Solucionario chapra primer capitulo 5ta edicion

Páginas: 8 (1760 palabras) Publicado: 27 de febrero de 2012
CHAPTER 1

1.1 For body weight:

[pic]

TW = 1.5%

For total body water:

[pic]

IW = 55%

1.2
[pic]

[pic]
[pic]

Therefore, the final temperature is 20 + 10.50615 = 30.50615oC.

1.3 This is a transient computation. For the period from ending June 1:

Balance = Previous Balance + Deposits – Withdrawals

Balance =1512.33 + 220.13 – 327.26 = 1405.20

The balances for the remainder of the periods can be computed in a similar fashion as tabulated below:

|Date |Deposit |Withdrawal |Balance |
|1-May | | | \$ 1512.33 |
| | \$ 220.13 | \$ 327.26 | |
|1-Jun | | | \$1405.20 |
| | \$ 216.80 | \$ 378.61 | |
|1-Jul | | | \$ 1243.39 |
| | \$ 450.25 | \$ 106.80 | |
|1-Aug | | | \$ 1586.84 |
| | \$ 127.31 | \$ 350.61 | |
|1-Sep | | |\$ 1363.54 |

1.4 [pic]

[pic]

1.5 [pic]

[pic]

[pic]

[pic]

1.6 [pic]

jumper #1: [pic]

jumper #2: [pic]

[pic]

[pic]

[pic]

[pic]

1.7 You are given the following differential equation with the initial condition, v(t = 0) = v(0),

[pic]

The most efficient way to solve this iswith Laplace transforms

[pic]

Solve algebraically for the transformed velocity

[pic] (1)

The second term on the right of the equal sign can be expanded with partial fractions

[pic]

Combining the right-hand side gives

[pic]

By equating like terms in the numerator, the following must hold

[pic]

The firstequation can be solved for A = mg/c. According to the second equation, B = –A. Therefore, the partial fraction expansion is

[pic]

This can be substituted into Eq. 1 to give

[pic]

Taking inverse Laplace transforms yields

[pic]

or collecting terms

[pic]

The first part is the general solution and the second part is the particularsolution for the constant forcing function due to gravity.

1.8 At t = 10 s, the analytical solution is 44.87 (Example 1.1). The relative error can be calculated with

[pic]

The numerical results are:

|step |v(10) |absolute |
| | |relative error |
|2 |47.9690 |6.90% |
|1|46.3639 |3.32% |
|0.5 |45.6044 |1.63% |

The error versus step size can then be plotted as

[pic]

Thus, halving the step size approximately halves the error.

1.9 (a) You are given the following differential equation with the initial condition, v(t = 0) = 0,

[pic]

Multiply both sides by m/c([pic]

Define [pic]

[pic]

Integrate by separation of variables,

[pic]

A table of integrals can be consulted to find that

[pic]

Therefore, the integration yields

[pic]

If v = 0 at t = 0, then because tanh–1(0) = 0, the constant of integration C = 0 and the solution is

[pic]

This result can thenbe rearranged to yield

[pic]

(b) Using Euler’s method, the first two steps can be computed as

[pic]

[pic]

The computation can be continued and the results summarized and plotted as:

|t |v |dv/dt |
|0 |0 |9.8 |
|2 |19.6 |8.53075 |
|4 |36.6615...

Regístrate para leer el documento completo.