Question

The oil level in a tank is 2 m above the ground. The tank cover is air tight and the air pressure above the oil surface is 150 kPa gage. The specific gravity of the oil is 0.88 a. Determine the flow velocity at the nozzle outflow. Assume inviscid, irrotational flow. In your discussion, compare this velocity to your values in Problem 5 of Homework 1. b. Determine the maximum height to which the oil stream could rise. (Answer: 19.4 m)

Answer 1

Answer:

a) 24.692 m/s

b) 19.4 m

Explanation:

To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:

$\frac{P1}{pg}+\frac{V1^2}{2g}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}+Z2$

We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:

$\frac{P1}{pg}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}$

a) The velocity (V2) is:

$\frac{P1}{pg}+Z1=\frac{V2^2}{2g}$

$(\frac{P1}{pg}+Z1)(2g)=V2^2$

$V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}$

Substituting the known values we can get the velocity at the out:

Atmospheric pressure= 101000 Pa

Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3

$V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}$

$V2=24.692 m/s$

b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:

$\frac{P2}{pg}+\frac{V2^2}{2g}+Z2=\frac{P3}{pg}+\frac{V3^2}{2g}+Z3$

We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:

$\frac{V2^2}{2g}=\frac{P3}{pg}+Z3$

$Z3=\frac{V2^2}{2g}-\frac{P3}{pg}$

Substituting the known values, the height (Z3) is:

$Z3=\frac{(24.692 m/s)^2}{2(9.81 m/s2)}-\frac{101000 Pa}{(9.81 m/s)(880 kg/m3)}$

Z3=Maximum Height=19.376=19.4 m