Question

A pressurized tank of water has a 10-cm-diameter orifice at the bottom, where water discharges to the atmosphere. The water level is 2.5 m above the outlet. The tank air pressure above the water level is 250 kPa (absolute) while the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine the initial discharge rate of water from the tank.

Answer 1

Answer:

The velocity is 18.68m/s

Explanation:

Bernoulli's equation is applicable for stream line flow of a fluid. The flow must be steady and uniform flow. The Bernoulli's equation between inlet and outlet is written as:

P1/pg + V1/2g + Z1 = P2/pg + V2^2 + Z2

Where V1 and V2 are velocity of fluid at point 1 and 2b. The diameter of the tank too will be larger than that of the nozzle. Hence the velocity at point 1 will be 0.V1= 0

Substituting the values in to the equation

250 ×10^3/1000g + 0/g + 2.5 = 100×10^3/1000g + V2^2/2g + 0

250 + 2.5g = 100 + V2^2/2

250 + (2.5 × 9.8) = 100 V2^2/2

250 + 23.525- 100 = V2^2/2

174.525 = V2^2/2

Cross multiply

174.525 × 2 = V2^2

V2 = 349.05

V2 = Sqrt(349.05)

V2 = 18.68m/s

Answer 2

Answer: 0.15m³/s

Explanation:

To solve this question, we are going to have to apply Bernoulli's equation.

[(P1/ρg) + (V1²/2g) + Z1] = [(P2/ρg) + (V2²/2g) + Z2]

V1 and V2 represent the velocities of the fluid at levels 1 and 2 respectively.

Note that, the diameter of the tank is very much larger than that of the nozzle, as such, it can be neglected. V1 = 0

Z2 also is 0

Substituting in the equation, we have

[(250*10^3/1000g) + (0/2g) + 2.5] = [(100*10^3/1000g) + (V2²/2g) + 0]

250/g + 2.5 = 100/g + V2²/2g

250 + 2.5g = 100 + V2²/2

Using g = 9.8

250 + 2.5*9.8 = 100 + V2²/2

250 + 24.5 = 100 + V2²/2

274.5 - 100 = V2²/2

174.5 = V2²/2

349 = V2²

V2 = 18.68m/s

V = Q/A

A = πd²/4

A = [π*(10*10^-2)²]/4

A = 7.85*10^-3m²

Q = VA

Q = 18.68 * 7.85*10^-3

Q = 0.15m³/s