lecture notes, me 340 fall 2013 32Lesson 6 – Liquid-level systemsLet h(t)denote the liquid level at time t of a tank of cross-sectionalarea A. Suppose that the flow rate into the tank is qi(t)and that theflow rate out of the tank is qo(t). By the principle of conservation ofmassddt(ρAh)= ρqi− ρqo(147)since an increase in the amount of mass inside the tank must beaccompanied by a larger flow rate of mass per unit time into the tankthan out of the tank.Figure 37: A single-tank liquid-levelsystem.Moreover, assume that the mass flow rate out of the tank througha valve is proportional to the pressure drop across the valve:ρqo=∆pRo(148)where Rois the fluid resistance of the output valve. Then, if the pres-sure drop equals the difference between the ambient atmosphericpressure paand the pressure at the bottom of the tank pa+ ρgh, itfollows that ∆ p = ρgh and thusddt(ρAh)= (149)This can be written asdhdt(t)+1Th(t)= (150)where the equalsT = (151)If h(0)=h0, thenh(t)= (152)lecture notes, me 340 fall 2013 33If we consider qi(t) as the input and h(t) as the output, then forh(0)=0, the system is a convolution with transfer functionH(s)=T(153)Example 44. The impulse response of the single-tank liquid level systemconsidered above is the inverse Laplace transform of the transfer function,namelyL−1(H(s))(t)=1Ae−t/T(154)which also equals the free response (in the absence of input) given an initialcondition h(0)=1/A.Example 45. The step response of the single-tank liquid level system con-sidered above is the inverse Laplace transform of the functionTAs(sT + 1)=TA!1s−"(155)h(t)=TA#1 − e−tT$, t ≥ 0(156)In particular, h(t)→TA=Rogas t → ∞ corresponding to a steady-statelevel of the liquid in the tank as the amount of liquid flowing into the tankequals that flowing out of the tank.We replace the open inlet with a pump that raises the ambientpressure by δp(t), and an inlet valve with resistance Riconnected tothe bottom of the tank.Figure 38: A single-tank liquid-levelsystem with a pump.In this case,ddt(ρAh)= ρqi− ρqo(157)whereρqi=δp − ρghRi(158)andρqo= (159)lecture notes, me 340 fall 2013 34so thatdhdt(t)+1Th(t)= (160)where the time constant equalsT = (161)If h(0)=h0, thenh(t)= (162)Example 46. We consider the two-tank liquid-level system in Fig. 7.14 inthe course text. The first tank corresponds to an integrator with feedback asshown in Fig. 39 below.Figure 39: A block diagram for the firsttank.Here, the scaled pressure differential across the pump is the input and theliquid height in the tank is the output. The corresponding transfer functionis given byH1(s)=α1s + β1(163)The second tank corresponds to an integrator with feedback as shown inFig. 40 below.Figure 40: A block diagram for thesecond tank.Here, the scaled flow rate at the inlet is the input and the liquid height inthe tank is the output. The corresponding transfer function is given byH2(s)=α2s + β2(164)lecture notes, me 340 fall 2013 35We couple the two tanks as shown in the figure by scaling the output ofthe first tank in order to obtain the input to the second tank and feed thisback from the first block diagram to the second block diagram.Figure 41: A block diagram for thecoupled two-tank liquid-level system.Thus, if the pressure differential is the input and the liquid level in thesecond tank is the output, then the transfer function equalsγH1(s)H2(s)=γα1α21(165)and the corresponding impulse response is given byγα1α2e−β1t∗ e−β2t(166)This is the transfer function of a double integrator with zero initial condi-tions and feedback followed by (or preceded by) amplification.Can you get oscillations from thissystem?If the tank has a cross-sectional area that varies with the height,then it can no longer be considered constant and the system is nolonger linear! As an example, for a conical tank, as in Fig. 7.23 of thecourse text, A =