A 4 N force is used to compress water in a syringe that has

A 4 N force is used to compress water in a syringe that has a plunger with an area of 1 cm². This pressure is released at the other end through an opening with a radius of 1 mm. Assume no height change, and neglect viscosity. If the water in is shot out horizontally from a height of 1 m, how far would it travel before hitting the ground?

Solution

Pressure at plunger end P₁ = F / A

P₁ = 4 / 1

P₁ = 4 N/cm² = 40000 Pa

P₂ = 0 (gage atmospheric pressure)

Area at opening A₂ = 3.14 / 4 * 1² = 0.785 mm² = 0.785*10^-6 m²

Area at plunger A₁ = 1 cm² = 1*10^-4 m²

By continuity equation, A1*V1 = A2*V2

1*10^-4 * V1 = 0.785*10^-6 * V2

V1 / V2 = 0.00785

Applying Bernoulli eqn, P1 + 1/2*rho*V1² + z1 = P2 + 1/2*rho*V2² + z2

We have z1 = z2

Hence, 40000 + 1/2*1000*V1² = 0 + 1/2*1000*V2²

V₂² - V₁² = 80

V₂² (1 - (V₁ / V₂)²) = 80

V₂² (1 - (0.00785)²) = 80

V₂ = 8.944 m/s

Time taken by fluid particle for vertical drop of 1 m is given by h = 1/2*g*t²

1 = 1/2* 9.81*t²

t = 0.4515 s

Horizontal distance covered by fluid particle in 0.4515 s given by s = V₂*t

s = 8.944 * 0.4515

s = 4.038 m