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Inessa05 [86]
3 years ago
7

La base de los tema relacionados a las ciencia de las ingeniería es?

Engineering
1 answer:
Yanka [14]3 years ago
5 0

Answer:

La ciencia y la ingeniería conciben el mundo como comprensible, con reglas que gobiernan su funcionamiento y que a través de un estudio cuidadoso y sistemático se puede evidenciar mediante patrones consistentes que permitan la oportunidad de examinar las características fundamentales que mejor describen los fenómenos.

Explanation:

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Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz
Naddik [55]

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Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s       and ρ= 850 kg/m³  

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ =  ρv

   =850 x 0.00062

   = 0.527 kg/m·s  

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

Q = \dfrac{\Delta P \pi D^4}{128 \mu L}

Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}

Q = 5.06 x 10⁻⁸ m³/s

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3 years ago
When you arrive at an intersection with a stop sign in your direction, if there is no marked stop
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C: Stop before entering the pedestrian crosswalk.

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2 years ago
An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter
slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg ,  1980.39 MJ/Kmol

Explanation:

molar mass of  ethanol (C2H5OH) = 46 g/mol

molar mass of   octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol =  0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value  ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg

HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0
2 years ago
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