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3 years ago

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A 350 m^3 retention pond that holds rainwater from a shopping mall is empty at the beginning of a rainstorm. The flow rate out o

f the retention pond must be restricted to 320 L/min to prevent downstream flooding from a 6-hour storm. What is the maximum flow rate (in L/min) into the pond from a 6-hr storm that will not flood it? a. 5,860 L/min b. 321 L/min c. 1.290 L/min d. 7.750 L/min

1 answer:

White raven [17]3 years ago

6 0

Answer:

correct option is c. 1.290 L/min

Explanation:

given data

retention volume = 350 m³

flow rate out = 320 L/min

time = 6 hour

to find out

maximum flow rate (x)

solution

we apply here Mass rate of accumulation formula that is

Mass rate of accumulation = mass rate of input - mass rate of output ................1

350 × 1000 L/m³ = x × ( 6hr × 60min ) - ( 320 L × 6 hr × 60 min )

3.5 × 10³ + 4.65 × 10³ = x × 360

x = 1290 L/min

so correct option is c. 1.290 L/min

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Answer:

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Explanation:

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Solving this algebraically would be time-consuming, but we can use matrices to make it easy.

$\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]$

First, we find the inverse of the coefficient matrix. This is messy to do by hand, so let's use a calculator:

$\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]$

Now we multiply by the solution matrix (again using a calculator):

$-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]$

So the cubic is:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

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4 years ago

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Answer:

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Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

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b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

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Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

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$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

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And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

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3 years ago