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navik [9.2K]
3 years ago
7

Electricity is generated in two forms namely………A. Alternating current and wave form B. Alternating current and basic current C.

Alternating current and direct current D. Direct current and indirect current E. Direct current and simple current 20. Which of the following liquid substance is used in electrolysis? A. fluid B. Flux C. Solder D. Solute E. Electrolyte pleaseee heeeelp
Engineering
2 answers:
Eva8 [605]3 years ago
5 0

Answer:

The correct options are;

C. Alternating current and direct current

E. Electrolyte

Explanation:

Electricity is generated in either AC or DC depending on the source of the energy which aids the electricity generation;

The most common method through which electricity is generated and through which the principle of electric generation was discovered by Michael Faraday, is the rotation of a wire loop or coil between the North and South poles of a magnet to produce Alternating Current

Electricity is also generated in Direct Current by solar cells which convert sunlight into electric energy by photovoltaic effect.

20. Electrolysis is a chemical reaction that is brought about by the application or passing of a direct electric current through an electrolyte, which is an ion containing liquid.

natulia [17]3 years ago
4 0
C and E

Hope this helps
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Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

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Harmonic excitation of motion is represent as
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