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igomit [66]
3 years ago
14

Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

this stream is 300 mols per minute, what is the molar flow rate of water vapor in the stream?
Engineering
1 answer:
Marizza181 [45]3 years ago
4 0

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H_{2}o ( YH_{2}o )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

 = 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH_{2}o = molar flow rate of water

Nair = molar  flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541  moles /min

Molar flowrate of water

=  n -  Nair

= 311.541 - 300 = 11.541 mol/min

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(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

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                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

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                           v*dv = - (0.1 + sin(x/0.8)) . dx

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                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

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- v = v_max when a = 0:

                           -0.1 = sin(x/0.8)

                             x = -0.8*0.1002

                             x = -0.080134 m

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                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

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