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A stream of humid air containing 2.50 mol% H2O(v) and the balance dry air is to be humidified to a water content of 10.0 mole% H20. For this purpose, liquid water is fed through a flowmeter and evaporated into the air stream. The flowmeter reading, R, is 105.0. The only available calibration data for the flowmeter are two points scribbled on a sheet of paper, indicating that readings R = 21.0 and R = 54.0 correspond to flow rates V = 42.0 ft/h and V = 100.9 ft/h, respectively. Water Flow Estimate the flow rate of the liquid water feed if the volumetric flow rate is a linear function of R. V= / 1 lb-mole/h Estimate the flow rate of the liquid water feed if the reading Ris a linear function of V . V= Ib-mole/h

Answer :

IthaloAbreu

Answer:

664 lb-mole/h

Explanation:

If the volumetric flow rate (V) is a linear function of R:

V(R) = a*R + b, where a and b are the linear constants. For the data given:

42.0 = 21a + b  x(-1)

100.9 = 54a + b

-42.0 = -21a -b

100.9 = 54a + b (summing both equations)

58.9 = 33a

a = 1.78

b + 21*1.78 = 42.0

b + 37.38 = 42.0

b = 4.62

So, V(R) = 1.78*R + 4.62

For R = 105.0, the flow rate is:

V = 1.78*105.0 + 4.62

V = 191.5 ft³/h

The water density is 62.428 lb/ft³, so the mass flow is:

m = 191.5 ft³/h x  62.428 lb/ft³

m = 11955 lb/h

The molar mass of water is 18 lb/lb-mole, so, the flow in lb-mole/h is:

v = (11955lb/h)/(18lb/lb-mole)

v = 664 lb-mole/h

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