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Anettt [7]
3 years ago
14

How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t

hat larger fire engine mounted pumps can deliver from 90-300 gallons per minute, and hoses can be as wide as 30 cm diameter What alternative estimate would that info give you?)
Engineering
1 answer:
Mrac [35]3 years ago
7 0

Answer:

z_{2} = 91.640\,m

Explanation:

The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.

\frac{P_{1}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}= z_{2}+\frac{P_{2}}{\rho\cdot g}

The velocity of water delivered by the fire hose is:

v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}

v_{1} = 0.267\,\frac{m}{s}

The maximum height is cleared in the Bernoulli's equation:

z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}

z_{2}= \frac{1\times 10^{6}\,Pa-101.325\times 10^{3}\,Pa}{(1000\,\frac{kg}{m^{3}} )\cdot(9.807\,\frac{m}{s^{2}} )} + \frac{(0.267\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}

z_{2} = 91.640\,m

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The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

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1 5segundos

Explanation:

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C

Explanation:

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(Energy ) i think .....

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Write the chemical equation of the polymerization reaction to synthesize PS
marin [14]

Explanation:

Styrene is  a vinyl monomer in which there is a carbon carbon double bond.

The polymerization of the styrene, which is initiated by using a free radical which reacts with the styrene and the compound thus forms react again and again to form polystyrene (PS).

The equation is shown below as:

\begin{matrix}& C_6H_5 \\&|\\n H_2C & =CH\end{matrix}        ⇒                \begin{matrix}&C_6H_5 \\&|\\ -[-H_2C & -CH-]-_n\end{matrix}

6 0
3 years ago
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