The pressure at the entry to the pipeline of 0.15 m diameter is 8.2 bar, and the flow rate at this section is…

Question

The pressure at the entry to the pipeline of 0.15 m diameter is 8.2 bar, and the flow rate at this section is 7.5 m 3 /min. The pipe diameter gradually increases to 0.3 m, and the levels rise by 3 m above the entrance. Determine the pressure at the location. Neglect losses.

ScanSolve AI Tutor · Accepted Answer

This problem involves applying Bernoulli's equation and the continuity equation to determine the pressure in a pipeline where both diameter and elevation change, neglecting losses.

Given:

At entry (Section 1):
- Diameter, $D_1 = 0.15 \text{ m}$
- Pressure, $P_1 = 8.2 \text{ bar}$
- Flow rate, $Q = 7.5 \text{ m}^3/\text{min}$
- Elevation, $z_1 = 0 \text{ m}$ (reference level)
At the location (Section 2):
- Diameter, $D_2 = 0.3 \text{ m}$
- Elevation, $z_2 = 3 \text{ m}$ (3 m above entrance)
Fluid is water, so density, $\rho = 1000 \text{ kg/m}^3$
Acceleration due to gravity, $g = 9.81 \text{ m/s}^2$

We will use the following equations:

Flow Rate: $Q = A \cdot V$ , where $A = \frac{\pi D^2}{4}$
Bernoulli's Equation (neglecting losses): $P_1 + \frac{1}{2}\rho V_1^2 + \rho g z_1 = P_2 + \frac{1}{2}\rho V_2^2 + \rho g z_2$

Convert given values to SI units.
- $P_1 = 8.2 \text{ bar} = 8.2 \times 10^5 \text{ Pa}$
- $Q = 7.5 \text{ m}^3/\text{min} = \frac{7.5}{60} \text{ m}^3/\text{s} = 0.125 \text{ m}^3/\text{s}$
Calculate the cross-sectional areas at both sections.
- Area at Section 1: $A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.15 \text{ m})^2}{4} = \frac{\pi \times 0.0225}{4} \approx 0.017671 \text{ m}^2$
- Area at Section 2: $A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.3 \text{ m})^2}{4} = \frac{\pi \times 0.09}{4} \approx 0.070686 \text{ m}^2$
Calculate the flow velocities at both sections using the flow rate.
- Velocity at Section 1: $V_1 = \frac{Q}{A_1} = \frac{0.125 \text{ m}^3/\text{s}}{0.017671 \text{ m}^2} \approx 7.0735 \text{ m/s}$
- Velocity at Section 2: $V_2 = \frac{Q}{A_2} = \frac{0.125 \text{ m}^3/\text{s}}{0.070686 \text{ m}^2} \approx 1.7684 \text{ m/s}$
Apply Bernoulli's equation to determine the pressure at Section 2.
Rearrange Bernoulli's equation to solve for $P_2$ :
$P_2 = P_1 + \frac{1}{2}\rho (V_1^2 - V_2^2) + \rho g (z_1 - z_2)$
Substitute the known values:
$P_2 = (8.2 \times 10^5 \text{ Pa}) + \frac{1}{2}(1000 \text{ kg/m}^3) ((7.0735 \text{ m/s})^2 - (1.7684 \text{ m/s})^2) + (1000 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(0 \text{ m} - 3 \text{ m})$ $P_2 = 820000 + 500 (49.0074 - 3.1272) + (9810)(-3)$ $P_2 = 820000 + 500 (45.8802) - 29430$ $P_2 = 820000 + 22940.1 - 29430$ $P_2 = 842940.1 - 29430$ $P_2 = 813510.1 \text{ Pa}$
Convert the final pressure to bar.
$P_2 = \frac{813510.1}{10^5} \text{ bar} \approx 8.1351 \text{ bar}$
The pressure at the location is $8.14 \text{ bar}$ .

The pressure at the entry to the pipeline of 0.15 m diameter is 8.2 bar, and the flow rate at this section is 7.5 m 3 /min. The pipe diameter gradually increases to 0.3 m, and the levels rise by 3 m above the entrance. Determine the pressure at the location. Neglect losses.

Answer

Step-by-step solution

Convert given values to SI units.

Calculate the cross-sectional areas at both sections.

Calculate the flow velocities at both sections using the flow rate.

Apply Bernoulli's equation to determine the pressure at Section 2.

Convert the final pressure to bar.

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