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faltersainse [42]
3 years ago
12

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

isture takes place. To what temperature must atmospheric air be cooled in order to have humidity ratio of 0.005 lb/lb?
Engineering
1 answer:
REY [17]3 years ago
6 0

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma                                  (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

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notka56 [123]

Answer:

For the two you haven't answered: (Drag greater than thrust, lift greater than weight) It will accelerate backwards (decelerate) and upwards

(Lift greater than weight, thrust greater than drag) accelerate upwards and forwards.

4 0
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}

     

7 0
3 years ago
PLS HELP ME
Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

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33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

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Explanation:

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Answer:

...................

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