A cylindrical tank containing oil with a density of 800 kg/m3. The depth is 3.0

Question

A cylindrical tank containing oil with a density of 800 kg/m3. The depth is 3.0 meters. Above the oil is air with some pressure. A U-tube is connected to the air-filled space in the tank with one leg and the other flows into the open air. The tube contains mercury with a density of 13600 kg/m3. Calculate both absolute and relative pressure on the bottom of the tank if the meniscus (mercury surface) of the leg opening in the open air is a / 100 mm above the other b / 100 mm below the other. The atmospheric pressure is 990 hPa.

Explanation / Answer

Pb--> pressure at the bottom of cylinder

Pa--> pressure of air in the cylindrical tube

P0--> atmospheric pressure

a) Pa= P0 + mgh

m--> density of mercury

h---> difference b/w two mercury surfaces

Pa = P0+13600*9.8*100*10^(-3)

=P0+13328

Pb =Pa+gd

--> density of oil

d=3m--> depth of the cylindrical tank

Pb= Pa+800*9.8*3

= Pa+36848

=1358.48 hPa absolute pressure

b) same procedure but h= -100mm

so we get Pb= P0+23520-13328=1091.92hPa

relative pressure depends on which you consider as standard pressure let it be Ps

then Pb - Ps gives the relative pressure

Ps as P0

a)then relative pressure =368.48hPa

b)relative pressure =101.92hPa

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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