什么是流体力学？
流体力学是研究流体内部和外部的力，以及流体的能量和运动的学科。对于任何涉及流体的工程问题——例如喷淋系统、水坝或常见的淡水和污水管道的设计——都需要流体力学的知识。流体力学包含两个子领域：流体静力学，研究静止的流体；流体动力学，研究运动的流体。然而，在详细讨论这两个领域之前，我们有必要深入研究几个基本但必要的概念和定义。
流体的定义
流体定义为一组分子，这些分子服从容器形状。因此，液体和气体都可以定义为流体。相比之下，固体往往保持其形状，因此在容器形状变化时不容易变形。很明显，水、油和水银都是流体的例子；相比之下，钢管和混凝土都是固体的明显例子。
流体的有用性质
流体具有一些在流体力学研究中至关重要的性质，其中最重要的是比重（γ）。γ 被定义为流体的重量除以其体积；因此，它的单位在 SI 中为 N m 3 {\displaystyle {\begin{matrix}{\frac {N}{m^{3}}}\end{matrix}}} l b f t 3 {\displaystyle {\begin{matrix}{\frac {lb}{ft^{3}}}\end{matrix}}}
w = M g {\displaystyle \mathbf {w} =Mg}
m s 2 {\displaystyle {\begin{matrix}{\frac {m}{s^{2}}}\end{matrix}}} f t s 2 {\displaystyle {\begin{matrix}{\frac {ft}{s^{2}}}\end{matrix}}}
γ = w V {\displaystyle {\boldsymbol {\gamma }}={\begin{matrix}{\frac {w}{V}}\end{matrix}}}
ρ = M V {\displaystyle {\boldsymbol {\rho }}={\begin{matrix}{\frac {M}{V}}\end{matrix}}}
k g m 3 {\displaystyle {\begin{matrix}{\frac {kg}{m^{3}}}\end{matrix}}} s l u g s f t 3 {\displaystyle {\begin{matrix}{\frac {slugs}{ft^{3}}}\end{matrix}}}
γ = ρ g = M g V {\displaystyle {\boldsymbol {\gamma }}={\boldsymbol {\rho }}g={\begin{matrix}{\frac {Mg}{V}}\end{matrix}}}
压力
流体上的压力或由流体引起的压力定义为所施加的力除以施加力的面积。从数学上来说，我们定义压力为
P = F A {\displaystyle P={\begin{matrix}{\frac {\mathbf {F} }{A}}\end{matrix}}}
压力的单位是帕斯卡（Pa）；1 帕斯卡等于 1 N m 2 {\displaystyle {\begin{matrix}{\frac {1N}{m^{2}}}\end{matrix}}}
流体静力学，如前所述，是研究静止流体的学科。例如，通过流体静力学概念可以推导出静止水体对阻挡它的水坝侧面的压力。
本节将讨论关于压力的几个方面，它们都与静止流体池中压力如何变化和作用的问题有关。例如，我们希望能够回答诸如：在 5 米深的水中，会有多少帕斯卡的压力？此外，我们希望能够对流体压力的作用方向做出结论；例如，作用在 5 米深潜水员身上的向上压力是否等于作用在该潜水员身上的向下压力？此外，潜水员上方的水量是否会影响潜水员所承受的压力？换句话说，半径为 1 厘米的 8 厘米高的玻璃杯底部压力是否等于半径为 8 厘米的 8 厘米高的玻璃杯底部压力？我们将从回顾基础物理学开始讨论。
（顺便说一句，通常认为“黑体”符号或字母表示向量。然而，在本例中，黑体用于强调，不一定表示向量。）
牛顿定律，再回顾
艾萨克·牛顿的第一和第二定律在数学形式上如下所示。
对于所有物体
F = M a {\displaystyle \mathbf {F} =M\mathbf {a} }
a {\displaystyle \mathbf {a} } a {\displaystyle \mathbf {a} }
∑ F x = ∑ F y = ∑ F z = 0 {\displaystyle \sum \mathbf {F_{x}} =\sum \mathbf {F_{y}} =\sum \mathbf {F_{z}} =0}
a {\displaystyle \mathbf {a} }
我们想象一个静止流体池中间任意一个流体立方体，如图 1 所示。我们给作用在这个立方体上的未知压力标上标签，如图 1 所示。让我们将标有“1”的压力设为“正”方向，我们将标有“2”的压力设为“负”方向。请注意，这些符号是任意的，如果互换，会产生完全相同的结果。我们假设流体不会因作用在其上的压力而压缩。
图 1 ：任意一个静止流体立方体 。因此，我们有六个压力： P 1 x , P 2 x , P 1 y , P 2 y , P 1 z {\displaystyle \mathbf {P_{1x}} ,\mathbf {P_{2x}} ,\mathbf {P_{1y}} ,\mathbf {P_{2y}} ,\mathbf {P_{1z}} } P 2 z {\displaystyle \mathbf {P_{2z}} } P A {\displaystyle \mathbf {P_{A}} }
P A {\displaystyle \mathbf {P_{A}} } P 1 x {\displaystyle \mathbf {P_{1x}} } P 1 x {\displaystyle \mathbf {P_{1x}} }
δ P X = ∂ P ∂ x δ X ( 1 2 ) {\displaystyle {\boldsymbol {\delta P_{X}}}={\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)}
δ P X {\displaystyle {\boldsymbol {\delta P_{X}}}} P 1 x {\displaystyle \mathbf {P_{1x}} } P 2 x {\displaystyle \mathbf {P_{2x}} } δ X ( 1 2 ) {\displaystyle {\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)}
δ P Y = ∂ P ∂ y δ Y ( 1 2 ) {\displaystyle {\boldsymbol {\delta P_{Y}}}={\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)}
δ P Z = ∂ P ∂ z δ Z ( 1 2 ) {\displaystyle {\boldsymbol {\delta P_{Z}}}={\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta Z}}\left({\frac {1}{2}}\right)}
按照通常的惯例，假设“右”为正，则 P 2 x {\displaystyle \mathbf {P_{2x}} } P A {\displaystyle \mathbf {P_{A}} }
P 2 x = δ P X + P A = ∂ P ∂ x δ X ( 1 2 ) + P A {\displaystyle \mathbf {P_{2x}} ={\boldsymbol {\delta P_{X}}}+P_{A}={\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A}}
P 1 x {\displaystyle \mathbf {P_{1x}} }
P 1 x = − δ P X + P A = − ∂ P ∂ x δ X ( 1 2 ) + P A {\displaystyle \mathbf {P_{1x}} =-{\boldsymbol {\delta P_{X}}}+P_{A}=-{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A}}
P 1 y = − δ P Y + P A = − ∂ P ∂ y δ Y ( 1 2 ) + P A {\displaystyle \mathbf {P_{1y}} =-{\boldsymbol {\delta P_{Y}}}+P_{A}=-{\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A}}
P 2 y = δ P Y + P A = ∂ P ∂ y δ Y ( 1 2 ) + P A {\displaystyle \mathbf {P_{2y}} ={\boldsymbol {\delta P_{Y}}}+P_{A}={\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A}}
P 1 Z = − δ P Z + P A = − ∂ P ∂ z δ Z ( 1 2 ) + P A {\displaystyle \mathbf {P_{1Z}} =-{\boldsymbol {\delta P_{Z}}}+P_{A}=-{\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta Z}}\left({\frac {1}{2}}\right)+P_{A}}
P 2 Z = δ P Z + P A = ∂ P ∂ z δ Z ( 1 2 ) + P A {\displaystyle \mathbf {P_{2Z}} ={\boldsymbol {\delta P_{Z}}}+P_{A}={\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta Z}}\left({\frac {1}{2}}\right)+P_{A}}
∑ F T = 0 {\displaystyle \sum \mathbf {F_{T}} =0}
F 1 x = ( δ Z δ Y ) P 1 X = δ Z δ Y ( − ∂ P ∂ x δ X ( 1 2 ) + P A ) {\displaystyle \mathbf {F_{1x}} =({\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}})\mathbf {P_{1X}} ={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {-\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A})}
F = P A {\displaystyle \mathbf {F} =PA}
δ Z δ Y = A {\displaystyle {\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}=A}
F 2 x = ( δ Z δ Y ) P 2 X = δ Z δ Y ( ∂ P ∂ x δ X ( 1 2 ) + P A ) {\displaystyle \mathbf {F_{2x}} =({\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}})\mathbf {P_{2X}} ={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A})}
∑ F X = 0 = δ Z δ Y ( ∂ P ∂ x δ X ( 1 2 ) + P A ) − δ Z δ Y ( − ∂ P
0 = δ Z δ Y ∂ P ∂ x δ X ( 1 2 ) + P A + δ Z δ Y ∂ P ∂ x δ X ( 1 2 ) − P A {\displaystyle 0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A}+{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)-P_{A}}
0 = δ Z δ Y ∂ P ∂ x δ X ( 1 2 ) + δ Z δ Y ∂ P ∂ x δ X ( 1 2 ) {\displaystyle 0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)}
0 = δ Z δ Y δ X ∂ P ∂ x = ∂ P ∂ x δ V {\displaystyle 0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}={\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta V}}}
δ {\displaystyle {\boldsymbol {\delta }}} δ V {\displaystyle {\boldsymbol {\delta V}}}
P A {\displaystyle \mathbf {P_{A}} }
0 = δ Z δ Y δ X ∂ P ∂ z = ∂ P ∂ z δ V {\displaystyle 0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\frac {\partial {\boldsymbol {P}}}{\partial z}}={\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta V}}}
起初，我们可能倾向于声称 Y 方向上的压力推导与 X 和 Z 方向上的压力推导基本相同。然而，仔细阅读的读者会注意到，在图 1 中，没有忽略流体的重量。流体的重量等于比重乘以体积
δ Z δ Y δ X γ = δ V γ = W {\displaystyle {\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\boldsymbol {\gamma }}={\boldsymbol {\delta V\gamma }}={\boldsymbol {W}}}
∑ F Y = W = δ Z δ Y δ X γ = δ Z δ Y ( ∂ P ∂ y δ Y ( 1 2 ) + P A ) − δ Z δ Y ( − ∂ P ∂ y δ Y ( 1 2 ) + P A ) {\displaystyle \sum \mathbf {F_{Y}} =W={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\boldsymbol {\gamma }}={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A})-{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {-\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A})}
δ Z δ Y δ X γ = δ Z δ Y δ X ∂ P ∂ y = ∂ P ∂ y δ V {\displaystyle {\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\boldsymbol {\gamma }}={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\frac {\partial {\boldsymbol {P}}}{\partial y}}={\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta V}}}
γ = ∂ P ∂ y {\displaystyle {\boldsymbol {\gamma }}={\frac {\partial {\boldsymbol {P}}}{\partial y}}}
∫ P 1 P 2 d P = ∫ y 1 y 2 γ d y {\displaystyle \int _{P_{1}}^{P_{2}}dP=\int _{y_{1}}^{y_{2}}{\boldsymbol {\gamma }}dy}
Δ P = ( y 2 − y 1 ) γ {\displaystyle {\boldsymbol {\Delta P}}=(y_{2}-y_{1}){\boldsymbol {\gamma }}}
1. 压强只随流体深度的变化而变化；没有其他尺寸变化会影响压强。
2. 由于静止流体中的一点必须没有净力作用在其上，因此压强从各个角度均匀地作用。
P = ( h ) γ {\displaystyle \mathbf {P} =(h){\boldsymbol {\gamma }}}
h {\displaystyle \mathbf {h} }
例题 1 ：求水管中的压强。
图 2 ：水箱和水管装置。
i ) {\displaystyle {\boldsymbol {i)}}} P A {\displaystyle \mathbf {P_{A}} }
解 :
P A {\displaystyle \mathbf {P_{A}} } h A {\displaystyle \mathbf {h_{A}} } γ {\displaystyle \mathbf {\gamma } } γ {\displaystyle \mathbf {\gamma } }
P A {\displaystyle \mathbf {P_{A}} } 2 ( 9.8 ) ( K N m 3 ) m {\displaystyle 2(9.8)\left({\frac {KN}{m^{3}}}\right)m}
P A {\displaystyle \mathbf {P_{A}} } 19.6 ( K N m 2 ) {\displaystyle 19.6\left({\frac {KN}{m^{2}}}\right)}
P A {\displaystyle \mathbf {P_{A}} } 19.6 K P a {\displaystyle 19.6\mathbf {KPa} }
i i ) {\displaystyle {\boldsymbol {ii)}}} P B {\displaystyle \mathbf {P_{B}} }
解 :
P B {\displaystyle \mathbf {P_{B}} } [ h A + h B ] γ {\displaystyle [\mathbf {h_{A}} +\mathbf {h_{B}} ]\mathbf {\gamma } }
P B {\displaystyle \mathbf {P_{B}} } 3 ( 9.8 ) ( K N m 3 ) m {\displaystyle 3(9.8)\left({\frac {KN}{m^{3}}}\right)m}
P B {\displaystyle \mathbf {P_{B}} } 29.4 ( K N m 2 ) {\displaystyle 29.4\left({\frac {KN}{m^{2}}}\right)}
P B {\displaystyle \mathbf {P_{B}} } 29.4 K P a {\displaystyle 29.4\mathbf {KPa} }
i i i ) {\displaystyle {\boldsymbol {iii)}}} P c {\displaystyle \mathbf {P_{c}} }
解 :
点C的绝对压力是大气压， P 0 {\displaystyle {\boldsymbol {P_{0}}}}
i v ) {\displaystyle {\boldsymbol {iv)}}} P b {\displaystyle \mathbf {P_{b}} }
解 :
P b {\displaystyle \mathbf {P_{b}} } [ h A + h B ] γ + P 0 {\displaystyle [\mathbf {h_{A}} +\mathbf {h_{B}} ]\mathbf {\gamma } +\mathbf {P_{0}} }
P b {\displaystyle \mathbf {P_{b}} } 3 ( 9.8 ) + 101 K P a ( K N m 3 ) m {\displaystyle 3(9.8)+101\mathbf {KPa} \left({\frac {KN}{m^{3}}}\right)m}
P b {\displaystyle \mathbf {P_{b}} } 130.4 ( K N m 2 ) {\displaystyle 130.4\left({\frac {KN}{m^{2}}}\right)}
P b {\displaystyle \mathbf {P_{b}} } 130.4 K P a {\displaystyle 130.4\mathbf {KPa} }
例题 2 :寻找不漏水的管道长度 'r' 的最小值。
图 3 : 寻找管道长度 'r'，以确保水不会从管道末端泄漏。 466 K P a {\displaystyle 466\mathbf {KPa} }
![{\displaystyle [\mathbf {h_{A}} +\mathbf {h_{B}} ]\mathbf {\gamma } }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b729a70e0635f53cf7deb0b8b68483f8cd80c92)
![{\displaystyle [\mathbf {h_{A}} +\mathbf {h_{B}} ]\mathbf {\gamma } +\mathbf {P_{0}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ea8a15639c2ded840bc829779b851e989aca7dc)
