什麼是流體力學？
流體力學是研究流體內部和外部的力，以及流體的能量和運動的學科。對於任何涉及流體的工程問題——例如噴淋系統、水壩或常見的淡水和汙水管道的設計——都需要流體力學的知識。流體力學包含兩個子領域：流體靜力學，研究靜止的流體；流體動力學，研究運動的流體。然而，在詳細討論這兩個領域之前，我們有必要深入研究幾個基本但必要的概念和定義。
流體的定義
流體定義為一組分子，這些分子服從容器形狀。因此，液體和氣體都可以定義為流體。相比之下，固體往往保持其形狀，因此在容器形狀變化時不容易變形。很明顯，水、油和水銀都是流體的例子；相比之下，鋼管和混凝土都是固體的明顯例子。
流體的有用性質
流體具有一些在流體力學研究中至關重要的性質，其中最重要的是比重（γ）。γ 被定義為流體的重量除以其體積；因此，它的單位在 SI 中為 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}}}
壓力
流體上的壓力或由流體引起的壓力定義為所施加的力除以施加力的面積。從數學上來說，我們定義壓力為
P = F A {\displaystyle P={\begin{matrix}{\frac {\mathbf {F} }{A}}\end{matrix}}}
壓力的單位是帕斯卡（Pa）；1 帕斯卡等於 1 N m 2 {\displaystyle {\begin{matrix}{\frac {1N}{m^{2}}}\end{matrix}}}
流體靜力學，如前所述，是研究靜止流體的學科。例如，透過流體靜力學概念可以推匯出靜止水體對阻擋它的水壩側面的壓力。
本節將討論關於壓力的幾個方面，它們都與靜止流體池中壓力如何變化和作用的問題有關。例如，我們希望能夠回答諸如：在 5 米深的水中，會有多少帕斯卡的壓力？此外，我們希望能夠對流體壓力的作用方向做出結論；例如，作用在 5 米深潛水員身上的向上壓力是否等於作用在該潛水員身上的向下壓力？此外，潛水員上方的水量是否會影響潛水員所承受的壓力？換句話說，半徑為 1 釐米的 8 釐米高的玻璃杯底部壓力是否等於半徑為 8 釐米的 8 釐米高的玻璃杯底部壓力？我們將從回顧基礎物理學開始討論。
（順便說一句，通常認為“黑體”符號或字母表示向量。然而，在本例中，黑體用於強調，不一定表示向量。）
牛頓定律，再回顧
艾薩克·牛頓的第一和第二定律在數學形式上如下所示。
對於所有物體
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} }
我們想象一個靜止流體池中間任意一個流體立方體，如圖 1 所示。我們給作用在這個立方體上的未知壓力標上標籤，如圖 1 所示。讓我們將標有“1”的壓力設為“正”方向，我們將標有“2”的壓力設為“負”方向。請注意，這些符號是任意的，如果互換，會產生完全相同的結果。我們假設流體不會因作用在其上的壓力而壓縮。
圖 1 ：任意一個靜止流體立方體 。因此，我們有六個壓力： 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)}
按照通常的慣例，假設“右”為正，則 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，我們可能傾向於聲稱 Y 方向上的壓力推導與 X 和 Z 方向上的壓力推導基本相同。然而，仔細閱讀的讀者會注意到，在圖 1 中，沒有忽略流體的重量。流體的重量等於比重乘以體積
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. 壓強只隨流體深度的變化而變化；沒有其他尺寸變化會影響壓強。
2. 由於靜止流體中的一點必須沒有淨力作用在其上，因此壓強從各個角度均勻地作用。
P = ( h ) γ {\displaystyle \mathbf {P} =(h){\boldsymbol {\gamma }}}
h {\displaystyle \mathbf {h} }
例題 1 ：求水管中的壓強。
圖 2 ：水箱和水管裝置。
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 :
點C的絕對壓力是大氣壓， 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} }
例題 2 :尋找不漏水的管道長度 'r' 的最小值。
圖 3 : 尋找管道長度 'r'，以確保水不會從管道末端洩漏。 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)
