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Chapter 5: Q5.3-21E (page 260)

Fluid Ejection.In the design of a sewage treatment plant, the following equation arises: 60-H=(77.7)H''+(19.42)(H')2;H(0)=H'(0)=0where H is the level of the fluid in an ejection chamber, and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h = 0.5 to approximate H(t)over theinterval [0, 5].

Short Answer

Expert verified

The result can get by the Runge-Kutta method.

Step by step solution

01

Transform the equation

Here the equation is60-H=(77.7)H''+(19.42)(H')2;H(0)=H'(0)=0.

The system can be written as:

x1=H(t)x2=H'=x'1H''=x'2

The transform equation:

x'1=x2x'2=[60-x1-(19.42)x22]77.7

The initial conditions are:

x1(0)=H(0)=0x2(0)=H'(0)=0

02

Apply the Runge-Kutta method.

Here h=0.5, N=10 steps, x1,0=0,x2,0=0 then;

x1,0=0,x2,0=0k1,1=hx2,0=0.5(0)=0k2,1=h60-x1,0-(19.42)x22,077.7=0.38610k1,2=0.09653k2,2=0.38144k1,3=0.09536k2,3=0.38124k1,4=0.19062k2,4=0.36732

03

Apply the recursive formula

t1=t0+h0+0.5=0.5x1(0.5)=x1,1=0.09573x2(0.5)=x2,1=0.37980

Apply the same procedure for n=1, 2,…., 9.

N

tn

x1,n=H(tn)

0

0

0

1

0.5

0.09573

2

1.0

0.37389

3

1.5

0.81045

4

2.0

1.37361

5

2.5

2.03111

6

3.0

2.7549

7

3.5

3.52322

8

4.0

4.319

9

4.5

5.13307

10

5.0

5.95554

This is the required result.

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