Siphon flow calculation

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How to calculate the length and flow rate of a siphon tube? In fact, there is water flow resistance and energy loss during siphon flow. It is necessary to know the length and arrangement of the siphon tube. As a beginner, it can be assumed to be an ideal situation without considering energy loss. Assuming the height difference from the water source surface to the outlet of the siphon tube is H, the Bernoulli equation from the water source surface to the outlet of the siphon tube is obtained as H1=V ^ 2/(2g), and the siphon flow rate is V=(2gH1) ^ (1/2). The siphon flow rate is Q=(3.14D ^ 2/4) (2gH1) ^ (1/2), and D is the inner diameter of the siphon tube. Assuming the highest point pressure is P and the height difference from the highest point of the siphon tube to the outlet is H2, the Bernoulli equation from the highest point to the outlet can be obtained as follows: H2+P/(pg)+V ^ 2/(2g)=V ^ 2/(2g). Therefore, P=- pgH2 (relative pressure, excluding atmospheric pressure, is negative, meaning absolute pressure is less than atmospheric pressure, indicating a certain vacuum state. Theoretically, the maximum vacuum value cannot exceed 10 meters of water column, that is, H2<10 meters of water column). The Bernoulli equation from the liquid level of the container to the highest point can also be obtained as: 0=H3+P/(pg)+V ^ 2/(2g) P=- pg [H3+V ^ 2/(2g)]=- pg [H3+H1]=- pgH2（ The answer is the same as above) Of course, one of the working conditions of a siphon is that the siphon must be filled with water before it is rubbed in the siphon hall, and the pipeline must not be filled with air (the part where air is easily introduced is at the top of the siphon, because the pressure here is lower than atmospheric pressure, while the inlet and outlet at both ends of the siphon are greater than atmospheric pressure, making it difficult to introduce air). Therefore, the wall of the siphon cannot have holes or cracks. Due to the resistance and energy loss of actual water flow, the allowable installation height of the siphon tube vertex is much less than 10 meters for noisy plants! Explanation: In this example, without considering the energy loss of water flow and the uniform cross-section of the siphon tube, it is concluded that it is independent of the cross-sectional area, tube length, and flow velocity. But in reality, there is energy loss due to water flow, and the calculation is much more complicated than the above.

2. Siphon flow rate of one inch pipe

The siphon flow rate of one inch pipe is not a fixed value, and its calculation needs to be combined with variables such as height difference, pipe diameter, pipe length, and resistance coefficient.

1. A simple theoretical algorithm is derived based on the law of conservation of energy, with the formula Q=(π D ²/4) × √ (2gH ₁). • Parameter description: D=0.025m (1 inch pipe diameter), g=9.8m/s ², H ₁ is the vertical height from the water surface to the outlet. • Example demonstration: When H ₁=5m, Q ≈ [3.14 × (0.025) ²]/4 × √ (2 × 9.8 × 5)=0.0024m ³/s. Formula Q=√ (H/(SL)) • Parameter extension: S=10.3n ²/d ⁵ When n=0.01

2, d=0.205m, H=120m, L=1800m, S ≈ 6.911, Q ≈√ (120/(6.911 × 1800))=0.0982m ³/s, these two methods can cover the calculation needs of most siphon scenarios. Prioritizing the us

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