Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) 
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
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I think will help
Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.
Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:
The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:
= moles of a gas / total number moles of gas
The rigid tank has total pressure of 1MPa.
molar mass = 14g/mol
mass in the tank = 2000g
number of moles in the tank: 
= 142.85mols
molar mass = 44g/mol
mass in the tank = 4000g
number of moles in the tank: 
= 90.91mols
Total number of moles: 142.85 + 90.91 = 233.76 mols
To calculate partial pressure:
For Nitrogen gas:
= 0.6
For Carbon Dioxide:
0.4
Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.
Answer:
Detailed solution is attached in the images below showing step wise solution and answer for each part individually.
