雷诺数

当液体流入通道时，它会与管道发生碰撞。工程师确保通过城市管道的液体流量尽可能一致。结果，一个称为雷诺数的数字可以预测液体的流动是平滑的还是湍流的。乔治斯托克爵士是第一个引入这个概念的人。后来被奥斯本·雷诺兹推广开来，这个数字被称为雷诺数。雷诺数是一个纯数，它指定有多少液体流入容器。因为空气和液体会运动，所以它们是流体，它们的运动是流动的。此外，我们将在本主题中研究称为雷诺数的流体流动参数以及雷诺数公式。流体是湍流还是层流取决于它的值。

雷诺数

它是惯性力与流体粘性力的比值。此外，惯性力是流动流的质量的动量。它衡量了在基本条件下调整移动流的速度有多困难。另一方面，粘性力是处理流动流体摩擦的力。此外，惯性力和粘性力非常相似。此外，它们共享相同的单位，这意味着雷诺数的单位更少。

The Reynolds number is a dimensionless quantity that determines whether a flow pattern is laminar or turbulent as it passes through a pipe. The ratio of inertial forces to viscous forces determines the Reynolds number.

在日常生活中，雷诺数具有多种潜力。它可用于描述管道中的液体流动、翼型流动或在流体中移动的物体。层流是一种流体流动，其中流体以可预测的模式平稳流动。另一方面，湍流本质上是不规则的并且涉及大量混合。

雷诺数公式

The formula for the Reynolds number is given as:

Re = (pVD) / u

Here,

Re is the Reynolds number, p is the density of fluid, V is the velocity of flow, D is the diameter of pipe and u is the viscosity of the fluid.

如果计算出的雷诺数很高，则称通过管道的流动是湍流。如果雷诺数较低，则称流动为层流。这些是可接受的数值，而层流和湍流通常在一个范围内分组。层流的雷诺数小于 2000 ，而湍流的雷诺数大于 2000 。

示例问题

问题 1：确定相对密度为 100 kg/m ³ 、粘度为 0.5 Ns/m ²的流体以 5 m/s 的速度通过 0.2 m 管道的流量。

解决方案：

The type of flow can be determined by the value of Reynolds number.

Given:

Velocity of fluid, V=5 m/s

Diameter of pipe, D= 0.2 m

Relative density of fluid, p=100 kg/m^3

Viscosity of fluid=0.5 Ns/m^2

The formula of Reynolds number is given as:

Re = pVD/u

Substitute all the values in the formula to calculate the Reynolds number.

Re = (100 kg/m³)(5 m/s)(0.2 m)/(0.5 Ns/m²)

    = 200

Since, the Reynolds number is less than 2000, the flow of liquid is laminar.

问题 2：假设我们正在研究通过管道流动的水。我们还想比较显示层流的两条管道的直径，即当水以直线平稳流动时。

此外，具有湍流的管道，其中水以混乱的方式流动，使流量预测具有挑战性。它还会产生振动，最终会导致流动系统磨损，从而导致其失效。

在我们的研究论文中，通过两条独立管道的水的雷诺数应为 2200。该组中的第一根管子的直径为 2.75 cm (0.0275 m)。此外，水的密度为 1,000 Kg/m^3。最重要的是，水的粘度为 0.0013 kg/(ms)。计算通过管道的水流速度以满足这些标准。

解决方案：

The formula of Reynolds number is given as:

Re = pVD/u

Substitute the given values to find out the velocity.

2200 = (1000 kg/m³)(0.0275 m)V/(0.00133 Pa⋅ s)

V ≈ 0.11 ms

Now we must use various pipes of various sizes with the same configuration, as well as compute the water velocity that must travel through it. Furthermore, when the water is running through the pipe at a calculated velocity, we will add a dye to it.

Also, if the dye water shows laminar flow, we’ve demonstrated how dynamic similitude works. Let’s finish the computation to estimate the water velocity that should flow through a smaller pipe.

Consider the pipe diameter be 0.005 m.

2200=(1000 kg/m³)(0.005 m)V/(0.00133 Pa⋅s)

V ≈ 0.59 ms

We arrive at the conclusion that water in a smaller pipe flows quicker than water in a larger conduit after the calculations. Also, because the Reynolds number is the same as in the preceding example, even when the water in the smaller pipe flows faster, the water still flows in a laminar pattern.

问题 3：层流的雷诺数应该是多少？

回答：

The fluid particles in laminar flow move in unmixing layers or streams that follow a smooth continuous course. The Reynolds number should be less than 2000 for laminar flow.

问题 4：湍流的雷诺数应该是多少？

回答：

The motion of fluid particles in turbulent flow is uneven, and it follows erratic and unpredictable patterns. The Reynolds number should be more than 2000 for turbulent flow.

问题 5：计算圆形管道中油流的雷诺数Re 。管道直径为60 mm，油的密度为910 kg/m ³ ，油的体积流量为60 L/min，油的动态粘度为50 m Pa s。

解决方案：

Given:

Diameter of the pipe, D = 60 mm = 0.06 m.

Density of the oil, p = 910 kg/m³

Volume flow rate, Q = 60 L/min = 0.01 m³/s.

Dynamic viscosity, u=50 m Pa s =0.05 Pa s

The area of the pipe is given as:

A = π(D/2)²

   = π(0.06/2)²

   = 0.0283 m²

The formula for volume flow rate is given as:

Q = Av

0.01 = 0.0283 × V

    V = 0.353 m/s

The formula of Reynolds number is given as:

Re = pVD/u

Substitute all the values to find the Reynolds number.

Re = (910 kg/m³)(0.353 m/s)(0.06 m)/(0.05 Pa s)

     = 386

Thus, the Reynolds number of the flow is 386.