperature T 1 . We have studied first-order systems, so we already know what the first-order step response looks like: at the time of disturbance t d , the output T ′ o  will depart from its initial zero value toward an ultimate value equal to the  product of gain K  1  and the step Δ T 1 . The time required depends on the magnitude of time constant τ : when time equal to one time constant has  passed (i.e., t = t d  + τ ) the outlet temperature will be about 63% of its way toward the long-term value. revised 2006 Mar 31 3  Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Heated Tank However, doing the formalities for two step disturbances and one steady input, ( ) ( ) 0TttUTTttUTT 'c2d2 '21d1 '1  =−Δ=−Δ=  (5.3-1) We take Laplace transforms of these input variables: 0)s(Te sT)s(Te sT)s(T 'cst2'2st1'1 2d1d =Δ=Δ=  −−  (5.3-2) Then we substitute (5.3-2) into the system model (5.2-1). 01sK esT1sK esT1sK )s(T 3st22st11'o 2d1d +τ+Δ+τ+Δ+τ=  −−  (5.3-3) We must invert each term; this is most easily done by processing the  polynomial first and then applying the time delay. Thus τ−− −=⎭⎬⎫⎩⎨⎧+τ t1 e1s11s1L (5.3-4) ( )  ⎟ ⎠ ⎞⎜⎝ ⎛ −−=⎭⎬⎫⎩⎨⎧⎟ ⎠ ⎞⎜⎝ ⎛ +τ τ−−−− )tt( 1dst1 1d1d e1ttUe s11s1L (5.3-5) and finally ( ) ( )  ⎟ ⎠ ⎞⎜⎝ ⎛ −−Δ+ ⎟ ⎠ ⎞⎜⎝ ⎛ −−Δ=  τ−−τ−− )tt( 2d22 )tt( 1d11 *o 2d1d e1ttUTK e1ttUTK T  (5.3-6) Figure 5.3-1 shows a sample calculation for opposing input disturbances.  Notice that the gains K  1  and K  2  are less than unity. This is because we dilute each disturbance with other streams: streams of matter (two input flows) and energy (through the heat transfer surface). From the plot, can you tell which gain is greater? revised 2006 Mar 31 4
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