Solucionario chapra primer capitulo 5ta edicion

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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]

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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...
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