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2 years ago

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A 12-m3 oxygen tank is at 17°C and 850 kPa absolute. The valve is opened, and some oxygen is released until the pressure in the

tank drops to 650 kPa. Calculate the mass of oxygen that has been released from the tank if the temperature in the tank does not change during the process.

1 answer:

sweet-ann [11.9K]2 years ago

3 0

Answer:

Released oxygen mass: 15.92 kg

Step-by-step explanation:

ideal gas law : P*V=nRT

P:pressure

V:volume

T:temperature

n:number of moles of gas

n [mol] = m [g] /M [u]

m : masa

M: masa molar = 15,999 u (oxygen)

R: ideal gas constant = 8.314472 cm^3 *MPa/K*mol =

grados K = °C + 273.15

P1*V*M/R*T = m1

P2*V*M/R*T = m2

masa released : m1-m2 = (P1-P2) * V*M/R*T

m2-m1 = 200 * 10^-3 MPa * 12 * 10^6 cm^3 * 15.999 u / 8.314472 (cm^3 * MPa/K *mol) * 290. 15 K

m2-m1= 38 397.6 * 10^3 u*mol / 2412.44 = 15916.5 g = 15.9165 kg

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Example 4.5 introduced the concept of time headway in traffic flow and proposed a particular distribution for X 5 the headway be

exis [7]

Answer:

a. k = 3

b. Cumulative distribution function X, $F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when headway exceeds 2 seconds = 0.125

Probability when headway is between 2 and 3 seconds = 0.088

d. Mean value of headway = 1.5

Standard deviation of headway = 0.866

e. Probability that headway is within 1 standard deviation of the mean value = 0.9245

Step-by-step explanation:

From the information provided,

Let X be the time headway between two randomly selected consecutive cars (sec).

The known distribution of time headway is,

$f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.$

a. Value of k.

Since the distribution of X is a valid density function, the total area for density function is unity. That is,

$\int\limits^{\infty}_{-\infty} f(x)dx=1$

So, the equation becomes,

$\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3$

b. For this problem, the cumulative distribution function is defined as :

$F(x) = \int\limits^1_{\infty} f(x)dx + \int\limits^x_1 f(x)dx$

Now,

$F(x) = 0 + \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 + \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}$

Therefore the cumulative distribution function X is,

$F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when the headway exceeds 2 secs.

Using cdf in part b, the required probability is,

$P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125$

Probability when headway is between 2 seconds and 3 seconds

Using the cdf in part b, the required probability is,

$P(2$

≅ 0.088

d. Mean value of headway,

$E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5$

And,

$E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3$

The standard deviation of headway is,

$= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254$

≅ 0.866

e. Probability that headway is within 1 standard deviation of the mean value

$P(\alpha - \beta < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X$

From part b, F(x) = 0, if x ≤ 1

$=1-(2.366)^{-3}\\=0.9245$

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2 years ago

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(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

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$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
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$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

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