A long horizontal hose of diameter 4.6 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose ?
i found .945 m/s
b) What is the pressure differential between the water in the hose and water in the nozzle ?
c) How long will it take to fill a tub of volume 150 liters with the hose ?
a) Law of continuinity:
A1*V1=A1*V2
v1= (1/4.6)^2 *20 you are correct.
b) v1^2/2-v2^2/2+ (P1-P2)/rho=0
P1-P2= rho(.945^2-20^2)*2 where rho= 1E3kg/m^3
= about 200E3kg/m*s^2= about 200E3Pascal
I ignored the .945^2, you need to work that. Check my thinking.
c. time= volume/flow= 150E-3m^3/PI*.005^2*20m/s
a) To determine the velocity of the water in the hose, we can apply the principle of continuity. According to this principle, the product of the cross-sectional area and velocity of a flowing fluid remains constant along a streamline.
Let's denote the diameter of the hose as D1 (4.6 cm) and the diameter of the nozzle as D2 (1 cm). The cross-sectional area of the hose (A1) is given by the formula:
A1 = π * (D1/2)^2 = π * (4.6/2)^2 = π * 2.3^2 = 16.619 cm^2
Similarly, the cross-sectional area of the nozzle (A2) is given by:
A2 = π * (D2/2)^2 = π * (1/2)^2 = π * 0.25 = 0.785 cm^2
Using the principle of continuity, we can equate the product of the area and the velocity for the hose and the nozzle:
A1 * V1 = A2 * V2
Substituting the values we have:
16.619 * V1 = 0.785 * 20
Simplifying further:
V1 = (0.785 * 20) / 16.619
Calculating this expression yields approximately 0.942 m/s.
b) To find the pressure differential between the water in the hose and the nozzle, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid along a streamline.
Bernoulli's equation can be stated as:
P1 + (1/2) * ρ * V1^2 + ρ * g * h1 = P2 + (1/2) * ρ * V2^2 + ρ * g * h2
Where:
P1 and P2 are the pressures at locations 1 (inside the hose) and 2 (at the nozzle), respectively.
ρ is the density of water, which is approximately 1000 kg/m^3.
V1 and V2 are the velocities at locations 1 and 2, respectively.
g is the acceleration due to gravity, which is approximately 9.81 m/s^2.
h1 and h2 are the heights at locations 1 and 2, respectively. In this case, we can assume both heights are the same, so h1 = h2 = 0.
Simplifying this equation for our scenario:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
Substituting the values we have:
P1 + (1/2) * 1000 * (0.942)^2 = P2 + (1/2) * 1000 * (20)^2
Simplifying further:
P1 + 445.644 = P2 + 20000
Therefore, the pressure differential (P2 - P1) is:
P2 - P1 = 20000 - 445.644 = 19554.356
So, the pressure in the hose is approximately 19554.356 Pa higher than the pressure at the nozzle.
c) To determine the time taken to fill a tub of volume 150 liters with the hose, we need to calculate the flow rate of the water through the hose.
The flow rate can be calculated using the formula:
Flow rate = A1 * V1
Where A1 is the cross-sectional area of the hose and V1 is the velocity of water in the hose we found earlier.
Substituting the values we have:
Flow rate = 16.619 * 0.942
Flow rate ≈ 15.646 L/s
Now, since we have the flow rate in liters per second, we can calculate the time taken to fill the tub using the formula:
Time = Volume / Flow rate
Converting 150 liters to cubic meters (1 L = 0.001 m^3):
Volume = 150 * 0.001 = 0.15 m^3
Substituting the values we have:
Time = 0.15 / 15.646
Time ≈ 0.0096 seconds
Therefore, it would take approximately 0.0096 seconds to fill a tub with a volume of 150 liters using the given hose.