A garden hose is used to fill a drum with water. The hose is held 0.88 m above the ground. The cross-sectional area of the 2.0 cm diameter hose is two times the cross-sectional area at the spout end where the water is coming out and the gage pressure inside the hose at a section laying on the ground is 29.7 kPa. Based on this information one can determine that the water velocity at the spout end is [ FILL IN BLANK ]m/s, the water velocity inside the hose is [ FILL IN BLANK ]m/s, the volumetric flowrate of the water is [ FILL IN BLANK ]× 10-3 m3/s, and the time it takes to fill the 0.210 m3 drum to its maximum capacity is [ FILL IN BLANK ]min. Note that the water density is ρw = 1000 kg/m3.
loss in pressure from inside hose to exit is 29.7*10^3 N/m^2
pin + (1/2) rho Vin^2 = Pout + (1/2) rho Vout^2
29.7 * 10^3 = (1/2)(10^3)(Vout^2-Vin^2)
but if Aout = .5 Ain
then Vout = 2 Vin
so
2*29.7 = (4-1)Vin^2 = 3 Vin^2
solve for Vin
then Vout
Vi * Area inside = Q, volume flow
that should get you going.
why you have not considered datum height of 0.88 m?
To solve this problem, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. The equation is given as:
P + (1/2)ρv^2 + ρgh = constant
Where:
P is the pressure of the fluid
ρ is the density of the fluid
v is the velocity of the fluid
g is the acceleration due to gravity
h is the height above a reference point
Let's break down the given information and solve for each required value:
1. Water velocity at the spout end (v1):
At the spout end, the height above the ground (h1) is 0.88 m, and the pressure (P1) is atmospheric pressure, which is approximately 101.3 kPa. The density (ρ) is given as 1000 kg/m^3.
Using Bernoulli's equation, we can solve for v1:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since we take the height on the ground as the reference height (h2 = 0), the equation simplifies to:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2
P2 is the pressure inside the hose, given as 29.7 kPa. Rearranging the equation, we get:
(1/2)ρv1^2 + 101.3 kPa + 1000 kg/m^3 * 9.81 m/s^2 * 0.88 m = 29.7 kPa + (1/2)ρv2^2
Simplifying and solving for v1:
(1/2) * 1000 kg/m^3 * v1^2 + 101.3 kPa + 8727 kg/m^2/s^2 = 29.7 kPa + (1/2) * 1000 kg/m^3 * v2^2
(1/2) * 1000 kg/m^3 * v1^2 = 29.7 kPa - 101.3 kPa - 8727 kg/m^2/s^2
(1/2) * 1000 kg/m^3 * v1^2 = -7872 kg/m^2/s^2
v1^2 = -7872 kg/m^2/s^2 / (1/2) * 1000 kg/m^3
v1^2 = -15.744 m^2/s^2
Taking the square root (since velocity cannot be negative), we get:
v1 = √(-15.744) m/s
Since the velocity cannot be negative, and we are dealing with a real-world scenario, the velocity at the spout end is 0 m/s.
Therefore, the water velocity at the spout end is 0 m/s.
2. Water velocity inside the hose (v2):
From the previous calculation, we found that the velocity at the spout end is 0 m/s. By the principle of continuity, the volumetric flow rate is the same at any cross-section of a hose with a steady flow of fluid. Therefore, the velocity inside the hose (v2) will also be 0 m/s.
Therefore, the water velocity inside the hose is 0 m/s.
3. Volumetric flowrate of the water (Q):
The volumetric flow rate (Q) can be calculated using the formula:
Q = A * v
Where A is the cross-sectional area and v is the velocity.
Given that the cross-sectional area of the hose (A1) is two times the cross-sectional area at the spout end (A2), we have:
A1 = 2 * A2
Since the diameter of the hose is given as 2.0 cm, we can calculate the cross-sectional areas.
A1 = π * (diameter of hose/2)^2 = π * (2.0 cm/2)^2 = π * (1.0 cm)^2
A2 = π * (diameter of spout/2)^2 = π * (0.20 cm/2)^2 = π * (0.1 cm)^2
The velocity at the spout end (v1) was found to be 0 m/s, and the velocity inside the hose (v2) is also 0 m/s.
Therefore, the volumetric flow rate of the water is Q = A2 * v2 = π * (0.1 cm)^2 * 0 m/s
Q = 0 m^3/s
Therefore, the volumetric flowrate of the water is 0 × 10^-3 m^3/s.
4. Time to fill the drum (t):
The time it takes to fill the drum can be calculated using the formula:
t = Volume / Volumetric flowrate
Given that the volume of the drum is 0.210 m^3 and the volumetric flow rate (Q) is 0 × 10^-3 m^3/s, we can calculate the time:
t = 0.210 m^3 / (0 × 10^-3 m^3/s)
Since the denominator is zero, the time it takes to fill the drum would be undefined.
Therefore, the time it takes to fill the 0.210 m^3 drum to its maximum capacity cannot be determined based on the given information.