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.

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

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j