Question

A tank containing diesel fuel(SG = 0.8) is open to the atmosphere at the top. A U-tube manometer is connected to the bottom of the tank. The depth of diesel fuel in the manometer is 1.2 m below the bottom of the tank.The manometer uses Mercury as the fluid and has a differential Mercury height of 2.3 m. What is the depth of diesel fuel in the tank?

Answer 1

h = 37.9 m

Step-by-step explanation:

Assume the height of fuel be h meter in tank

specific gravity of diesel is 0.8

specific gravity of mercury is 13.6

we know rhat density of water is 1000kg/m^3

so density of mercury is
`13.6 * 10^3 kg/m^3`

density of mercury is
`0.8* 10^3 kg/m^3`

Now equating PRESSURE at AA' position and tank

`\rho_d  g (h+ 1.2) = \rho_(hg) g2.3`

`h +1.2 = (13.6 * 10^3 * 2.3)/(0.8* 10^(3))`

h = 39.1 - 1.2

h = 37.9 m

Answer 2

h= 37.9 m

Step-by-step explanation:

Given that

SG = 0.8 for fuel so density of fluid will be 800 kg/m³.

We know that SG = 13.6 For Hg so density will be 13600 kg/m³.

Now by balancing the pressure

`\rho_d* g* h +\rho_d* g* 1.2 =\rho_(hg)* g* 2.3`

`800* 9.81* h+ 800* 9.81* 1.2 =13600* 9.81* 2.3`

`800* 9.81* h =13600* 9.81* 2.3-800* 9.81* 1.2`

`h=(297439.2)/(800* 9.81)`

h= 37.9 m