System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 7, Problem 7.32P
To determine

(a)

The pump resistance and the steady-state height.

Expert Solution
Check Mark

Answer to Problem 7.32P

The pump resistance is 600N.s/(kg.m2) and the steady-state height is 1.223m.

Explanation of Solution

Calculation:

The figure below shows the pump curve and the line for steady-state flow.

System Dynamics, Chapter 7, Problem 7.32P

Figure-(1)

Write the expression of the two resistances in series.

qm2=(1R1+R2)Δp.... (I)

Here, the outlet flow rate of the tank is qm2, pump resistance is R1, the resistance of the outlet pipe is R2 and the pressure rise in the pump is Δp.

Write the expression of mass conservation.

qm1qm2=(Δpρgh)R1(ρghR2).... (II)

Here, the outlet flow rate of the tank is qm2, the inlet flow rate of the tank is qm1, pump resistance is R1, the resistance of outlet pipe is R2, the density of the fluid is ρ, the acceleration due to gravity is g, the height of liquid in the tank is h and the pressure rise in the pump is Δp.

Write the expression of the height of the tank.

h=Δpρg(R2R1+R1).... (III)

Here, the pump resistance is R1, the resistance of outlet pipe is R2, the density of the fluid is ρ, the acceleration due to gravity is g, the height of liquid in the tank is h and the pressure rise in the pump is Δp.

Refer to Figure-(1) to obtain the pressure rise in the pump as 3×104N/m2 and outer flow rate as 30kg/sec.

Substitute 30kg/sec for qm2

3×104N/m2 for Δp and 400l/(m.s) for R2 in Equation (I).

(30kg/sec)=1R1+(400l/(m.sec))×(3×104N/m2)(30kg/sec)×(R1+400l/(m.sec))=(3×104N/m2)[R1+400l/(m.sec)]=(3×104N/m2)(30kg/sec)R1=(3×104N/m2)(30kg/sec)400l/(m.sec)

R1=(600N.s/(kg.m2))

Thus, the pump resistance is 600N.s/(kg.m2).

Substitute 0 for (qm1qm2) in Equation (II).

ΔpρghR1ρghR2=0ΔpR1ρghR1ρghR2=0ΔpR1ρgh[1R1+1R2]=0ρgh[1R1+1R2]=ΔpR1

ρgh[R1+R2R2]=Δp

Substitute 3×104N/m2 for Δp, (600N.s/(kg.m2)) for R1, (400N.s/(kg.m2)) for R2, 1000kg/m3 for ρ and 9.81m/sec2 for g in Equation (III).

h=(3×104N/m2(1000kg/m3)×(9.81m/sec2))((400N.s/(kg.m2))(600N.s/(kg.m2))+(400N.s/(kg.m2)))=(3.0581m)(0.4)=1.223m

Thus, the steady-state height is 1.223m.

To determine

(b)

The linearized dynamic model of the height deviation in the tank.

Expert Solution
Check Mark

Answer to Problem 7.32P

The linearized dynamic model of the height deviation in the tank is Addtδh=0.0595×δh.

Explanation of Solution

Calculation:

Write the expression of the linear model.

δqm1=1rδ(Δp).... (IV) Here, the small change in flow rate is δqm1, the slope of the line is (1r) and the small change in pressure is δ(Δp).

Write the expression of mass conservation.

ρgh[R1+R2R2]=Δp.... (V)

Here, the pump resistance is R1, the resistance of outlet pipe is R2, the density of the fluid is ρ, the acceleration due to gravity is g, the height of liquid in the tank is h and the pressure rise in the pump is Δp.

Write the expression of small deviation in mass flow rates of the mass conservation.

ρAddtδh=δqm1δqm2.... (VI)

Here, the area of the tank is A, the density of the fluid is ρ, the deviation in height is δh and the small change in flow rate is δqm1 and δqm2.

Write the expression of the outlet flow rate of the tank.

qm2=ρghR2.... (VII) Here, the outlet flow rate of the tank is δqm2. the resistance of outlet pipe is R2, the density of the fluid is ρ, the acceleration due to gravity is g and the height of liquid in the tank is h.

Write the expression of the deviation.

ρAddtδh=(1r)(R1+R2R2ρgδh)ρgR2δhAddtδh=(1r(R1+R2R2)+1R2)gδh.... (VIII).

Here, the area of the tank is A, the pump resistance is R1, the resistance of outlet pipe is R2, the slope is 1r, the acceleration due to gravity is g, the time interval is t and the deviation in height is δh.

Substitute R1+R2R2ρgδh for δ(Δp) in Equation (IV).

δqm1=(1r)(R1+R2R2ρgδh)

Substitute (1r)(R1+R2R2ρgδh) for δqm1 and ρgR2δh for δqm2 in Equation (VI).

ρAddtδh=(1r)(R1+R2R2ρgδh)ρgR2δhAddtδh=(1r(R1+R2R2)+1R2)gδh

Substitute (400N.s/(kg.m2)) for R2, (600N.s/(kg.m2)) for R1, 700 for r and 9.81m/sec2 for g in Equation (VIII).

Addtδh=[(1700)((600N.s/(kg.m2))+(400N.s/(kg.m2))(400N.s/(kg.m2)))+(1(400N.s/(kg.m2)))]×(9.81m/sec2)×δh=(0.00357+0.0025)×(9.81m/sec2)×δh=0.595×δh

Thus, the linearized dynamic model of the height deviation in the tank is Addtδh=0.0595×δh.

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Chapter 7 Solutions

System Dynamics

Ch. 7 - 7.11 Derive the expression for the capacitance of...Ch. 7 - Air flows in a certain cylindrical pipe 1 m long...Ch. 7 - Derive the expression for the linearized...Ch. 7 - Consider the cylindrical container treated in...Ch. 7 - A certain tank has a bottom area A = 20 m2. The...Ch. 7 - A certain tank has a circular bottom area A = 20...Ch. 7 - The water inflow rate to a certain tank was kept...Ch. 7 - Prob. 7.18PCh. 7 - Prob. 7.19PCh. 7 - In the liquid level system shown in Figure P7.20,...Ch. 7 - The water height in a certain tank was measured at...Ch. 7 - Derive the model for the system shown in Figure...Ch. 7 - (a) Develop a model of the two liquid heights in...Ch. 7 - Prob. 7.24PCh. 7 - Design a piston-type damper using an oil with a...Ch. 7 - Prob. 7.26PCh. 7 - 7.27 An electric motor is sometimes used to move...Ch. 7 - Prob. 7.28PCh. 7 - Prob. 7.29PCh. 7 - Figure P7.3O shows an example of a hydraulic...Ch. 7 - Prob. 7.31PCh. 7 - Prob. 7.32PCh. 7 - Prob. 7.33PCh. 7 - Prob. 7.34PCh. 7 - Prob. 7.35PCh. 7 - Prob. 7.36PCh. 7 - Prob. 7.37PCh. 7 - (a) Determine the capacitance of a spherical tank...Ch. 7 - Obtain the dynamic model of the liquid height It...Ch. 7 - Prob. 7.40PCh. 7 - Prob. 7.41PCh. 7 - Prob. 7.42PCh. 7 - Prob. 7.43PCh. 7 - Prob. 7.44PCh. 7 - Prob. 7.45PCh. 7 - The copper shaft shown in Figure P7.46 consists of...Ch. 7 - A certain radiator wall is made of copper with a...Ch. 7 - A particular house wall consists of three layers...Ch. 7 - A certain wall section is composed of a 12 in. by...Ch. 7 - Prob. 7.50PCh. 7 - Prob. 7.51PCh. 7 - A steel tank filled with water has a volume of...Ch. 7 - Prob. 7.53PCh. 7 - Prob. 7.54PCh. 7 - Prob. 7.55P
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