Consider a U-tube whose arms are open to the atmosphere. Now water is poured into the U-tube from one arm, and light oil (???? =
790 kg/m3) from the other. One arm contains 70-cm-high water, while the other arm contains both fluids with an oil-to-water height ratio of 6. Determine the height of each fluid in that arm.
1 answer:
Answer:
Oil high= 73.14cm
Water high= 12.19cm
Explanation:
See the attached images.
The first important concept is that in an open tube at both ends the atmospheric pressure balances itself.
The pressure on both sides of the tube is the same in the equilibrium. The pressure of each fluid depends on its height and density.
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Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
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The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is: =<em>0,231 kg O₂/ kg air</em>
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I hope it helps!