System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 7, Problem 7.53P
To determine

The thermal resistance R1 of the water in the tank.

Expert Solution & Answer
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Answer to Problem 7.53P

The thermal resistance R1 of the water in the tank is 2.11504×105sec°F/ft-lb.

Explanation of Solution

Calculation:

Write the expression for the rate of heat transfer in terms of total thermal resistance.

dQdt=1R1(T1T0) (I)

Here, the rate of heat transfer is dQdt, the total thermal resistance is R1, the tank s water temperature is T1, the temperature of the air surrounding the tank is T0 and the time is t.

Write the expression for the thermal capacitance.

C1=ρVc

Here, the thermal capacitance is C1, the mass density is ρ, the volume of the water in the tank is V and the specific heat of the water at the room temperature is c.

Write the expression for the rate of heat transfer in terms of thermal resistance.

dQdt=ρVcdT1dt (II)

Substitute ρVc for C1 in Equation (II).

dQdt=C1dT1dt (III)

Equate the values from the Equation (I) and Equation (III) for the conservation of heat energy.

C1dT1dt=1R1(T1T0) (IV)

Write the expression for the temperature.

θ1=T1T0 (V)

Differentiate the above expression.

dθ1dt=ddt(T1T0)dθ1dt=dT1dt

Substitute 1.94slug/ft3 for ρ, 1000ft3 for V and 25000ft-lb/slug-°F for c in Equation (I).

C1=(1.94slug/ft3)(1000ft3)(25000ft-lb/slug-°F)=(1940slug)(25000ft-lb/slug-°F)=4.85×107ft-lb/°F

Substitute 70°F for T1 in Equation (V).

θ1=T1(70°F)

Substitute 4.85×107ft-lb/°F for C1 and 70°F for T0 in Equation (IV).

(4.85×107ft-lb/°F)dT1dt=1R1[T1(70°F)]

Substitute dθ1dt for dT1dt and θ1 for T1(70°F) in above expression.

(4.85×107ft-lb/°F)dθ1dt=1R1[θ1](4.85×107ft-lb/°F)×dθ1dt=θ1R1dθ1θ1=dtR1×(4.85×107ft-lb/°F)

Integrate the above expression.

dθ1θ1=dtR1×(4.85×107ft-lb/°F)ln(θ1)=tR1×(4.85×107ft-lb/°F)+C

Substitute θ1 for T1(70°F) in above expression.

ln[T1(70°F)]=tR1×(4.85×107ft-lb/°F)+C (VI)

Apply boundary conditions.

At t=0sec ; T1=90°F.

Substitute 90°F for T1 at t=0sec in Equation (VI).

ln[(90°F)(70°F)]=0secR1×(4.85×107ft-lb/°F)+Cln(20°F)=0+CC=ln(20°F)

Substitute ln(20°F) for C in Equation (VI).

ln[T1(70°F)]=tR1×(4.85×107ft-lb/°F)+ln(20°F)ln[T1(70°F)]ln(20°F)=tR1×(4.85×107ft-lb/°F)ln[T1(70°F)(20°F)]=tR1×(4.85×107ft-lb/°F)R1=t(4.85×107ft-lb/°F)ln[(20°F)T1(70°F)] (VII)

Substitute 90°F for T1 and 0sec for t in Equation (VII).

R1=(0sec)(4.85×107ft-lb/°F)ln[(20°F)(90°F)(70°F)]=00

The above results are not defined.

Substitute 82°F for T1 and 500sec for t in Equation (VII).

R1=(500sec)(4.85×107ft-lb/°F)ln[(20°F)(82°F)(70°F)]=(2.018159×105)sec°F/ft-lb(2.01816×105)sec°F/ft-lb

Substitute 77°F for T1 and 1000sec for t in Equation (VII).

R1=(1000sec)(4.85×107ft-lb/°F)ln[(20°F)(77°F)(70°F)]=(1.964×105)sec°F/ft-lb

Substitute 75°F for T1 and 1500sec for t in Equation (VII).

R1=(1500sec)(4.85×107ft-lb/°F)ln[(20°F)(75°F)(70°F)]=(2.23097×105)sec°F/ft-lb

Substitute 73°F for T1 and 2000sec for t in Equation (VII).

R1=(2000sec)(4.85×107ft-lb/°F)ln[(20°F)(73°F)(70°F)]=(2.17367×105)sec°F/ft-lb

Substitute 72°F for T1 and 2500sec for t in Equation (VII).

R1=(2500sec)(4.85×107ft-lb/°F)ln[(20°F)(72°F)(70°F)]=(2.23863×105)sec°F/ft-lb

Substitute 71°F for T1 and 3000sec for t in Equation (VII).

R1=(3000sec)(4.85×107ft-lb/°F)ln[(20°F)(71°F)(70°F)]=(2.06479×105)sec°F/ft-lb

Substitute 70°F for T1 and 4000sec for t in Equation (VII).

R1=(4000sec)(4.85×107ft-lb/°F)ln[(20°F)(70°F)(70°F)]=(4000sec)(4.85×107ft-lb/°F)ln[0]

The above results are not defined.

The Table (1) shows the calculated values of thermal resistance R1.

S. No (i) t(sec) T1(°F) R1(sec°F/ft-lb)
1 0 90
2 500 82 2.01816×105
3 1000 77 1.964×105
4 1500 75 2.23097×105
5 2000 73 2.17367×105
6 2500 72 2.23863×105
7 3000 71 2.06479×105
8 4000 70

Table-(1)

Calculate the average of the thermal resistance from the above table.

(R1)avg=i=27(R1)i6=[{(2.01816×105sec°F/ft-lb)+(1.964×105sec°F/ft-lb)+(2.23097×105sec°F/ft-lb)+(2.17367×105sec°F/ft-lb)+(2.23863×105sec°F/ft-lb)+(2.06479×105sec°F/ft-lb)}6]=2.115036×105sec°F/ft-lb2.11504×105sec°F/ft-lb

Thus, the thermal resistance R1 of the water in the tank is 2.11504×105sec°F/ft-lb.

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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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