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See More xteamemail8892 Follow Advertisement See More Go explore. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.If you wish to opt out, please close your SlideShare account. Learn more. You can change your ad preferences anytime. Save so as not to loseStress Strain Modulus of ElasticitySolutions Manual for Engineering With Excel 4th Edition by Larsen. Full Download. Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.comV: 3000 litersThe cold water flowing in will tend t. If the effluent is at the top of the tank (as pictured), warm water will preferentially leave the tank, and the time to cooPiston hL Manometer Transducer hL Manometer Pressure. Setting Reading Output ReadingReadingCalibration DataPressure (atm). Transducer CalibrationPressurePower. Setting Thermometer Average Stdev PredictedThermocoupleData. PredictedTemp. ResistanceIt is an old RTD using the older grade of platinum.Diam. (cm) Circ. (cm)Exp. Value: 3.375. Percent Error: 7.4. Note: There was a conscious effort not to beDiameter (cm)As Budgeted. Actual. Research:As Budgeted Actual OverActualLoad (lbs) Deflection (mm)Linear trendline works, but does not fit data set. Logarithmic trendline does not fit data set. Power trendline fits data set well.The predicted deflection at 13400 lb load varies quite a bit with type of trendline used.Load (lbs)Full Download. 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Shed the societal and cultural narratives holding you back and let step-by-step Engineering with Excel textbook solutions reorient your old paradigms. NOW is the time to make today the first day of the rest of your life. Unlock your Engineering with Excel PDF (Profound Dynamic Fulfillment) today. YOU are the protagonist of your own life. Let Slader cultivate you that you are meant to be! Please reload the page. The site may not work properly if you don't update your browser. If you do not update your browser, we suggest you visit old reddit. Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts Log in sign up User account menu 1 Solutions Manual for Engineering With Excel 4th Edition by Larsen. Dynamics and Control of Robotic Systems offers If the ideal gas law were not valid, one would use an appropriate equation of state instead of Eq. 3. 2.6 a) Assumptions: 1. Each compartment is perfectly mixed. 2. ? and C are constant. 3. No heat losses to ambient. Two new equations: Component material balances on each compartment.Each controller provides an additional equation. Also, the flow out of the tank is now a manipulated variable being adjusted by the controller. So, we have 4 parameters: C, ?, Tsp, Vsp 6 variables: V, T, wi, Ti, Q, w 4 equations 2. Because V is constant, the mass balance for the tank is. Therefore, both Eqs. 1 and 3 will have to be solved simultaneously. Since concentration does not appear in Eq. 1, we would anticipate no effect on the h response.The new steady state would be 1 ft. (d) the inflow rate is returned to its original value for 16 minutes (24 2-22 The graph should show an exponential decrease to the previous steady state of 0.5 ft. The initial value should coincide with the final value from part (c). Putting all the graphs together would look like this: 2-23 2.18 Parameters (fixed by design process): m, C, me, Ce, he, Ae. CVs: T and Te. Input variables (disturbance): w, Ti. The magnitude of the effect is greater for changing V than for changing zF.Mathematica) to solve for y(t).A pulse can be described by the sum of two step functions.The solution is formulated as a function of t, f, A, and tw. The following script will solve the problem (note that only 4 of the 5 possible initial conditions on y and its derivatives are included, otherwise the problem is overspecified). K M ? s ?1 s From Table 3.1, the step response is: y(t ).Rewriting the above equations in terms of deviation variables, and noting that mC dTw dTw.P1 ) R1 1 w2 ? ( P1 ? P2 ) R2 1 w3 ? ( P2 ? Ph ) R3 w1 ? (5) (6) (7) Degrees of freedom:.Pc (t ) ? ( ? ) P1 ? P2 RT dt R1 R1 R2 R2 (10) Substitute (6) and (7) into (9): MV2 dP2 1 1 ? ( P1 ? P2 ) ? ( P2 ? Ph ) RT dt R2 R3 MV2 dP2 1 1 1 1. P1 ? ( ? ) P2 ? Ph (t ) RT dt R2 R2 R3 R3 (11) dP1.Note that 4.12 (a) First write the steady-state equations: 0.Because T'e ( s ) is the intermediate variable, remove it. Then rearranging gives:.The effect of Q' on T' can be derived by assuming that Ti is constant at its nominal steady-state value, Ti.Substituting (2) into (1) and simplifying gives: wt L dh. Substituting (4) into (3) yields a nonlinear dynamic model for the tank with qi and q as inputs: dh 1 ? (qi ? q ) dt 2 L ( D ? h)h To linearize this equation about the operating point (h.T '? s? K dT ?; mc p ? UA ?T ? Ta ? Because the gain is small and the time (c) Q.As shown in Fig. S6.1c, the system is stable but oscillations show up because of pure imaginary roots. 6.2 (a) Standard form: G.If ?1 ? K ?? ? ? 2 K1 ? ? ?? ? ? ? K1 K1? ? K 2 K1 ? 0, the zero is in right half plane. K1? ? K 2 Two possibilities: 1. K10 e) Gain is negative if K1 0. This is the only possibility. This approximate TF is exactly the same as would have been obtained using a plug flow assumption for the transfer line. Thus we conclude that investing a lot of effort into obtaining an accurate dynamic model for the transfer line is not worthwhile in this case. P1 P1 ? P2 ? ? RT1 dt Ra Rb (1) V2 M dP2 P1. Rb ) ?2 ? V2 M Rb Rc RT2 ( Rb. Rc ) Substituting for P1?( s ) from Eq. 3 into 4, 6-14 (3) (4). The first method requires solution of a system of 6-18 algebraic equations to obtain the coefficients of the four partial fractions. The residues (listed under r) are exactly the coefficients of the corresponding poles; in other words, the coefficients that would have been obtained via a manual partial fraction expansion.F0 c0 ? RF2 c2 ? F1c1 ? Vkc1 dt dc V 2 ? F1c1 ? ?1 ? R ? F2c2 ? Vkc2 dt Subtracting the steady-state equation and substituting deviation variables yields: dc ' V 1. F0c0' ? RF2c2' ? F1c1' ? Vkc1' dt dc ' V 2 ? F1c1' ? ?1 ? R ? F2c2' ? Vkc2' dt (b) The transfer function model can be derived based on Laplace transform: VsC1'.Mass balance for the tank is dV.Because TF ? T ? 0 (heating) and Ts. To simplify the transfer function gain, we can substitute UpT (Ts. T ) ? ? ?CqF (TF ? T ) h from the steady-state relation. Liquid level accumulates any changes in qF, increasing for positive changes and vice-versa.Liquid level is independent of Ts and steam pressure Ps. T ? q transfer function is second-order due to the interaction with liquid level; it is the product of an integrator and a first-order process. T ? qF transfer function is second-order due to the interaction with liquid level; it has numerator dynamics since qF affects T directly as well if TF ? T. T ? Ts transfer function is first-order because there is no interaction with liquid level.T ? qF: for TF ? T, T decreases initially (inverse response) and then increases. After long times, T increases like a ramp function. T ? q: T decreases, eventually at a constant rate. The single-tank process is described by the following equation in deviation variables: 6-29 dh' 1 '.Summing up: The process designer would like to have A1. However, the process response should be checked for typical changes in wi (t ) to make sure that h1 does not overflow. If it does, area A1 needs to be increased until it is not a problem. Note that ?2 ? ?3 when A1 ? A, thus a careful study (simulations) should be made before designing the partitioned tank. Otherwise, leave wellenough alone and use the non-partitioned tank. 6.23 The process transfer function is Y ( s) K. In fact, since 0.1 0. System II (air-to-close valve): as the signal to the control valve increases, the flow through the valve decreases. Kv 8-4 b) System I: Flow rate too high.For an air-to-close control valve, the controller output p should increase. Thus as h increases p decreases.For an air-to-close control valve, the controller output should decrease. Thus as h increases p decreases.For an air-to-open control valve, the controller output p should decrease. Thus as h increases p decreases.Configuration (b): As h increases, we want to increase q, the exit flow rate. For an air-to-open control valve, the controller output should increase. Thus as h increases p increases. Thus a steady state is reached with a value of y that is independent of the value of y sp. Use of these control algorithms is inadvisable if offset is a concern. As long as the sign on the error stays the same (i.e., if the measurement does not cross the set point), the integral component will continue to change monotonically. If the measurement crosses the set point, the error term will change sign and the integral component will begin to change in the other direction. Thus, it will no longer be monotonic. 8-11 8.12 a) False. The controller output could saturate or the controller could be in the manual mode. b) False. Even with integral control action, offset can occur if the controller output saturates. Or the controller could be in the manual mode. 8.13 First consider qualitatively how h2 responds to a change in q2. From physical considerations, it is clear that if q2 increases, h2 will increase. Thus, if h2 is increasing, we want q2 to decrease, and vice versa. Since the q2 control valve is airto-open, the level controller output p should decrease in order to have q2 decrease. In summary, if h2 increases we want p to decrease; thus a reverse-acting controller is required. 8.14 First consider qualitatively how solute mass fraction x responds to a change in steam flow rate, S. From physical considerations, it is clear that if S increases, x will also increase. Thus, if x is increasing, we want S to decrease, and vice versa. For a failopen (air-to-close) control valve, the controller output p should increase in order to have S decrease. In summary, if x increases we want S to decrease, which requires an increase in controller output p; thus a direct-acting controller is required. 8.15 First consider qualitatively how exit temperature Th2 responds to a change in cooling water flow rate, wc. From physical considerations, it is clear that if wc decreases, Th2 will increase. Thus, if Th2 is decreasing, we want wc to decrease, and vice versa. But in order to specify the controller action, we need to know if the control valve is fail open or fail close. Based on safety considerations, the control valve should be fail open (air-to-close). Otherwise, the very hot liquid stream could become even hotter and cause problems (e.g., burst the pipe or generate a two phase flow). For an air-to-close control valve, the temperature controller output p should increase in order to have wc decrease. The only way to decrease x is to increase w2 (but x can never be less than x2). Therefore, w2 should be increased when x increases, in order to have x decrease. If the control valve is fail open (air-to-close), then the composition controller output signal p should decrease. Thus a reverse-acting controller should be selected. Conversely, for a fail close (airto-open) control valve, a direct-acting controller should be used. Setting valve 3 as fail open prevents pressure build up in the vessel. Valve 4 should be failopen to evacuate the system and help keep pressure low. Valve 5 should be fail-close to prevent any additional pressure build-up. Setting valve 2 as fail open (air-to-close) will allow the steam chest to be evacuated, setting valve 3 as fail close (air-to-open) prevents vapor from escaping the vessel. Setting valve 4 as fail open (air-to-close) allows liquid to leave, preventing vapor build up. Setting valve 4 as fail close (air-to-open) prevents pressure buildup.Setting valve 2 as fail open will allow the steam chest to be evacuated. Setting valve 3 as fail open prevents pressure buildup in drum. Setting valve 4 as fail close prevents liquid from escaping. Setting valve 5 as fail close prevents liquid build-up in drum 9-2 9.3 Note: This exercise is best understood after the material in Ch. 11 has been considered.Chapter 11), closed-loop stability could be adversely affected.Thus, this change will not affect closed-loop stability. Because the performance of the closed-loop system depends on the gains of each individual element (cf. Chapter 11), closed-loop stability could be adversely affected d) For this process, changing the feed flow rate could affect both its steady-state gain and its dynamic characteristics (e.g., time constant and time delay). Because the performance of the closed-loop system depends on the gains of each individual element (cf.The selection of air-to-close vs.For the three controllers: i. Concentration controller: As the product concentration xB increases, we want the steam pressure, Ps to increase. Thus, the controller should be direct-acting. 9-6 ii. Level controller: As the liquid level h increases, we want the product flow rate B to increase. Thus, the controller should be direct-acting. iii. Pressure controller: As the pressure P increases, we want the solvent flow rate D to increase. Thus, the controller should be reverse-acting. 9. 7 Because the system dynamic behavior would be described using deviation variables, the dynamic characteristic can be analyzed by considering that the input terms (not involving x) can be considered to be constant, and thus deviations are zero. The starting form is the linear homogeneous ODE: M d 2x dx. Thus, the pressure drop will be very large and flow control will be ineffective. Configuration II: This parallel configuration will be effective because the large control valve can be adjusted to provide the nominal flow rate, while the small control valve can be used to regulate the flow rate. Then, Tm? ( s ) 1 ? T ?( s) ( s ? 1)(0.1s ? 1) where T is the temperature being measured and Tm is the measured value. Therefore, the long detection time would not be acceptable if the hazardous gas is CO. 10.2 (a) Start with a mass balance on the tank. These concerns can be addressed by the following instrumentation. 1. Leak detection: sensors for hazardous gases should be located in the vicinity of the flash drum. 2. Over pressurization: Use a high pressure switch (PSH) to shut off the feed when a high pressure occurs. 3. Liquid inventory: Use a low level switch (LSL) to shut down the pump if a low level occurs. 4. Liquid entrainment: Use a high level alarm to shut off the feed if the liquid level becomes too high. This SIS system is shown in Fig. S10.3 with conventional control loops for pressure and liquid level. But if the pressure in the column exceeds a specified limit, the high pressure switch (PSH) activates an alarm (PAH) and causes the valve to open fully, thus reducing the pressure in the tank. 10.5 Define k as the number of sensors that are working properly. Then P2 and the related reliability, R2, can be calculated as (cf. Then P3 and the related reliability, R3, can be calculated as (cf.Hence Gc(s) does need to be changed, and retuning is required. ii. Km changes. The close loop transfer functions changes, hence Gc(s) needs to be adjusted to compensate for changes in Gm and Km. The PI controller should be retuned. iii. Km remains unchanged when zero is adjusted. The controller does not need to be retuned. Normally proportional control does not eliminate offset, but it does for this integrating process. (d) Using Eq. 5-50 or taking the inverse Laplace transform of the response given above we get y.K c Assume that the level transmitter and the control valve have negligible dynamics. Then, Gm ( s) ? K m Gv ( s ) ? K v 11-8 The block diagram for this control system is the same as in Fig.11.8. Hence Eqs. 11-26 and 11-29 can be used for closed-loop responses to setpoint and load changes, respectively. The transfer functions G p (s ) and G d (s ) are as given in Eqs. 11-66 and 11-67, respectively.As is evident from the block diagrams above, the feedback loop contains, in addition to Gc, only a first-order process in part a), but a second-orderplus-time-delay process in part b). G3 ( D ? Y1 ) ? G2 K c E Y ? G3 D ? G3 G1 K c E ? G2 K c E Y ? G3 D ? (G3G1 K c ? G2 K c ) E E ? ? K mY Y ? G3 D ? K c (G3G1 ? G2 ) K mY G3 Y ? D 1 ? K c (G3G1 ? G2 ) K m b) Characteristic Equation: 1.Thus Gm(s) changes and the characteristic equation is affected. Stability limits would be expected to change.Hence Gm(s) remains unchanged and stability limits do not change.Gc G p Gv G m ? 1 ? K c s 3 ? 3s 2 ? 3s ? 1 ? 5K c ? aK c s 5 ? as ? ( s ? 1) 3 ( s ? 1) 3 s 3 ? 3s 2 ? (3 ? aK c ) s ? 1 ? 5K c s ? ( s ? 1) 3 A necessary condition for stability is all the coefficients of the numerator are positive. K c (1 ? 1 ) is substituted into 1. GmG p GcGv 4s and we get: 1. G m G p Gc Gv ? 1 ? 2 1 ? 4(1 ? ) (4s ? 1)(s ? 1) 4s 2s( s ? 1) ? Kc 2s 2 ? 2s ? Kc ? 2s( s ? 1) 2s( s ? 1) To find the stability region, the roots of the numerator polynomial should be on the right half plane. Hint: You do not need to obtain the analytical response y(t) to answer the above questions. Use the standard second order model expressed in terms of. Kc E Figure S11.26a Step response to unit step change with proportional control. As evidenced by the sketch, there is offset for this controller.The controller setting for (a) and (d) are essentially identical. The characteristic equation is: 1. The PID controller allows the controlled variable to reach the new set point more quickly than the PI controller, due to its larger Kc value. This large Kc allows an initially larger response from the controller during times from 1 to 4 minutes.Gv G p G m ? then G? 2.6 ? 10 ?4 187.5 32e ? s 0.0833s ? 1 1 ? 4.71s 1.56e ? s (4.71s ? 1)(0.0833s ? 1) For process with a dominant time constant. It is well known that the presence of a large time delay in a feedback control loop limits its performance.Hence the controller does not require retuning.By using Simulink, more accurate values can be obtained by trial and error. That is, operate temporarily in an open loop mode by switching the controller to the manual mode. This change provides a constant controller output and a constant manipulated input. If oscillations persist, they must be due to external disturbances. If the oscillations vanish, they were caused by the feedback loop. 12.15 The sight glass has confirmed that the liquid level is rising. Because the controller output is saturated, the controller is working fine. Hence, either the feed flow is higher than recorded, or the liquid flow is lower than recorded, or both. Because the flow transmitters consist of orifice plates and differential pressure transmitters, a plugged orifice plate could lead to a higher recorded flow.Thus, the Simulink diagram must be modified by copying the RAT loop for xB and adding it to the xD portion of the diagram. Also, the parameters for the relay block must be changed. Clearly, it is easier to control a liquid level by manipulating a large exit stream, rather than a small stream. The bypass fraction f has a dynamic effect on x4 but has no steady-state effect because it also does not change the relative amounts of materials that are blended. Thus, w2 is the best choice. 13.3 Both the steady-state and dynamic behaviors need to be considered. From a steady-state perspective, the reflux stream temperature TR would be a poor choice because it is insensitive to changes in xD, due to the small nominal value of 5 ppm. For example, even a 100 change in from 5 to 10 ppm would result in a negligible change in TR. Similarly, the temperature of the top tray would be a poor choice. An intermediate tray temperature would be more sensitive to changes in the tray composition but may not be representative of xD. Ideally, the tray location should be selected to be the highest tray in the column that still has the desired degree of sensitivity to composition changes. The choice of an intermediate tray temperature offers the advantage of early detection of feed disturbances and disturbances that originate in the stripping (bottom) section of the column. However, it would be slow to respond to disturbances originating in the condenser or in the reflux drum. Consequently, pressure control is easier when the liquid level is low and more difficult when the level is high. By contrast, for the conventional process design in Fig. 13.2, the liquid level has a very small effect on the pressure control loop. Thus, the flooded condenser is more difficult to control because the level and pressure control loops are more interacting, than they are for the conventional process design in Fig. 13.2. 13.5 (a) The larger the tank, the more effective it will be in “damping out” disturbances in the reactor exit stream. A large tank capacity also provides a large feed inventory for the distillation column, which is desirable for periods where the reactor is shut down. Skryť