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3 years ago

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un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti

lizando factor de compresivilidad

1 answer:

777dan777 [17]3 years ago

7 0

Ecfñnokg pinogdf gabn Etta r

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In the combination of resistors above, consider the 1.50 µΩ and 0.75 µΩ. How can you classify the connection between these two r

Airida [17]

Answer: they are connected in series.

Explanation:

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3 years ago

A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and

Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B = 164.6 < 85.42°Ω

C = 2.061 * 10^-3 < 90.32° s

D = 0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = $\sqrt{\frac{z}{y} }$ = $\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }$

= $\sqrt{79837.128< 4.899^o}$ = 282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate gl

gl = $\sqrt{zy} * d$

d = 500

z = 0.03 i + j 0.35

y = j4.4*10^-6 S/km

gl = $\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500$

= $\sqrt{1.5456*10^{-6} < 175.1^0} * 500$

= 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )

C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where : cos h (gl) = $\frac{e^{gl} + e^{-gl} }{2}$

sin h (gl) = $\frac{e^{gl}-e^{-gl} }{2}$

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2 years ago

What was a campaign belief in the 1980 presidential election? Carter called for a stronger national defense. Carter promised to

andreyandreev [35.5K]

Answer:

C

Explanation:

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3 years ago

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What is an example of a class 2 lever?

9966 [12]

Answer:

A wheelbarrow, a bottle opener, and an oar are examples of second class levers

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3 years ago

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1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

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3 years ago