Benoulli's Equation Problem

[aliz_khanz]

Homework Statement

A garden hose having internal diameter of 1.75cm is connected to a lawn sprinkler that consists merely of an enclosure with 24 holes each , each 0.05cm in diameter. If the water in the hose has a speed of 2m/s, at what speed does it leave the sprinkler holes ?

Homework Equations

The Attempt at a Solution

I think the 24 holes will be multiplied with 0.05 in order to get the total speed and then later dividing it by 24 to get a single hole speed . But we don't know the density? Stumped !

 

[Cygni]

Perhaps You have to look up at the density of water? $$\rho$$ = 999.7026 kg/m³ at T = +10 C

 

[kerimek]

There are a few equations that can be used to solve this problem, including Bernoulli's equation and the continuity equation. First, we can use the continuity equation, which states that the mass flow rate of a fluid is constant throughout a pipe or system. This means that the mass flow rate of water entering the hose (ρ1A1V1) is equal to the mass flow rate of water exiting the sprinkler (ρ2A2V2), where ρ is the density of water, A is the cross-sectional area of the hose or sprinkler hole, and V is the velocity of the water.

Next, we can use Bernoulli's equation, which states that the total energy of a fluid (pressure energy, kinetic energy, and potential energy) remains constant throughout a pipe or system. Since the water is at the same height throughout the system, we can ignore the potential energy term. This means that the pressure energy (P1 + 1/2ρV1^2) at the hose entrance is equal to the pressure energy (P2 + 1/2ρV2^2) at the sprinkler holes.

Using these two equations, we can solve for the velocity at the sprinkler holes (V2). We know the velocity at the hose entrance (V1 = 2m/s), the cross-sectional area of the hose (A1 = π(0.875cm)^2), and the number of holes (n = 24). We also know the diameter of each hole (d = 0.05cm), so we can calculate the cross-sectional area of one hole (A2 = π(0.025cm)^2).

Plugging these values into the continuity equation, we get:

ρ1A1V1 = ρ2A2V2

ρ2 = ρ1(A1V1)/(A2V2)

Since we don't know the density of water, we can plug in the values for the density of water at room temperature (1g/cm^3) and solve for V2:

1g/cm^3 = (1g/cm^3)(π(0.875cm)^2(2m/s))/((24)(π(0.025cm)^2)V2)

V2 = 0.05m/s

Therefore, the water leaves the sprinkler holes at a speed of 0.05m/s. This is a much slower velocity