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Papessa [141]
3 years ago
5

Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new

fire station be located such that the maximum rectilinear travel distance is minimized? Assume that the total value of the homes in each sub-division is equal. The centroid locations (in miles) are as follows:
Subdivsion x-coordinate y-coordinate Total Vaue (in millions)
A 20 15 50
B 25 25 120
C 13 32 100
D 25 14 250
E 4 21 300
F 18 8 75
Engineering
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))     where i=1 to N-1 and N=6

C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))

A area of the polygon can be found by the following formulaA=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i}) where i=1 to N-1

A=\frac{1}{2}[ (x_{1}  y_{2} -x_{2}  y_{1})+ (x_{2}  y_{3} -x_{3}  y_{2})+(x_{3}  y_{4} -x_{4}  y_{3})+(x_{4}  y_{5} -x_{5}  y_{4})+(x_{5}  y_{6} -x_{6}  y_{5})]

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(x_{2}+x_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(x_{3}+x_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(x_{4}+x_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(x_{5}+x_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{x} =15.36

For Cy

C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(y_{2}+y_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(y_{3}+y_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(y_{4}+y_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(y_{5}+y_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{y} =22.55

So coordinates for the fire station should be (15.36,22.55)

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The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphe
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Answer:

a. The time required for the tank to empty halfway is presented as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

b. The time it takes for the tank to empty the remaining half is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time 't', is presented as follows;

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}

\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} =  \dfrac{v_2}{2 \cdot g}

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh

The time for the tank to drop halfway is given as follows;

\int\limits^{t_1}_0 {} \,  dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_1  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) =  { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   =   { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)The time required for the tank to empty halfway, t₁, is given as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

\int\limits^{t_2}_0 {} \,  dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_2  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time, t, to empty the tank is given as follows;

t = t_1 + t_2 =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} } =  \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

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