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

Answer:

h= 37.9 m

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

$\rho_d\times g\times h +\rho_d\times g\times 1.2 =\rho_{hg}\times g\times 2.3$

$800\times 9.81\times h+ 800\times 9.81\times 1.2 =13600\times 9.81\times 2.3$

$800\times 9.81\times h =13600\times 9.81\times 2.3-800\times 9.81\times 1.2$

$h=\dfrac{297439.2}{800\times 9.81}$

h= 37.9 m

Answer 2

Answer:

h = 37.9 m

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 $13.6 \times 10^3 kg/m^3$

density of mercury is $0.8\times 10^3 kg/m^3$

Now equating PRESSURE at AA' position and tank

$\rho_d g (h+ 1.2) = \rho_{hg} g2.3$

$h +1.2 = \frac{13.6 \times 10^3 \times 2.3}{0.8\times 10^{3}}$

h = 39.1 - 1.2

h = 37.9 m