The static pressure in a horizontal pipeline is 4.3 x 10^4 Pa, the total pressure is 4.7 x 10^4 Pa, and the area of cross-section is 20cm^2. The fluid may be considered to be incompressible and non-viscous and has a density of 10^3 kg/m^3. Calculate (i) the flow velocity in the pipeline, (ii) the volume flow rate in the pipeline.
The difference between the total pressure and the static pressure (which is what you measure with a standard Pitot tube), is (1/2)(rho)V^2.
"rho" is the density of the fluid.
(i) Use that equation to solve for the flow velocity, V. Assume it is constant across the cross section, (even through it will be less in wall boundary layers)
(ii) The volume flow rate is the product of V and the cross sectional area A, which will have to me in square meters if you want the flow rate in m^3/s.
To calculate the flow velocity in the pipeline, we can use Bernoulli's equation, which states that the total pressure in a fluid system remains constant along a streamline. In this case, we can equate the static pressure, total pressure, and dynamic pressure to find the flow velocity.
(i) Flow Velocity Calculation:
Bernoulli's equation:
P + (1/2)ρv^2 = constant
P = Static Pressure = 4.3 x 10^4 Pa
ρ = Density = 10^3 kg/m^3
Using the given values and rearranging the equation, we can solve for the flow velocity (v):
4.7 x 10^4 + (1/2)(10^3)v^2 = 4.3 x 10^4
(1/2)(10^3)v^2 = 4.7 x 10^4 - 4.3 x 10^4
(1/2)(10^3)v^2 = 4 x 10^3
v^2 = (4 x 10^3) / (1/2)(10^3)
v^2 = 8 x 10^3
v = √(8 x 10^3)
v ≈ 89.4 m/s
Therefore, the flow velocity in the pipeline is approximately 89.4 m/s.
(ii) Volume Flow Rate Calculation:
The volume flow rate (Q) is the product of the cross-sectional area (A) and the flow velocity (v).
Given:
A = 20 cm^2 = 20 x 10^(-4) m^2 (converted from cm^2 to m^2)
v = 89.4 m/s
Using the given values, we can calculate the volume flow rate (Q):
Q = A x v
Q = (20 x 10^(-4)) x 89.4
Q ≈ 1.788 m^3/s
Therefore, the volume flow rate in the pipeline is approximately 1.788 m^3/s.
To calculate the flow velocity in the pipeline, you can use Bernoulli's equation, which states that the sum of the static pressure, dynamic pressure, and gravitational potential energy per unit volume of a fluid along a streamline is constant.
In this case, we can neglect the gravitational potential energy since the pipeline is horizontal. Therefore, we can write Bernoulli's equation as:
P + 0.5ρv^2 = constant
Where:
P is the total pressure
ρ is the density of the fluid
v is the flow velocity
We are given:
P = 4.7 x 10^4 Pa
ρ = 10^3 kg/m^3
Let's solve for the flow velocity (v):
4.7 x 10^4 + 0.5 * 10^3 * v^2 = 4.3 x 10^4
Subtracting 4.3 x 10^4 from both sides:
0.4 x 10^4 = 0.5 * 10^3 * v^2
Divide both sides by 0.5 * 10^3:
0.4 x 10^4 / (0.5 * 10^3) = v^2
8 = v^2
Taking the square root of both sides, we get:
v = √8 = 2√2 ≈ 2.83 m/s
So, the flow velocity in the pipeline is approximately 2.83 m/s.
To calculate the volume flow rate in the pipeline, you can multiply the cross-sectional area of the pipeline (A) by the flow velocity (v).
We are given:
A = 20 cm^2 = 0.002 m^2
v = 2.83 m/s
Let's calculate the volume flow rate (Q):
Q = A * v
= 0.002 * 2.83
= 0.00566 m^3/s
So, the volume flow rate in the pipeline is approximately 0.00566 m^3/s.