A vertical tube has diameter 0.06 m at its bottom end from which water...
Mass of water flowing in one second
Let V. be the velocity of flow of water at the bottom . then the mass of the water flowing per second =
Let v
2 be the velocity of flow of water at the point where the radius of the tube is 0.02m. The mass of the water flowing per second is 0.03
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A vertical tube has diameter 0.06 m at its bottom end from which water...
Given data:
Diameter of the tube at the bottom end = 0.06 m
Rate of water flow = 1.3 kg/minute
Pressure at the end = 0.9 mm of Hg
Diameter of the tube at a height of 3 m = 0.04 m
We need to find the pressure at that point.
Assumptions:
1. The flow of water is steady.
2. The density of water remains constant throughout the tube.
3. The tube is vertical, so there is no change in height of the water column due to elevation.
Explanation:
Step 1: Convert units:
Convert the pressure at the end from mm of Hg to Pascal.
1 mm Hg = 133.3223684 Pascal
0.9 mm Hg = 0.9 * 133.3223684 Pascal = 119.99013156 Pascal
Step 2: Calculate the velocity of water at the bottom end:
Using the equation of continuity, we can relate the velocities and areas at different points in the tube.
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas and v1 and v2 are the velocities at points 1 and 2, respectively.
Given:
Diameter at the bottom end = 0.06 m
Radius at the bottom end = 0.06/2 = 0.03 m
Area at the bottom end = πr1^2 = π(0.03)^2 = 0.00282743339 m^2
Rate of water flow = 1.3 kg/minute
Mass flow rate (m_dot) = 1.3 kg/minute = 1.3/60 kg/s
Density of water (ρ) = mass/volume = m_dot/(A1v1)
v1 = m_dot / (ρA1) = (1.3/60) / (1000 * 0.00282743339) = 0.07627118644 m/s
Step 3: Calculate the pressure at a height of 3 m:
Using the Bernoulli's equation between the bottom end and the point at 3 m,
P1 + ρgh1 + (1/2)ρv1^2 = P2 + ρgh2 + (1/2)ρv2^2
Since the tube is vertical, the velocity at the top end is zero, v2 = 0.
Also, h1 = 0 and h2 = 3 m.
P1 + (1/2)ρv1^2 = P2 + ρgh2
P2 = P1 + (1/2)ρv1^2 - ρgh2
Given:
P1 = 119.99013156 Pascal
ρ = 1000 kg/m^3 (density of water)
v1 = 0.07627118644 m/s
g = 9.8 m/s^2 (acceleration due to gravity)
h2 = 3 m
Substituting the values, we get:
P2 = 119.99013156 + (1/2)(1000)(0.07627118644^2
A vertical tube has diameter 0.06 m at its bottom end from which water...
Mass of water flowing in one second
Let V. be the velocity of flow of water at the bottom . then the mass of the water flowing per second =
Let v
2 be the velocity of flow of water at the point where the radius of the tube is 0.02m. The mass of the water flowing per second is 0.03