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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 95. The only available calibration data for the flowmeter are two points scribbled on a sheet of paper, indicating that readings R = 15 and R = 50 correspond to flow rates V = 40 m/h and V = 96 m/h, respectively.

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batolisis

Answer: hello your question has some missing data attached below is the missing data

answer:

i) volumetric flow rate = 168 m^3/h = 5932.86 ft^3/h

ii) n = 582.4 Ib mole/hr

Explanation:

Given that the volumetric flow rate is a Linear function of R

V = ∝R + β

where; R1 = 15 , V1 = 40 m/h , R2 = 50, V2 = 96 m/h  input values into equation above

40 = 15∝ + β ----- ( 1 )

96 = 50∝ + β ----- ( 2 )

resolve equations 1 and 2 simultaneously

56 = 35∝ + 0

∴ ∝ = 56 / 35 = 1.6

back to equation 1:  40 = 15(1.6) + β    ∴ β = 40 - 24 = 16

hence ; V = 1.6(95) + 16 = 168 m^3/h = 5932.86 ft^3/h

lets assume room temperature

density of water = 62.4 Ib/ft^3

molecular weight of water = 18

therefore n = ( 62.4 / 18 ) * 168

                  = 582.4 Ib mole/hr

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