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svetoff [14.1K]
4 years ago
9

A power of 100 kW (105 W) is delivered to the other side of a city by a pair of power lines, between which the voltage is 12,000

V. (a) Use the formula P 5 IV to show that the current in the lines is 8.3 A. (b) If each of the two lines has a resistance of 10 Ω, show that there is a 83-V change of voltage along each line.
Engineering
1 answer:
vivado [14]4 years ago
3 0

Answer:

I = 8.3 Amp

potential drop = 83 V

Explanation:

Power = 100 KW

V = 12,000 V

R = 10 ohms

a)

Calculate current I in each wire:

P = I*V

I = P / V

I = 100 / 12 = 8.333 A

b)

Calculate potential drop in each wire:

V = I*R

V = (8.3) * (10)

V = 83 V

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2 years ago
The in situ moisture content of a soil is 18% and the moist (total) unit weight is 105 pcf. The soil is to be excavated and tran
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Answer: Volume= 1.16 yd³

Explanation:

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First we find out the dry unit weight

Dry unit weight= Moist unit weight / (1+ moisture content)

Dry unit weight= \frac{105}{1.18}

Dry unit weight= 88.983 pcf

Now, to find the volume ( in terms of each cubic yard i.e 1 yd³ we have

Volume = Dry unit weight compacted / Dry unit weight (in-situ) * 1 yd³

Volume= \frac{103.5}{88.983} * 1

Volume= 1.16 yd³

6 0
4 years ago
The Acme threading tool forms an inc luded angle of how many degrees? A. 30 B. 55 C. 29 D. 60
Schach [20]

Answer:

(C) 29°

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4 years ago
For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh
Paul [167]

Answer:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere dp= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction \in =  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas \rho = 0.15 mol/dm ⁻³

viscosity of methane gas \mu = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  0.1 5 *\dfrac{16}{0.1^3}

Density =  2400

Density \rho_f =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

Re = \dfrac{dV \rho}{\mu}

Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}

Re=2276.317705

For Re > 1000

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\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}

\Delta P=8575.755212*2.5

\Delta = 21439.38803 \ Pa

To atm ; we have

\Delta P = \dfrac{21439.38803 }{101325}

\Delta P =0.2115903087  \ atm

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

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