Water flows at 0.25L/s through a 9.0-m-long garden hose 2.0cm in diameter that is lying flat on the ground.The temperature of the water is 20degC. What is the gauge pressure of the water where it enters the hose? in Pa
To find the gauge pressure of the water where it enters the hose, we can use the Bernoulli's equation, which relates the pressure, velocity, and height of the fluid in a flowing system.
Bernoulli's equation can be written as:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Here,
P₁ = gauge pressure at the entry of the hose (unknown)
P₂ = atmospheric pressure (assumed to be 1 atm)
ρ = density of water (assumed to be constant at 1000 kg/m³)
v₁ = velocity of water at the entry of the hose (unknown)
v₂ = velocity of water at the exit of the hose (0 m/s since the hose is lying flat on the ground)
g = acceleration due to gravity (assumed to be 9.8 m/s²)
h₁ = height of the water surface at the entry of the hose (assumed to be the same as the height at the exit)
Since the hose is lying flat, the height of the water surface, h₁, and h₂ is the same, so they will cancel out in the equation.
Also, since the water is entering the hose horizontally, the velocity at the entry of the hose, v₁, will be the same as the velocity of the water flow, which is given as 0.25 L/s.
Now, let's convert the given values to the appropriate units:
v₁ = 0.25 L/s = 0.25 × 10⁻³ m³/s
ρ = 1000 kg/m³
g = 9.8 m/s²
Substituting these values into the equation:
P₁ + ½(1000 kg/m³)(0.25 × 10⁻³ m³/s)² = 1 atm
Now, let's solve the equation to find P₁:
P₁ + ½(1000 kg/m³)(6.25 × 10⁻⁷ m⁶/s²) = 1 atm
P₁ + 3.125 × 10⁻⁴ Pa = 1 atm
Since 1 atm is approximately equal to 1.013 × 10⁵ Pa, we can rewrite the equation as:
P₁ + 3.125 × 10⁻⁴ Pa ≈ 1.013 × 10⁵ Pa
P₁ ≈ 1.013 × 10⁵ Pa - 3.125 × 10⁻⁴ Pa
P₁ ≈ 1.013 × 10⁵ Pa
Therefore, the gauge pressure of the water where it enters the hose is approximately 1.013 × 10⁵ Pa.
To find the gauge pressure of the water where it enters the hose, we can use the Bernoulli's equation. Bernoulli's equation relates the pressure, height, and velocity of a fluid flowing in a pipe. Here's how we can approach the problem:
Step 1: Convert the given flow rate from liters per second to cubic meters per second.
0.25 L/s = 0.25 * 10^-3 m^3/s
Step 2: Calculate the cross-sectional area of the hose.
The diameter of the hose is given as 2.0 cm. We can convert it to meters by dividing by 100:
Diameter = 2.0 cm = 0.02 m
Since the hose is circular, the cross-sectional area can be calculated using the formula:
A = π * (diameter/2)^2 = π * (0.02/2)^2 = 0.00031415 m^2
Step 3: Calculate the velocity of water in the hose.
The flow rate, Q, is given by Q = A * v, where A is the cross-sectional area and v is the velocity.
Rearranging the equation, v = Q / A
Plugging in the values:
v = (0.25 * 10^-3 m^3/s) / (0.00031415 m^2)
Step 4: Calculate the gauge pressure using Bernoulli's equation.
Bernoulli's equation states: P + (1/2)ρv^2 + ρgh = constant
Since the hose is lying flat on the ground, the height difference is zero, so the ρgh term can be ignored. In addition, we can assume the atmospheric pressure is negligible compared to the gauge pressure, so we can ignore it.
Simplifying the equation, we get:
P + (1/2)ρv^2 = constant
Rearranging the equation for the gauge pressure P, we have:
P = constant - (1/2)ρv^2
We need to find the gauge pressure where water enters the hose, so we consider that point as reference and take the constant term to be zero.
Plugging in the values:
P = - (1/2)ρv^2
The density of water, ρ, is approximately 1000 kg/m^3.
P = - (1/2) * (1000 kg/m^3) * (v^2)
Substituting the calculated value of v into the equation, we can calculate the gauge pressure.
Step 5: Calculate the gauge pressure.
P = - (1/2) * (1000 kg/m^3) * (v^2)
Now we can substitute the value of v calculated in step 3 into the equation to find the gauge pressure.
Use the formula
Q=pi*R^4*P/8*n*L
n= viscosity of the fluid
L= length
R= radius
THANKS
After conversions Q = 2.5*10-4; if I plug this into my equation:
p= [(8nL)*Q]/pi*R^4
p=[(8*(1.0*10-3 Pa*s)*9m)*2.5*10-4m3/s]/pi*0.01m4
p= 1.8*10-5/3.14*10-8 p = 573.2 Pa