Water flows steadily along a uniform flow tube of cross-section 30cm^2. The static pressure is 1.20 x 10^5 Pa and the total pressure is 1.28 x 10^5 Pa. Calculate the flow velocity and the mass of water per second flowing past a section of the tube. (density of water= 1000 kg/m^3)

Question

Water flows steadily along a uniform flow tube of cross-section 30cm^2. The static pressure is 1.20 x 10^5 Pa and the total pressure is 1.28 x 10^5 Pa. Calculate the flow velocity and the mass of water per second flowing past a section of the tube. (density of water= 1000 kg/m^3)

Answer 1

To calculate the flow velocity and mass of water per second flowing past a section of the tube, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing along a streamline. Bernoulli's equation is given as:

P + 1/2ρv^2 + ρgh = constant

Where:
P = pressure
ρ = density of the fluid
v = velocity of the fluid
g = acceleration due to gravity
h = height (or elevation) of the fluid

In this problem, the tube is horizontal, so there is no change in height. Therefore, the equation simplifies to:

P + 1/2ρv^2 = constant

To solve for the flow velocity (v), we can rearrange the equation:

1/2ρv^2 = constant - P

Substituting the given values:
Constant = total pressure = 1.28 x 10^5 Pa
P = static pressure = 1.20 x 10^5 Pa

We can now solve for v:

1/2ρv^2 = 1.28 x 10^5 Pa - 1.20 x 10^5 Pa

Hence,

1/2ρv^2 = 8 x 10^3 Pa

Now, substitute the value of the density of water:

1/2(1000 kg/m^3)v^2 = 8 x 10^3 Pa

To convert the flow area from cm^2 to m^2, divide by 10,000:

(30 cm^2 / 10,000) = 0.003 m^2

Now, substitute the values into the equation:

1/2(1000 kg/m^3)v^2 = 8 x 10^3 Pa

v^2 = (8 x 10^3 Pa)/(0.003 m^2) = 2.667 x 10^6 m^2/s^2

Taking the square root of both sides, we get:

v = √(2.667 x 10^6 m^2/s^2) = 1633.86 m/s (approx.)

Now, to calculate the mass flow rate (m_dot) of water per second flowing past the section of the tube, we can use the equation:

m_dot = ρAv

Where:
m_dot = mass flow rate
ρ = density of water
A = cross-sectional area of the tube
v = velocity of the water

Substitute the given values into the equation:

m_dot = (1000 kg/m^3)(0.003 m^2)(1633.86 m/s)

m_dot = 4.90158 kg/s (approx.)

Therefore, the flow velocity is approximately 1633.86 m/s, and the mass of water per second flowing past the section of the tube is approximately 4.90158 kg/s.

Water flows steadily along a uniform flow tube of cross-section 30cm^2. The static pressure is 1.20 x 10^5 Pa and the total pressure is 1.28 x 10^5 Pa. Calculate the flow velocity and the mass of water per second flowing past a section of the tube. (density of water= 1000 kg/m^3)

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To calculate the flow velocity and mass of water per second flowing past a section of the tube, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing along a streamline. Bernoulli's equation is given as: