Steam flows into a tank at entrance 1 and liquid water flows out at exit 2. The

Question

Steam flows into a tank at entrance 1 and liquid water flows out at exit 2. The conditions at one and two are:

(entrance 1, steam)

V1=10m/s

A1=10^-2  m^2

P1=20 bar

T1=600 C

(exit 2, liquid water)

m2= 6 kg/s

P2= 10 bar

T2= 150 C

a)What is the volumetric flow rate of steam at the entrance

b)Determine the time rate of change of mass within the tank

c)Sketch a graph of how the mass in the tank varies with time. At t= infinity , will the tank volume be full, empty, or at some intermediate value?(No calculations required for part c)

Explanation / Answer

Volumetric flow at the entrance = area * velocity = 10^-1 m^3 / s. OR     0.1 m^3 / s

6kg of water flows out each second at entrance 2.

No. of moles of water geting out = 6000 / 18 = 333.3 moles

P2 = 10 bar = 1000000 Pa , T2 = 423 K, R = 8.314 J / mol / K

So, volume of water getting out, V2 = nRT2 / P2 = 1.172 m^3 / sec OR 1172 litres per second.

Water enters the tank as vapour.

Moles entering the tank = 100 L / 22.414 L = 4.461 moles

P1 = 20 bar = 2000000 Pa, T1 = 873 K

So, equivalent volume of water entering tank, V1 = 4.461 * 8.314 * 873 / 2000000 = 0.016 m^3 / sec OR 16 litres / sec

NOTE- THIS VOLUME IS NOT THE VOLUME OF GAS ENTERING, BUT THE VOLUME OF WATER THAT THE WHOLE GAS ENTERING FORMS INTO.

IN OTHER WORDS, 100 LITRES OF WATER VAPOUR IS EQUIVALENT TO 16 LITRES OF WATER.

THIS IS DONE TO EASE THE CALCULATIONS.

From above part, it is seen that whenn 1172 litres go out, only 16 litres get inside.

Thus, after some time, the tank will become empty.

So sorry, I couldn't draw the graph, but it would have a positive intercept of the initial volume in the tank.

The slope would be negative with a value of (1172 - 16) = 1156

The slope is, thus, -1156.

THANK YOU. DO RATE.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.