The pressure in a section of horizontal pipe with a diameter of 2 cm is 143 kPa. Water flows through the pipe at 2.80 L/ s. If the pressure at a certain point is to be reduced to 103 kPa by constricting a section of the pipe, what should the diameter of the constricted section be? Assume laminar nonviscous flow. Express your answer in meters to four decimal places.
To solve this problem, we can apply the principle of conservation of mass and Bernoulli's equation. Here's how you can find the diameter of the constricted section:
1. Start by converting the flow rate from liters per second to cubic meters per second since the formula requires the latter unit.
- 1 L = 0.001 m^3 (conversion factor)
- Flow rate = 2.80 L/s * 0.001 m^3/L = 0.0028 m^3/s
2. Apply the principle of conservation of mass to determine the velocity of water at the initial (unconstricted) section of the pipe.
- The formula for flow rate (Q) is given by Q = A * V, where A is the cross-sectional area and V is the velocity.
- Rearranging the formula, we get V = Q / A.
- The cross-sectional area (A) of a pipe is given by the formula A = π * r^2, where r is the radius.
- Since the diameter (d) is given as 2 cm, the radius (r) is half of that, so r = 1 cm = 0.01 m.
- The initial cross-sectional area (A_initial) is then A_initial = π * (0.01 m)^2 = 0.00031416 m^2.
- Therefore, the initial velocity (V_initial) can be calculated as V_initial = 0.0028 m^3/s / 0.00031416 m^2 ≈ 8.914 m/s.
3. Use Bernoulli's equation to relate the pressure and velocity between the known and unknown sections.
- Bernoulli's equation states P1 + 1/2 * ρ * V1^2 + ρ * g * h1 = P2 + 1/2 * ρ * V2^2 + ρ * g * h2.
- Since the pipe is horizontal, there is no change in height (h1 = h2), and the equation simplifies to P1 + 1/2 * ρ * V1^2 = P2 + 1/2 * ρ * V2^2.
- Since the flow is assumed nonviscous and laminar, we can ignore any frictional losses or changes in potential energy (ρ * g * h terms).
- Rearranging the equation, we get V2^2 = V1^2 + 2 * (P1 - P2) / ρ.
- Here, ρ is the density of water, which is approximately 1000 kg/m^3.
- Substituting the known values, V2^2 = (8.914 m/s)^2 + 2 * (143 kPa - 103 kPa) / 1000 kg/m^3.
4. Calculate the velocity (V2) at the constricted section of the pipe by taking the square root of V2^2 from Step 3.
5. Apply the principle of conservation of mass again to find the cross-sectional area (A_constricted) at the constricted section by rearranging the formula A_constricted = Q / V2.
6. Calculate the radius (r_constricted) at the constricted section by taking the square root of A_constricted / π.
7. Finally, calculate the diameter (d_constricted) at the constricted section by doubling the radius (d_constricted = 2 * r_constricted).
By following these steps, you should be able to determine the diameter of the constricted section of the pipe.