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.

Question

Asked on June 27, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-27T12:06:24.131Z

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 m Pressure, P_1 = 8.2 bar Flow rate, Q = 7.5 m^3/min Elevation, z_1 = 0 m (reference level) At the location (Section 2): Diameter, D_2 = 0.3 m Elevation, z_2 = 3 m (3 m above entrance) Fluid is water, so density, = 1000 kg/m^3 Acceleration due to gravity, g = 9.81 m/s^2 We will use the following equations: 1. Flow Rate: Q = A · V, where A = ( D^2)/(4) 2. Bernoulli's Equation (neglecting losses): P_1 + (1)/(2) V_1^2 + g z_1 = P_2 + (1)/(2) V_2^2 + g z_2 Step 1: Convert given values to SI units. P_1 = 8.2 bar = 8.2 × 10^5 Pa Q = 7.5 m^3/min = (7.5)/(60) m^3/s = 0.125 m^3/s Step 2: Calculate the cross-sectional areas at both sections. Area at Section 1: A_1 = ( D_1^2)/(4) = (0.15 m)^24 = ( × 0.0225)/(4) ≈ 0.017671 m^2 Area at Section 2: A_2 = ( D_2^2)/(4) = (0.3 m)^24 = ( × 0.09)/(4) ≈ 0.070686 m^2 Step 3: Calculate the flow velocities at both sections using the flow rate. Velocity at Section 1: V_1 = (Q)/(A_1) = 0.125 m^3/s0.017671 m^2 ≈ 7.0735 m/s Velocity at Section 2: V_2 = (Q)/(A_2) = 0.125 m^3/s0.070686 m^2 ≈ 1.7684 m/s Step 4: Apply Bernoulli's equation to determine the pressure at Section 2. Rearrange Bernoulli's equation to solve for P_2: P_2 = P_1 + (1)/(2) (V_1^2 - V_2^2) + g (z_1 - z_2) Substitute the known values: P_2 = (8.2 × 10^5 Pa) + (1)/(2)(1000 kg/m^3) ((7.0735 m/s)^2 - (1.7684 m/s)^2) + (1000 kg/m^3)(9.81 m/s^2)(0 m - 3 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 Pa Step 5: Convert the final pressure to bar. P_2 = (813510.1)/(10^5) bar ≈ 8.1351 bar The pressure at the location is 8.14 bar. Send me the next one 📸

Accepted Answer · 2026-06-27T12:06:24.131Z

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 m Pressure, P_1 = 8.2 bar Flow rate, Q = 7.5 m^3/min Elevation, z_1 = 0 m (reference level) At the location (Section 2): Diameter, D_2 = 0.3 m Elevation, z_2 = 3 m (3 m above entrance) Fluid is water, so density, = 1000 kg/m^3 Acceleration due to gravity, g = 9.81 m/s^2 We will use the following equations: 1. Flow Rate: Q = A · V, where A = ( D^2)/(4) 2. Bernoulli's Equation (neglecting losses): P_1 + (1)/(2) V_1^2 + g z_1 = P_2 + (1)/(2) V_2^2 + g z_2 Step 1: Convert given values to SI units. P_1 = 8.2 bar = 8.2 × 10^5 Pa Q = 7.5 m^3/min = (7.5)/(60) m^3/s = 0.125 m^3/s Step 2: Calculate the cross-sectional areas at both sections. Area at Section 1: A_1 = ( D_1^2)/(4) = (0.15 m)^24 = ( × 0.0225)/(4) ≈ 0.017671 m^2 Area at Section 2: A_2 = ( D_2^2)/(4) = (0.3 m)^24 = ( × 0.09)/(4) ≈ 0.070686 m^2 Step 3: Calculate the flow velocities at both sections using the flow rate. Velocity at Section 1: V_1 = (Q)/(A_1) = 0.125 m^3/s0.017671 m^2 ≈ 7.0735 m/s Velocity at Section 2: V_2 = (Q)/(A_2) = 0.125 m^3/s0.070686 m^2 ≈ 1.7684 m/s Step 4: Apply Bernoulli's equation to determine the pressure at Section 2. Rearrange Bernoulli's equation to solve for P_2: P_2 = P_1 + (1)/(2) (V_1^2 - V_2^2) + g (z_1 - z_2) Substitute the known values: P_2 = (8.2 × 10^5 Pa) + (1)/(2)(1000 kg/m^3) ((7.0735 m/s)^2 - (1.7684 m/s)^2) + (1000 kg/m^3)(9.81 m/s^2)(0 m - 3 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 Pa Step 5: Convert the final pressure to bar. P_2 = (813510.1)/(10^5) bar ≈ 8.1351 bar The pressure at the location is 8.14 bar. Send me the next one 📸

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.

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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.

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