An on-farm diesel fuel tank is on a stand 5 feet above the ground. the tank is vented to the atmosphere (the air in the tank is at atmospheric pressure). if the diesel level in the tank is 2.1 feet above the release point. if the density of diesel is 6.98 pounds/gallon, what is the gauge pressure of the fuel at the nozzle? what is the absolute pressure?

gauge pressure = psi

absolute pressure = psi

Pg= 0.766 psi

Pabs= 15.27 psi

Explanation:

Given

h= 2.1 ft

1 ft = 0.3048 m

2.1 ft = 0.64 m

h= 0.64 m

ρ=6.98  pounds/gallon

1 pounds/gallon = 119.82 kg/m³

ρ = 825.6 kg/m³

The gauge pressure Pg

Pg= ρ g h

Pg=  825.6  x 10 x 0.64     ( take g =10 m/s²)

Pg= 5283.84 Pa

We know that

1 Pa= 0.000145 psi

Pg= 0.766 psi

Lets take atmospheric pressure Patm

Patm= 100 KPa

Pg= 5.283 KPa

Absolute pressure = Atmospheric pressure + Gauge pressure

Pabs= 100 + 5.283

Pabs= 105.283 KPa

We know that

1 KPa= 0.145 Psi

Pabs= 15.27 psi

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft