一个封闭水箱中水面上方是压缩空气,表压为 190 kPa。水从水面以下 4 m 处直径 25 mm 的孔口流入大气。忽略损失,求 流量 Q。 取 g = 9.81 m/s²。
pair = 190 kPa (表压)h = 4 md = 25 mmρ = 1000 kg/m³g = 9.81 m/s²
求: Q (L/s)
显示提示
- 从水面(表压 p,静止)到射流(大气压)应用伯努利方程。
- p/ρ + gh = ½v²,所以 v = √(2(p/ρ + gh)),p 以 Pa 计。
- p/ρ + gh = 229.2 m²/s².
显示答案和分步解答
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
从液面到射流的伯努利方程
v = √(2·(190000/1000 + 9.81·4)) = 21.41 m/s
流量
Q = A·v = 0.0004909·21.41 = 0.01051 m³/s = 10.51 L/s