A1 流体性质、量纲与黏性# A1.1 重要公式、符号与关系#
内容 公式 符号与关系 密度与重度 ρ = m / V \rho=m/V ρ = m / V γ = ρ g \gamma=\rho g γ = ρ g ρ \rho ρ γ \gamma γ 压缩系数 β p = − 1 V d V d p \beta_p=-\dfrac{1}{V}\dfrac{\mathrm dV}{\mathrm dp} β p = − V 1 d p d V 压强升高时体积减小，故带负号 体积模量 K = 1 β p = − V d p d V K=\dfrac1{\beta_p}=-V\dfrac{\mathrm dp}{\mathrm dV} K = β p 1 = − V d V d p K K K 牛顿内摩擦定律 τ = μ d u d y \tau=\mu\dfrac{\mathrm du}{\mathrm dy} τ = μ d y d u τ \tau τ μ \mu μ 运动黏度 ν = μ / ρ \nu=\mu/\rho ν = μ / ρ [ μ ] = M L − 1 T − 1 [\mu]=ML^{-1}T^{-1} [ μ ] = M L − 1 T − 1 [ ν ] = L 2 T − 1 [\nu]=L^2T^{-1} [ ν ] = L 2 T − 1 平板线性剪切 F = τ A = μ A U δ F=\tau A=\mu A\dfrac{U}{\delta} F = τ A = μ A δ U 适用于油膜内速度近似线性 毛细上升 h = 4 σ cos θ ρ g d h=\dfrac{4\sigma\cos\theta}{\rho gd} h = ρ g d 4 σ cos θ σ \sigma σ d d d 汽化条件 p a b s ≤ p v p_{\rm abs}\le p_v p abs ≤ p v p v p_v p v
关系图：
velocity gradient d u d y → × μ τ → × A F , μ = ρ ν . \text{velocity gradient}\ \frac{\mathrm du}{\mathrm dy}
\xrightarrow{\times\mu}\tau
\xrightarrow{\times A}F,
\qquad
\mu=\rho\nu. velocity gradient d y d u × μ τ × A F , μ = ρ ν . **做题识别：**看到“薄油膜、匀速、平板、线性速度分布”就优先写 τ = μ U / δ \tau=\mu U/\delta τ = μU / δ K = − V Δ p / Δ V K=-V\Delta p/\Delta V K = − V Δ p /Δ V
A1-1 体积模量# 【21–22 真题】
中文题目： 某液体的绝对压强由 1.0 × 10 6 P a 1.0\times10^6\ \mathrm{Pa} 1.0 × 1 0 6 Pa 2.0 × 10 6 P a 2.0\times10^6\ \mathrm{Pa} 2.0 × 1 0 6 Pa 1000 c m 3 1000\ \mathrm{cm^3} 1000 c m 3 995 c m 3 995\ \mathrm{cm^3} 995 c m 3 K K K
English version: The absolute pressure of a liquid increases from 1.0 × 10 6 P a 1.0\times10^6\ \mathrm{Pa} 1.0 × 1 0 6 Pa 2.0 × 10 6 P a 2.0\times10^6\ \mathrm{Pa} 2.0 × 1 0 6 Pa 1000 c m 3 1000\ \mathrm{cm^3} 1000 c m 3 995 c m 3 995\ \mathrm{cm^3} 995 c m 3 K K K
答案与讲解 采用小变形近似：
K = − V Δ p Δ V = − 1000 ( 2.0 − 1.0 ) × 10 6 995 − 1000 = 2.0 × 10 8 P a . K=-V\frac{\Delta p}{\Delta V}
=-1000\frac{(2.0-1.0)\times10^6}{995-1000}
=2.0\times10^8\ \mathrm{Pa}. K = − V Δ V Δ p = − 1000 995 − 1000 ( 2.0 − 1.0 ) × 1 0 6 = 2.0 × 1 0 8 Pa . K = 2.0 × 10 8 P a \boxed{K=2.0\times10^8\ \mathrm{Pa}} K = 2.0 × 1 0 8 Pa 负号用于抵消 Δ V < 0 \Delta V<0 Δ V < 0 K K K
A1-2 动力黏度与运动黏度的量纲# 【23–24 真题改编】
中文题目： 写出动力黏度 μ \mu μ ν \nu ν
English version: State the SI units and dimensions of dynamic viscosity μ \mu μ ν \nu ν
答案与讲解 由 τ = μ d u / d y \tau=\mu\,\mathrm du/\mathrm dy τ = μ d u / d y
[ μ ] = M L − 1 T − 2 T − 1 = M L − 1 T − 1 , [\mu]=\frac{ML^{-1}T^{-2}}{T^{-1}}=ML^{-1}T^{-1}, [ μ ] = T − 1 M L − 1 T − 2 = M L − 1 T − 1 , 故
μ : P a ⋅ s = N ⋅ s / m 2 = k g / ( m ⋅ s ) . \boxed{\mu:\ \mathrm{Pa\cdot s}=\mathrm{N\cdot s/m^2}=\mathrm{kg/(m\cdot s)}}. μ : Pa ⋅ s = N ⋅ s/ m 2 = kg/ ( m ⋅ s ) . 又 ν = μ / ρ \nu=\mu/\rho ν = μ / ρ
[ ν ] = M L − 1 T − 1 M L − 3 = L 2 T − 1 , [\nu]=\frac{ML^{-1}T^{-1}}{ML^{-3}}=L^2T^{-1}, [ ν ] = M L − 3 M L − 1 T − 1 = L 2 T − 1 , ν : m 2 / s , μ = ρ ν . \boxed{\nu:\ \mathrm{m^2/s}},\qquad \boxed{\mu=\rho\nu}. ν : m 2 /s , μ = ρ ν . A1-3 斜面油膜上的匀速木块# 【课件例题改编】
中文题目： 一木块质量 m = 5 k g m=5\ \mathrm{kg} m = 5 kg A = 0.40 × 0.45 m 2 A=0.40\times0.45\ \mathrm{m^2} A = 0.40 × 0.45 m 2 tan θ = 5 / 12 \tan\theta=5/12 tan θ = 5/12 U = 1.0 m / s U=1.0\ \mathrm{m/s} U = 1.0 m/s δ = 0.10 m m \delta=0.10\ \mathrm{mm} δ = 0.10 mm
English version: A wooden block of mass 5 k g 5\ \mathrm{kg} 5 kg 0.40 × 0.45 m 2 0.40\times0.45\ \mathrm{m^2} 0.40 × 0.45 m 2 tan θ = 5 / 12 \tan\theta=5/12 tan θ = 5/12 1.0 m / s 1.0\ \mathrm{m/s} 1.0 m/s 0.10 m m 0.10\ \mathrm{mm} 0.10 mm
答案与讲解 匀速说明沿斜面合力为零：
m g sin θ = τ A = μ U δ A . mg\sin\theta=\tau A=\mu\frac{U}{\delta}A. m g sin θ = τ A = μ δ U A . 由 tan θ = 5 / 12 \tan\theta=5/12 tan θ = 5/12 sin θ = 5 / 13 \sin\theta=5/13 sin θ = 5/13
μ = m g sin θ δ A U = 5 × 9.81 × ( 5 / 13 ) × 10 − 4 0.40 × 0.45 × 1 = 1.05 × 10 − 2 P a ⋅ s . \mu=\frac{mg\sin\theta\,\delta}{AU}
=\frac{5\times9.81\times(5/13)\times10^{-4}}{0.40\times0.45\times1}
=1.05\times10^{-2}\ \mathrm{Pa\cdot s}. μ = A U m g sin θ δ = 0.40 × 0.45 × 1 5 × 9.81 × ( 5/13 ) × 1 0 − 4 = 1.05 × 1 0 − 2 Pa ⋅ s . μ ≈ 0.0105 P a ⋅ s \boxed{\mu\approx0.0105\ \mathrm{Pa\cdot s}} μ ≈ 0.0105 Pa ⋅ s A1-4 两层液体中的剪应力平衡# 【23–24 真题改编】
中文题目： 一薄板以速度 U U U μ 1 = 0.12 P a ⋅ s \mu_1=0.12\ \mathrm{Pa\cdot s} μ 1 = 0.12 Pa ⋅ s μ 2 = 0.04 P a ⋅ s \mu_2=0.04\ \mathrm{Pa\cdot s} μ 2 = 0.04 Pa ⋅ s H = 8 m m H=8\ \mathrm{mm} H = 8 mm
English version: A thin plate moves horizontally at speed U U U μ 1 = 0.12 P a ⋅ s \mu_1=0.12\ \mathrm{Pa\cdot s} μ 1 = 0.12 Pa ⋅ s μ 2 = 0.04 P a ⋅ s \mu_2=0.04\ \mathrm{Pa\cdot s} μ 2 = 0.04 Pa ⋅ s 8 m m 8\ \mathrm{mm} 8 mm
答案与讲解 设上、下油膜厚度分别为 h 1 , h 2 h_1,h_2 h 1 , h 2
τ 1 = μ 1 U h 1 , τ 2 = μ 2 U h 2 . \tau_1=\mu_1\frac{U}{h_1},\qquad
\tau_2=\mu_2\frac{U}{h_2}. τ 1 = μ 1 h 1 U , τ 2 = μ 2 h 2 U . 令 τ 1 = τ 2 \tau_1=\tau_2 τ 1 = τ 2
h 1 h 2 = μ 1 μ 2 = 3 , h 1 + h 2 = 8 m m . \frac{h_1}{h_2}=\frac{\mu_1}{\mu_2}=3,
\qquad h_1+h_2=8\ \mathrm{mm}. h 2 h 1 = μ 2 μ 1 = 3 , h 1 + h 2 = 8 mm . 因此
h 1 = 6 m m , h 2 = 2 m m . \boxed{h_1=6\ \mathrm{mm},\quad h_2=2\ \mathrm{mm}}. h 1 = 6 mm , h 2 = 2 mm . 黏度较大的液层需要更厚，才能使速度梯度较小并保持剪应力相等。
A1-5 旋转圆盘的黏性转矩# 【教材例题改编】
中文题目： 直径 D = 0.10 m D=0.10\ \mathrm m D = 0.10 m δ = 1.0 m m \delta=1.0\ \mathrm{mm} δ = 1.0 mm μ = 0.10 P a ⋅ s \mu=0.10\ \mathrm{Pa\cdot s} μ = 0.10 Pa ⋅ s 600 r / m i n 600\ \mathrm{r/min} 600 r/min
English version: A disk of diameter 0.10 m 0.10\ \mathrm m 0.10 m 600 r p m 600\ \mathrm{rpm} 600 rpm 1.0 m m 1.0\ \mathrm{mm} 1.0 mm 0.10 P a ⋅ s 0.10\ \mathrm{Pa\cdot s} 0.10 Pa ⋅ s
答案与讲解 半径 r r r u = ω r u=\omega r u = ω r τ = μ ω r / δ \tau=\mu\omega r/\delta τ = μ ω r / δ d A = 2 π r d r \mathrm dA=2\pi r\,\mathrm dr d A = 2 π r d r
d T = τ d A r = 2 π μ ω δ r 3 d r . \mathrm dT=\tau\,\mathrm dA\,r
=\frac{2\pi\mu\omega}{\delta}r^3\,\mathrm dr. d T = τ d A r = δ 2 π μ ω r 3 d r . 积分得
T = π μ ω R 4 2 δ . T=\frac{\pi\mu\omega R^4}{2\delta}. T = 2 δ π μ ω R 4 . R = 0.05 m R=0.05\ \mathrm m R = 0.05 m ω = 2 π ( 600 / 60 ) = 62.83 r a d / s \omega=2\pi(600/60)=62.83\ \mathrm{rad/s} ω = 2 π ( 600/60 ) = 62.83 rad/s
T = 0.0617 N ⋅ m , P = T ω = 3.88 W . \boxed{T=0.0617\ \mathrm{N\cdot m}},
\qquad
\boxed{P=T\omega=3.88\ \mathrm W}. T = 0.0617 N ⋅ m , P = T ω = 3.88 W . A1-6 温度对黏性的影响# 【24–25 真题】
中文题目： 温度升高时，液体和气体的动力黏度分别如何变化？解释其微观原因。
English version: How does dynamic viscosity change with increasing temperature for liquids and gases? Explain the microscopic reasons.
答案与讲解 liquid viscosity decreases; gas viscosity increases . \boxed{\text{liquid viscosity decreases; gas viscosity increases}.} liquid viscosity decreases; gas viscosity increases . 液体黏性主要来自分子间吸引力；升温后分子间距增大，吸引作用减弱。气体黏性主要来自分子动量交换；升温后分子运动更剧烈，跨层动量交换增强。
A1-7 毛细上升# 【教材习题改编】
中文题目： 内径 d = 1.0 m m d=1.0\ \mathrm{mm} d = 1.0 mm σ = 0.0728 N / m \sigma=0.0728\ \mathrm{N/m} σ = 0.0728 N/m θ = 0 ∘ \theta=0^\circ θ = 0 ∘ ρ = 1000 k g / m 3 \rho=1000\ \mathrm{kg/m^3} ρ = 1000 kg/ m 3
English version: A clean glass capillary tube of internal diameter 1.0 m m 1.0\ \mathrm{mm} 1.0 mm σ = 0.0728 N / m \sigma=0.0728\ \mathrm{N/m} σ = 0.0728 N/m θ = 0 ∘ \theta=0^\circ θ = 0 ∘ ρ = 1000 k g / m 3 \rho=1000\ \mathrm{kg/m^3} ρ = 1000 kg/ m 3
答案与讲解 竖直方向表面张力与液柱重量平衡：
π d σ cos θ = ρ g π d 2 4 h . \pi d\sigma\cos\theta=\rho g\frac{\pi d^2}{4}h. π d σ cos θ = ρ g 4 π d 2 h . 所以
h = 4 σ cos θ ρ g d = 4 × 0.0728 1000 × 9.81 × 10 − 3 = 2.97 × 10 − 2 m . h=\frac{4\sigma\cos\theta}{\rho gd}
=\frac{4\times0.0728}{1000\times9.81\times10^{-3}}
=2.97\times10^{-2}\ \mathrm m. h = ρ g d 4 σ cos θ = 1000 × 9.81 × 1 0 − 3 4 × 0.0728 = 2.97 × 1 0 − 2 m . h = 29.7 m m \boxed{h=29.7\ \mathrm{mm}} h = 29.7 mm A1-8 汽化压强与空化# 【教材概念题改编】
中文题目： 某处水流绝对压强为 2.0 k P a 2.0\ \mathrm{kPa} 2.0 kPa 2.34 k P a 2.34\ \mathrm{kPa} 2.34 kPa
English version: At a point in a water flow, the absolute pressure is 2.0 k P a 2.0\ \mathrm{kPa} 2.0 kPa 2.34 k P a 2.34\ \mathrm{kPa} 2.34 kPa
答案与讲解 由于
p a b s = 2.0 k P a < p v = 2.34 k P a , p_{\rm abs}=2.0\ \mathrm{kPa}<p_v=2.34\ \mathrm{kPa}, p abs = 2.0 kPa < p v = 2.34 kPa , 液体可能局部汽化，存在空化风险。
汽化压强以绝对真空为零点，因此必须与绝对压强 比较。表压采用当地大气压为零点，零点不同，不能直接比较。
A2 静水压强、相对平衡与测压# A2.1 重要公式、符号与关系#
内容 公式 说明 静止流体平衡 ∇ p = ρ f \nabla p=\rho\mathbf f ∇ p = ρ f 重力场中 d p / d z = − ρ g \mathrm dp/\mathrm dz=-\rho g d p / d z = − ρ g 静水基本式 p + ρ g z = C p+\rho gz=C p + ρ g z = C 同一种、静止、不可压缩液体内成立 深度式 p = p 0 + ρ g h p=p_0+\rho gh p = p 0 + ρ g h h h h 表压 p g = p a b s − p a p_g=p_{\rm abs}-p_a p g = p abs − p a gauge pressure 真空度 p v ∗ = p a − p a b s = − p g p_v^{*}=p_a-p_{\rm abs}=-p_g p v ∗ = p a − p abs = − p g vacuum pressure 测压管水头 z + p / ( ρ g ) z+p/(\rho g) z + p / ( ρ g ) piezometric head 匀加速容器 d p = ρ ( − a x d x − g d z ) \mathrm dp=\rho(-a_x\mathrm dx-g\mathrm dz) d p = ρ ( − a x d x − g d z ) 自由面满足 d p = 0 \mathrm dp=0 d p = 0 绕竖轴匀速转动 p = p 0 + ρ g ( z 0 − z ) + 1 2 ρ ω 2 ( r 2 − r 0 2 ) p=p_0+\rho g(z_0-z)+\frac12\rho\omega^2(r^2-r_0^2) p = p 0 + ρ g ( z 0 − z ) + 2 1 ρ ω 2 ( r 2 − r 0 2 ) 自由面为旋转抛物面 差压计 ( z A + p A ρ g ) − ( z B + p B ρ g ) = ( ρ m ρ − 1 ) Δ h \left(z_A+\frac{p_A}{\rho g}\right)-\left(z_B+\frac{p_B}{\rho g}\right)=\left(\frac{\rho_m}{\rho}-1\right)\Delta h ( z A + ρ g p A ) − ( z B + ρ g p B ) = ( ρ ρ m − 1 ) Δ h 两侧管内为同一被测液体
压强换算链：
p a b s = p a + p g , p g < 0 ⇒ p v ∗ = − p g . p_{\rm abs}=p_a+p_g,
\qquad
p_g<0\Rightarrow p_v^{*}=-p_g. p abs = p a + p g , p g < 0 ⇒ p v ∗ = − p g . **测压计通用走法：**沿液柱从 A 走到 B，向下加 ρ g Δ h \rho g\Delta h ρ g Δ h ρ g Δ h \rho g\Delta h ρ g Δ h
A2-1 自由落体容器# 【24–25 真题】
中文题目： 一个盛水的敞口容器作自由落体。忽略表面张力，求容器内部各点的表压及壁面压强分布。
English version: An open container filled with water is in free fall. Neglect surface tension. Determine the gauge pressure and the pressure distribution on the walls.
答案与讲解 随容器一起运动的参考系中，有效重力
g e f f = g − a = g − g = 0. g_{\rm eff}=g-a=g-g=0. g eff = g − a = g − g = 0. 因此 ∇ p = 0 \nabla p=0 ∇ p = 0
p a b s = p a , p g = 0 everywhere . \boxed{p_{\rm abs}=p_a,\qquad p_g=0\quad\text{everywhere}.} p abs = p a , p g = 0 everywhere . 壁面表压也为零。
A2-2 绝对压强、表压与真空度# 【教材概念题改编】
中文题目： 当地大气压为 98 k P a 98\ \mathrm{kPa} 98 kPa − 25 k P a -25\ \mathrm{kPa} − 25 kPa
English version: The local atmospheric pressure is 98 k P a 98\ \mathrm{kPa} 98 kPa − 25 k P a -25\ \mathrm{kPa} − 25 kPa
答案与讲解 p a b s = p a + p g = 98 − 25 = 73 k P a . p_{\rm abs}=p_a+p_g=98-25=73\ \mathrm{kPa}. p abs = p a + p g = 98 − 25 = 73 kPa . 真空度
p v ∗ = p a − p a b s = 25 k P a . p_v^{*}=p_a-p_{\rm abs}=25\ \mathrm{kPa}. p v ∗ = p a − p abs = 25 kPa . p a b s = 73 k P a , p v ∗ = 25 k P a \boxed{p_{\rm abs}=73\ \mathrm{kPa},\quad p_v^{*}=25\ \mathrm{kPa}} p abs = 73 kPa , p v ∗ = 25 kPa A2-3 静水压强分布# 【教材例题改编】
中文题目： 密闭水箱自由液面上的表压为 40 k P a 40\ \mathrm{kPa} 40 kPa 3.0 m 3.0\ \mathrm m 3.0 m
English version: The gauge pressure above the free surface in a closed water tank is 40 k P a 40\ \mathrm{kPa} 40 kPa 3.0 m 3.0\ \mathrm m 3.0 m
答案与讲解 p = p 0 + ρ g h = 40 × 10 3 + 1000 × 9.81 × 3 = 69.43 k P a . p=p_0+\rho gh
=40\times10^3+1000\times9.81\times3
=69.43\ \mathrm{kPa}. p = p 0 + ρ g h = 40 × 1 0 3 + 1000 × 9.81 × 3 = 69.43 kPa . p ρ g = 69.43 × 10 3 1000 × 9.81 = 7.08 m . \frac{p}{\rho g}=\frac{69.43\times10^3}{1000\times9.81}=7.08\ \mathrm m. ρ g p = 1000 × 9.81 69.43 × 1 0 3 = 7.08 m . p = 69.4 k P a , p / ( ρ g ) = 7.08 m H 2 O \boxed{p=69.4\ \mathrm{kPa},\quad p/(\rho g)=7.08\ \mathrm{mH_2O}} p = 69.4 kPa , p / ( ρ g ) = 7.08 m H 2 O A2-4 水平匀加速容器的自由液面# 【教材例题改编】
中文题目： 一敞口矩形水箱沿 x x x a = 3.0 m / s 2 a=3.0\ \mathrm{m/s^2} a = 3.0 m/ s 2 d z / d x \mathrm dz/\mathrm dx d z / d x
English version: An open rectangular tank accelerates uniformly in the positive x x x 3.0 m / s 2 3.0\ \mathrm{m/s^2} 3.0 m/ s 2 d z / d x \mathrm dz/\mathrm dx d z / d x
答案与讲解 在随箱参考系中：
d p = − ρ a d x − ρ g d z . \mathrm dp=-\rho a\,\mathrm dx-\rho g\,\mathrm dz. d p = − ρ a d x − ρ g d z . 自由面上 d p = 0 \mathrm dp=0 d p = 0
d z d x = − a g = − 3.0 9.81 = − 0.306. \frac{\mathrm dz}{\mathrm dx}=-\frac{a}{g}
=-\frac{3.0}{9.81}=-0.306. d x d z = − g a = − 9.81 3.0 = − 0.306. d z / d x = − 0.306 \boxed{\mathrm dz/\mathrm dx=-0.306} d z / d x = − 0.306 沿加速度方向液面下降，后侧液面升高。
A2-5 匀速旋转容器# 【21–22 真题改编】
中文题目： 一敞口圆筒绕竖直轴以角速度 ω = 6 r a d / s \omega=6\ \mathrm{rad/s} ω = 6 rad/s r = 0.30 m r=0.30\ \mathrm m r = 0.30 m
English version: An open cylindrical container rotates steadily about its vertical axis at ω = 6 r a d / s \omega=6\ \mathrm{rad/s} ω = 6 rad/s r = 0.30 m r=0.30\ \mathrm m r = 0.30 m
答案与讲解 自由面满足
z − z 0 = ω 2 r 2 2 g . z-z_0=\frac{\omega^2r^2}{2g}. z − z 0 = 2 g ω 2 r 2 . 代入：
Δ z = 6 2 × 0.30 2 2 × 9.81 = 0.165 m . \Delta z=\frac{6^2\times0.30^2}{2\times9.81}=0.165\ \mathrm m. Δ z = 2 × 9.81 6 2 × 0.3 0 2 = 0.165 m . Δ z = 0.165 m \boxed{\Delta z=0.165\ \mathrm m} Δ z = 0.165 m 自由面为旋转抛物面；同一水平面上，压强随 r r r
A2-6 U 形水银压力计# 【教材习题改编】
中文题目： 水管中 A 点连接开口水银压力计。A 点比水—汞界面高 0.20 m 0.20\ \mathrm m 0.20 m 0.35 m 0.35\ \mathrm m 0.35 m ρ H g = 13600 k g / m 3 \rho_{Hg}=13600\ \mathrm{kg/m^3} ρ H g = 13600 kg/ m 3
English version: Point A in a water pipe is connected to an open mercury manometer. Point A is 0.20 m 0.20\ \mathrm m 0.20 m 0.35 m 0.35\ \mathrm m 0.35 m ρ H g = 13600 k g / m 3 \rho_{Hg}=13600\ \mathrm{kg/m^3} ρ H g = 13600 kg/ m 3
答案与讲解 从开口汞面向下到界面：压强增加 ρ H g g ( 0.35 ) \rho_{Hg}g(0.35) ρ H g g ( 0.35 ) ρ w g ( 0.20 ) \rho_wg(0.20) ρ w g ( 0.20 )
p A = ρ H g g ( 0.35 ) − ρ w g ( 0.20 ) . p_A=\rho_{Hg}g(0.35)-\rho_wg(0.20). p A = ρ H g g ( 0.35 ) − ρ w g ( 0.20 ) . p A = 13600 × 9.81 × 0.35 − 1000 × 9.81 × 0.20 = 44.73 k P a . p_A=13600\times9.81\times0.35-1000\times9.81\times0.20
=44.73\ \mathrm{kPa}. p A = 13600 × 9.81 × 0.35 − 1000 × 9.81 × 0.20 = 44.73 kPa . p A = 44.7 k P a ( g a u g e ) \boxed{p_A=44.7\ \mathrm{kPa\ (gauge)}} p A = 44.7 kPa ( gauge ) A2-7 水银差压计与测压管水头差# 【课件例题改编】
中文题目： 水管两断面 A、B 用水银差压计连接，水银液面差为 Δ h = 40 m m \Delta h=40\ \mathrm{mm} Δ h = 40 mm H p A − H p B H_{pA}-H_{pB} H p A − H pB 13.6 13.6 13.6
English version: Two sections A and B of a water pipe are connected by a mercury differential manometer. The mercury-level difference is 40 m m 40\ \mathrm{mm} 40 mm H p A − H p B H_{pA}-H_{pB} H p A − H pB 13.6 13.6 13.6
答案与讲解 对同一种被测液体：
H p A − H p B = ( ρ H g ρ w − 1 ) Δ h = ( 13.6 − 1 ) × 0.040. H_{pA}-H_{pB}
=\left(\frac{\rho_{Hg}}{\rho_w}-1\right)\Delta h
=(13.6-1)\times0.040. H p A − H pB = ( ρ w ρ H g − 1 ) Δ h = ( 13.6 − 1 ) × 0.040. H p A − H p B = 0.504 m \boxed{H_{pA}-H_{pB}=0.504\ \mathrm m} H p A − H pB = 0.504 m 注意这里求的是 z + p / ( ρ g ) z+p/(\rho g) z + p / ( ρ g )
A2-8 静水压强悖论# 【24–25 真题】
中文题目： 四个敞口容器底面积均为 A A A h h h
English version: Four open vessels have the same bottom area A A A h h h
答案与讲解 底部处表压均为 p = ρ g h p=\rho gh p = ρ g h
F b = ρ g h A . \boxed{F_b=\rho ghA}. F b = ρ g h A . 桌面对“容器 + 液体”整体的支持力等于总重量，通常不等于 F b F_b F b
上宽下窄：侧壁对液体有向下分力，F b F_b F b
上窄下宽：侧壁对液体有向上分力，F b F_b F b
A3 静水总压力、浮力与稳定# A3.1 重要公式、符号与关系#
内容 公式 说明 平面总压力 F = ρ g h c A F=\rho gh_cA F = ρ g h c A h c h_c h c 压力中心 y D = y c + I c A y c y_D=y_c+\dfrac{I_c}{Ay_c} y D = y c + A y c I c y y y 竖直深度 h D = y D sin θ h_D=y_D\sin\theta h D = y D sin θ θ \theta θ 曲面水平分力 F H = ρ g h c A v F_H=\rho gh_cA_v F H = ρ g h c A v 等于曲面竖直投影所受静水力 曲面竖直分力 F V = ρ g V p F_V=\rho gV_p F V = ρ g V p 等于压力体内液体重量，方向由压力体判定 合力 F = F H 2 + F V 2 F=\sqrt{F_H^2+F_V^2} F = F H 2 + F V 2 作用线由两分力交点确定 浮力 F B = ρ g V d i s p F_B=\rho gV_{disp} F B = ρ g V d i s p 作用于排水体积形心，即浮心 B 漂浮平衡 G = F B G=F_B G = F B ρ b V g = ρ f V d i s p g \rho_bVg=\rho_fV_{disp}g ρ b V g = ρ f V d i s p g 稳心半径 B M = I w / V d i s p BM=I_w/V_{disp} BM = I w / V d i s p I w I_w I w 初稳性 G M = K B + B M − K G GM=KB+BM-KG GM = K B + BM − K G G M > 0 GM>0 GM > 0 G M < 0 GM<0 GM < 0
做题主线：
平面：找 A , h c , I c A,h_c,I_c A , h c , I c F F F
曲面：拆成 F H , F V F_H,F_V F H , F V ρ g h c A \rho gh_cA ρ g h c A
漂浮：先用重量=浮力求排水体积，再做 G M GM GM
A3-1 竖直矩形闸门# 【教材例题改编】
中文题目： 一竖直矩形闸门宽 b = 2.0 m b=2.0\ \mathrm m b = 2.0 m H = 3.0 m H=3.0\ \mathrm m H = 3.0 m
English version: A vertical rectangular gate is 2.0 m 2.0\ \mathrm m 2.0 m 3.0 m 3.0\ \mathrm m 3.0 m
答案与讲解 形心水深 h c = H / 2 = 1.5 m h_c=H/2=1.5\ \mathrm m h c = H /2 = 1.5 m A = b H = 6 m 2 A=bH=6\ \mathrm{m^2} A = b H = 6 m 2
F = ρ g h c A = 1000 × 9.81 × 1.5 × 6 = 88.29 k N . F=\rho gh_cA=1000\times9.81\times1.5\times6
=88.29\ \mathrm{kN}. F = ρ g h c A = 1000 × 9.81 × 1.5 × 6 = 88.29 kN . 矩形上缘在自由面时，三角形压强分布的合力作用点在 2 H / 3 2H/3 2 H /3
F = 88.3 k N , h D = 2.0 m . \boxed{F=88.3\ \mathrm{kN},\qquad h_D=2.0\ \mathrm m}. F = 88.3 kN , h D = 2.0 m . A3-2 倾斜矩形平板# 【教材习题改编】
中文题目： 一矩形平板宽 1.5 m 1.5\ \mathrm m 1.5 m 2.0 m 2.0\ \mathrm m 2.0 m 30 ∘ 30^\circ 3 0 ∘ 1.0 m 1.0\ \mathrm m 1.0 m
English version: A rectangular plate is 1.5 m 1.5\ \mathrm m 1.5 m 2.0 m 2.0\ \mathrm m 2.0 m 30 ∘ 30^\circ 3 0 ∘ 1.0 m 1.0\ \mathrm m 1.0 m
答案与讲解 形心水深
h c = 1.0 + 2.0 2 sin 30 ∘ = 1.5 m . h_c=1.0+\frac{2.0}{2}\sin30^\circ=1.5\ \mathrm m. h c = 1.0 + 2 2.0 sin 3 0 ∘ = 1.5 m . F = ρ g h c A = 1000 × 9.81 × 1.5 × ( 1.5 × 2 ) = 44.15 k N . F=\rho gh_cA=1000\times9.81\times1.5\times(1.5\times2)
=44.15\ \mathrm{kN}. F = ρ g h c A = 1000 × 9.81 × 1.5 × ( 1.5 × 2 ) = 44.15 kN . 从自由面与板延长线交点沿板量，y c = 1 / sin 30 ∘ + 1 = 3 m y_c=1/\sin30^\circ+1=3\ \mathrm m y c = 1/ sin 3 0 ∘ + 1 = 3 m
I c = b L 3 12 = 1.0 m 4 , I_c=\frac{bL^3}{12}=1.0\ \mathrm{m^4}, I c = 12 b L 3 = 1.0 m 4 , y D = y c + I c A y c = 3 + 1 3 × 3 = 3.111 m . y_D=y_c+\frac{I_c}{Ay_c}=3+\frac{1}{3\times3}=3.111\ \mathrm m. y D = y c + A y c I c = 3 + 3 × 3 1 = 3.111 m . 竖直深度
h D = y D sin 30 ∘ = 1.556 m . h_D=y_D\sin30^\circ=1.556\ \mathrm m. h D = y D sin 3 0 ∘ = 1.556 m . F = 44.1 k N , h D = 1.56 m \boxed{F=44.1\ \mathrm{kN},\quad h_D=1.56\ \mathrm m} F = 44.1 kN , h D = 1.56 m A3-3 圆形闸门的压力中心# 【23–24 真题改编】
中文题目： 直径 d = 2.0 m d=2.0\ \mathrm m d = 2.0 m h c = 2.0 m h_c=2.0\ \mathrm m h c = 2.0 m
English version: A vertical circular gate of diameter 2.0 m 2.0\ \mathrm m 2.0 m 2.0 m 2.0\ \mathrm m 2.0 m
答案与讲解 A = π d 2 4 = π m 2 , I c = π d 4 64 = π 4 m 4 . A=\frac{\pi d^2}{4}=\pi\ \mathrm{m^2},
\qquad
I_c=\frac{\pi d^4}{64}=\frac{\pi}{4}\ \mathrm{m^4}. A = 4 π d 2 = π m 2 , I c = 64 π d 4 = 4 π m 4 . F = ρ g h c A = 1000 × 9.81 × 2 π = 61.64 k N . F=\rho gh_cA=1000\times9.81\times2\pi=61.64\ \mathrm{kN}. F = ρ g h c A = 1000 × 9.81 × 2 π = 61.64 kN . h D = h c + I c A h c = 2 + π / 4 π × 2 = 2.125 m . h_D=h_c+\frac{I_c}{Ah_c}
=2+\frac{\pi/4}{\pi\times2}=2.125\ \mathrm m. h D = h c + A h c I c = 2 + π × 2 π /4 = 2.125 m . F = 61.6 k N , h D = 2.125 m \boxed{F=61.6\ \mathrm{kN},\quad h_D=2.125\ \mathrm m} F = 61.6 kN , h D = 2.125 m A3-4 圆弧闸门的水平、竖直分力# 【21–22 真题改编】
中文题目： 一圆弧闸门半径 r = 4 m r=4\ \mathrm m r = 4 m L = 5 m L=5\ \mathrm m L = 5 m 30 ∘ 30^\circ 3 0 ∘ 1 m 1\ \mathrm m 1 m 2 m 2\ \mathrm m 2 m
English version: A cylindrical curved gate has radius 4 m 4\ \mathrm m 4 m 5 m 5\ \mathrm m 5 m 30 ∘ 30^\circ 3 0 ∘ 1 m 1\ \mathrm m 1 m 2 m 2\ \mathrm m 2 m
答案与讲解 竖直投影面积 A v = 5 × 2 = 10 m 2 A_v=5\times2=10\ \mathrm{m^2} A v = 5 × 2 = 10 m 2 h c = 1 + 1 = 2 m h_c=1+1=2\ \mathrm m h c = 1 + 1 = 2 m
F H = ρ g h c A v = 1000 × 9.8 × 2 × 10 = 196 k N . F_H=\rho gh_cA_v=1000\times9.8\times2\times10=196\ \mathrm{kN}. F H = ρ g h c A v = 1000 × 9.8 × 2 × 10 = 196 kN . 由圆弧上方压力体体积求得（对应原题几何）
F V = ρ g V p = 61.8 k N . F_V=\rho gV_p=61.8\ \mathrm{kN}. F V = ρ g V p = 61.8 kN . F = 196 2 + 61.8 2 = 205.5 k N . F=\sqrt{196^2+61.8^2}=205.5\ \mathrm{kN}. F = 19 6 2 + 61. 8 2 = 205.5 kN . F H = 196 k N , F V = 61.8 k N , F = 205.5 k N \boxed{F_H=196\ \mathrm{kN},\quad F_V=61.8\ \mathrm{kN},\quad F=205.5\ \mathrm{kN}} F H = 196 kN , F V = 61.8 kN , F = 205.5 kN 曲面题的关键是：F H F_H F H F V F_V F V
A3-5 铰接闸门的启门力矩# 【教材习题改编】
中文题目： 一竖直矩形闸门宽 b = 1.0 m b=1.0\ \mathrm m b = 1.0 m H = 2.0 m H=2.0\ \mathrm m H = 2.0 m
English version: A vertical rectangular gate is 1.0 m 1.0\ \mathrm m 1.0 m 2.0 m 2.0\ \mathrm m 2.0 m
答案与讲解 总压力
F = ρ g H 2 ( b H ) = 1 2 ρ g b H 2 = 19.62 k N . F=\rho g\frac{H}{2}(bH)=\frac12\rho gbH^2
=19.62\ \mathrm{kN}. F = ρ g 2 H ( b H ) = 2 1 ρ g b H 2 = 19.62 kN . 压力中心距自由面 2 H / 3 = 4 / 3 m 2H/3=4/3\ \mathrm m 2 H /3 = 4/3 m
M = F 2 H 3 = 19.62 × 4 3 = 26.16 k N ⋅ m . M=F\frac{2H}{3}=19.62\times\frac43=26.16\ \mathrm{kN\cdot m}. M = F 3 2 H = 19.62 × 3 4 = 26.16 kN ⋅ m . M = 26.2 k N ⋅ m \boxed{M=26.2\ \mathrm{kN\cdot m}} M = 26.2 kN ⋅ m A3-6 漂浮体浸没比例# 【教材概念题改编】
中文题目： 密度为 750 k g / m 3 750\ \mathrm{kg/m^3} 750 kg/ m 3
English version: A homogeneous wooden block of density 750 k g / m 3 750\ \mathrm{kg/m^3} 750 kg/ m 3
答案与讲解 漂浮平衡：
ρ b V g = ρ w V d i s p g . \rho_bVg=\rho_wV_{disp}g. ρ b V g = ρ w V d i s p g . 因此
V d i s p V = ρ b ρ w = 750 1000 = 0.75. \frac{V_{disp}}{V}=\frac{\rho_b}{\rho_w}=\frac{750}{1000}=0.75. V V d i s p = ρ w ρ b = 1000 750 = 0.75. 75 % of the volume is submerged . \boxed{75\%\text{ of the volume is submerged}.} 75% of the volume is submerged . A3-7 完全浸没物体的稳定性# 【教材概念题改编】
中文题目： 一个完全浸没且处于平衡的物体，重心 G 位于浮心 B 的上方。判断其平衡稳定性，并解释受小扰动后的力矩方向。
English version: A fully submerged body is in equilibrium, with its center of gravity G above its center of buoyancy B. Determine the stability of the equilibrium and explain the moment produced after a small angular disturbance.
答案与讲解 完全浸没时，浮心相对物体固定。小角度转动后，重力仍通过 G 向下，浮力通过 B 向上。若 G 在 B 上方，两力形成使偏转进一步增大的力偶。
The equilibrium is unstable . \boxed{\text{The equilibrium is unstable}.} The equilibrium is unstable . 若 G 在 B 下方则稳定；G 与 B 重合则为随遇平衡。
A3-8 漂浮驳船的初稳性# 【教材习题改编】
中文题目： 一矩形驳船长 L = 6 m L=6\ \mathrm m L = 6 m B = 2 m B=2\ \mathrm m B = 2 m T = 0.8 m T=0.8\ \mathrm m T = 0.8 m K G = 0.90 m KG=0.90\ \mathrm m K G = 0.90 m
English version: A rectangular barge is 6 m 6\ \mathrm m 6 m 2 m 2\ \mathrm m 2 m 0.8 m 0.8\ \mathrm m 0.8 m 0.90 m 0.90\ \mathrm m 0.90 m
答案与讲解 排水体积
V = L B T = 6 × 2 × 0.8 = 9.6 m 3 . V= LBT=6\times2\times0.8=9.6\ \mathrm{m^3}. V = L BT = 6 × 2 × 0.8 = 9.6 m 3 . 矩形水线面对纵轴的二次矩
I w = L B 3 12 = 6 × 2 3 12 = 4.0 m 4 . I_w=\frac{LB^3}{12}=\frac{6\times2^3}{12}=4.0\ \mathrm{m^4}. I w = 12 L B 3 = 12 6 × 2 3 = 4.0 m 4 . B M = I w V = 0.4167 m , K B = T 2 = 0.40 m . BM=\frac{I_w}{V}=0.4167\ \mathrm m,
\qquad KB=\frac{T}{2}=0.40\ \mathrm m. BM = V I w = 0.4167 m , K B = 2 T = 0.40 m . G M = K B + B M − K G = 0.40 + 0.4167 − 0.90 = − 0.0833 m . GM=KB+BM-KG=0.40+0.4167-0.90=-0.0833\ \mathrm m. GM = K B + BM − K G = 0.40 + 0.4167 − 0.90 = − 0.0833 m . G M < 0 , unstable \boxed{GM<0,\ \text{unstable}} GM < 0 , unstable 需要降低重心或增大水线面惯性矩（如增宽）。
A4 流体运动描述、加速度与连续性# A4.1 重要公式、符号与关系#
内容 公式 说明 欧拉描述 u = u ( x , y , z , t ) \mathbf u=\mathbf u(x,y,z,t) u = u ( x , y , z , t ) Eulerian description：观察空间固定点 拉格朗日描述 r = r ( a , b , c , t ) \mathbf r=\mathbf r(a,b,c,t) r = r ( a , b , c , t ) Lagrangian description：跟踪流体质点 质点加速度 D u D t = ∂ u ∂ t + ( u ⋅ ∇ ) u \dfrac{D\mathbf u}{Dt}=\dfrac{\partial\mathbf u}{\partial t}+(\mathbf u\cdot\nabla)\mathbf u D t D u = ∂ t ∂ u + ( u ⋅ ∇ ) u local + convective acceleration 流线方程 d x u = d y v = d z w \dfrac{\mathrm dx}{u}=\dfrac{\mathrm dy}{v}=\dfrac{\mathrm dz}{w} u d x = v d y = w d z 固定时刻的速度切线 迹线方程 d x d t = u \dfrac{\mathrm dx}{\mathrm dt}=u d t d x = u d y d t = v \dfrac{\mathrm dy}{\mathrm dt}=v d t d y = v d z d t = w \dfrac{\mathrm dz}{\mathrm dt}=w d t d z = w 同一流体质点的运动轨迹 一般连续方程 ∂ ρ ∂ t + ∇ ⋅ ( ρ u ) = 0 \dfrac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf u)=0 ∂ t ∂ ρ + ∇ ⋅ ( ρ u ) = 0 质量守恒 不可压缩连续方程 ∇ ⋅ u = 0 \nabla\cdot\mathbf u=0 ∇ ⋅ u = 0 ∂ u / ∂ x + ∂ v / ∂ y + ∂ w / ∂ z = 0 \partial u/\partial x+\partial v/\partial y+\partial w/\partial z=0 ∂ u / ∂ x + ∂ v / ∂ y + ∂ w / ∂ z = 0 恒定总流连续方程 ρ 1 v 1 A 1 = ρ 2 v 2 A 2 \rho_1v_1A_1=\rho_2v_2A_2 ρ 1 v 1 A 1 = ρ 2 v 2 A 2 不可压缩时 Q = v A = Q=vA= Q = v A =
四个高频判断：
恒定流动：∂ ( ) / ∂ t = 0 \partial()/\partial t=0 ∂ ( ) / ∂ t = 0 当地加速度 为零；
均匀流动：沿流向速度不变，所以迁移加速度 为零；
恒定流中仍可能有迁移加速度；
流线与迹线只在恒定流中重合。
A4-1 当地加速度与迁移加速度# 【24–25 真题】
中文题目： 填空：恒定流动的 ______ 加速度为零；均匀流动的 ______ 加速度为零。
English version: Fill in the blanks: In steady flow, the ______ acceleration is zero; in uniform flow, the ______ acceleration is zero.
答案与讲解 local acceleration; convective acceleration \boxed{\text{local acceleration; convective acceleration}} local acceleration; convective acceleration 恒定流只排除了时间变化，空间变化仍可存在；均匀流排除了沿程空间变化。
A4-2 恒定流中的流线与迹线# 【教材概念题改编】
中文题目： 说明流线和迹线的定义，并判断二者在非恒定流中是否一定重合。
English version: Define a streamline and a pathline, and state whether they necessarily coincide in unsteady flow.
答案与讲解
**Streamline：**某一时刻处处与当地速度矢量相切的曲线；
**Pathline：**某一指定流体质点随时间运动形成的轨迹。
在非恒定流中，速度场随时间变化，同一质点经过的位置与同一时刻不同质点的速度切线一般不同，因此
streamlines and pathlines do not generally coincide in unsteady flow . \boxed{\text{streamlines and pathlines do not generally coincide in unsteady flow}.} streamlines and pathlines do not generally coincide in unsteady flow . 恒定流中二者重合。
A4-3 求流线# 【教材例题改编】
中文题目： 二维恒定速度场为 u = x u=x u = x v = − y v=-y v = − y
English version: For the two-dimensional steady velocity field u = x u=x u = x v = − y v=-y v = − y
答案与讲解 流线方程
d y d x = v u = − y x . \frac{\mathrm dy}{\mathrm dx}=\frac{v}{u}=-\frac{y}{x}. d x d y = u v = − x y . 分离变量：
d y y = − d x x ⇒ ln ∣ y ∣ = − ln ∣ x ∣ + C . \frac{\mathrm dy}{y}=-\frac{\mathrm dx}{x}
\Rightarrow \ln|y|=-\ln|x|+C. y d y = − x d x ⇒ ln ∣ y ∣ = − ln ∣ x ∣ + C . 因此
x y = C . \boxed{xy=C}. x y = C . 流线是一族矩形双曲线。
A4-4 求迹线# 【教材例题改编】
中文题目： 对 A4-3 的速度场，某质点在 t = 0 t=0 t = 0 ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 )
English version: For the velocity field in A4-3, a fluid particle is at ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) t = 0 t=0 t = 0
答案与讲解 d x d t = x ⇒ x = x 0 e t , \frac{\mathrm dx}{\mathrm dt}=x
\Rightarrow x=x_0e^t, d t d x = x ⇒ x = x 0 e t , d y d t = − y ⇒ y = y 0 e − t . \frac{\mathrm dy}{\mathrm dt}=-y
\Rightarrow y=y_0e^{-t}. d t d y = − y ⇒ y = y 0 e − t . 所以
x = x 0 e t , y = y 0 e − t . \boxed{x=x_0e^t,\qquad y=y_0e^{-t}}. x = x 0 e t , y = y 0 e − t . 消去 t t t
x y = x 0 y 0 = C , xy=x_0y_0=C, x y = x 0 y 0 = C , 与流线方程一致。
A4-5 判断速度场是否满足不可压缩连续方程# 【24–25 真题】
中文题目： 对空间不可压缩流体，给定
u = x 2 + x y − y 2 , v = x 2 + y 2 , w = 0. u=x^2+xy-y^2,\qquad v=x^2+y^2,\qquad w=0. u = x 2 + x y − y 2 , v = x 2 + y 2 , w = 0. 判断该速度场是否可能存在。
English version: For an incompressible fluid, the velocity field is
u = x 2 + x y − y 2 , v = x 2 + y 2 , w = 0. u=x^2+xy-y^2,\qquad v=x^2+y^2,\qquad w=0. u = x 2 + x y − y 2 , v = x 2 + y 2 , w = 0. Determine whether this velocity field is physically admissible.
答案与讲解 不可压缩流必须满足
∂ u ∂ x + ∂ v ∂ y + ∂ w ∂ z = 0. \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0. ∂ x ∂ u + ∂ y ∂ v + ∂ z ∂ w = 0. 这里
∂ u ∂ x = 2 x + y , ∂ v ∂ y = 2 y . \frac{\partial u}{\partial x}=2x+y,
\qquad
\frac{\partial v}{\partial y}=2y. ∂ x ∂ u = 2 x + y , ∂ y ∂ v = 2 y . 因此
∇ ⋅ u = 2 x + 3 y ≠ 0 \nabla\cdot\mathbf u=2x+3y\ne0 ∇ ⋅ u = 2 x + 3 y = 0 （不可能在整个流场恒为零），故
the flow field is not admissible for an incompressible fluid . \boxed{\text{the flow field is not admissible for an incompressible fluid}.} the flow field is not admissible for an incompressible fluid . A4-6 由连续方程补出第三个速度分量# 【课件例题改编】
中文题目： 不可压缩三维流动中，u = 2 x u=2x u = 2 x v = − y v=-y v = − y w = 0 w=0 w = 0 z = 0 z=0 z = 0 w ( x , y , z ) w(x,y,z) w ( x , y , z )
English version: In a three-dimensional incompressible flow, u = 2 x u=2x u = 2 x v = − y v=-y v = − y w = 0 w=0 w = 0 z = 0 z=0 z = 0 w ( x , y , z ) w(x,y,z) w ( x , y , z )
答案与讲解 连续方程：
2 − 1 + ∂ w ∂ z = 0 ⇒ ∂ w ∂ z = − 1. 2-1+\frac{\partial w}{\partial z}=0
\Rightarrow \frac{\partial w}{\partial z}=-1. 2 − 1 + ∂ z ∂ w = 0 ⇒ ∂ z ∂ w = − 1. 积分：
w = − z + C ( x , y ) . w=-z+C(x,y). w = − z + C ( x , y ) . 由 w ( x , y , 0 ) = 0 w(x,y,0)=0 w ( x , y , 0 ) = 0 C = 0 C=0 C = 0
w = − z . \boxed{w=-z}. w = − z . A4-7 矩形明渠连续方程# 【24–25 真题】
中文题目： 矩形人工渠道宽 B = 2.1 m B=2.1\ \mathrm m B = 2.1 m h 1 = 0.5 m h_1=0.5\ \mathrm m h 1 = 0.5 m h 2 = 0.3 m h_2=0.3\ \mathrm m h 2 = 0.3 m v 1 = 3.0 m / s v_1=3.0\ \mathrm{m/s} v 1 = 3.0 m/s Q Q Q v 2 v_2 v 2
English version: A rectangular artificial channel is 2.1 m 2.1\ \mathrm m 2.1 m 0.5 m 0.5\ \mathrm m 0.5 m 0.3 m 0.3\ \mathrm m 0.3 m 3.0 m / s 3.0\ \mathrm{m/s} 3.0 m/s Q Q Q v 2 v_2 v 2
答案与讲解 Q = v 1 A 1 = v 1 B h 1 = 3.0 × 2.1 × 0.5 = 3.15 m 3 / s . Q=v_1A_1=v_1Bh_1
=3.0\times2.1\times0.5=3.15\ \mathrm{m^3/s}. Q = v 1 A 1 = v 1 B h 1 = 3.0 × 2.1 × 0.5 = 3.15 m 3 /s . v 2 = Q B h 2 = 3.15 2.1 × 0.3 = 5.0 m / s . v_2=\frac{Q}{Bh_2}
=\frac{3.15}{2.1\times0.3}=5.0\ \mathrm{m/s}. v 2 = B h 2 Q = 2.1 × 0.3 3.15 = 5.0 m/s . Q = 3.15 m 3 / s , v 2 = 5.0 m / s \boxed{Q=3.15\ \mathrm{m^3/s},\qquad v_2=5.0\ \mathrm{m/s}} Q = 3.15 m 3 /s , v 2 = 5.0 m/s A4-8 计算欧拉加速度# 【教材例题改编】
中文题目： 二维恒定流动速度场为 u = a x u=ax u = a x v = − a y v=-ay v = − a y a a a
English version: For the two-dimensional steady velocity field u = a x u=ax u = a x v = − a y v=-ay v = − a y a a a
答案与讲解 当地加速度为零。迁移加速度：
a x = u ∂ u ∂ x + v ∂ u ∂ y = ( a x ) ( a ) + ( − a y ) ( 0 ) = a 2 x , a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}
=(ax)(a)+(-ay)(0)=a^2x, a x = u ∂ x ∂ u + v ∂ y ∂ u = ( a x ) ( a ) + ( − a y ) ( 0 ) = a 2 x , a y = u ∂ v ∂ x + v ∂ v ∂ y = ( a x ) ( 0 ) + ( − a y ) ( − a ) = a 2 y . a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}
=(ax)(0)+(-ay)(-a)=a^2y. a y = u ∂ x ∂ v + v ∂ y ∂ v = ( a x ) ( 0 ) + ( − a y ) ( − a ) = a 2 y . a x = a 2 x , a y = a 2 y \boxed{a_x=a^2x,\qquad a_y=a^2y} a x = a 2 x , a y = a 2 y 这说明“恒定”不等于“无加速度”。
A5 势流、流函数与流网# A5.1 重要公式、符号与关系#
内容 直角坐标 极坐标 无旋条件 ∂ v / ∂ x − ∂ u / ∂ y = 0 \partial v/\partial x-\partial u/\partial y=0 ∂ v / ∂ x − ∂ u / ∂ y = 0 1 r ∂ ( r u θ ) ∂ r − 1 r ∂ u r ∂ θ = 0 \dfrac1r\dfrac{\partial(ru_\theta)}{\partial r}-\dfrac1r\dfrac{\partial u_r}{\partial\theta}=0 r 1 ∂ r ∂ ( r u θ ) − r 1 ∂ θ ∂ u r = 0 速度势 u = ∂ ϕ / ∂ x u=\partial\phi/\partial x u = ∂ ϕ / ∂ x v = ∂ ϕ / ∂ y v=\partial\phi/\partial y v = ∂ ϕ / ∂ y u r = ∂ ϕ / ∂ r u_r=\partial\phi/\partial r u r = ∂ ϕ / ∂ r u θ = ( 1 / r ) ∂ ϕ / ∂ θ u_\theta=(1/r)\partial\phi/\partial\theta u θ = ( 1/ r ) ∂ ϕ / ∂ θ 流函数 u = ∂ ψ / ∂ y u=\partial\psi/\partial y u = ∂ ψ / ∂ y v = − ∂ ψ / ∂ x v=-\partial\psi/\partial x v = − ∂ ψ / ∂ x u r = ( 1 / r ) ∂ ψ / ∂ θ u_r=(1/r)\partial\psi/\partial\theta u r = ( 1/ r ) ∂ ψ / ∂ θ u θ = − ∂ ψ / ∂ r u_\theta=-\partial\psi/\partial r u θ = − ∂ ψ / ∂ r 拉普拉斯方程 ∇ 2 ϕ = 0 \nabla^2\phi=0 ∇ 2 ϕ = 0 ∇ 2 ψ = 0 \nabla^2\psi=0 ∇ 2 ψ = 0 对二维不可压缩势流成立 Cauchy–Riemann ϕ x = ψ y \phi_x=\psi_y ϕ x = ψ y ϕ y = − ψ x \phi_y=-\psi_x ϕ y = − ψ x 势线与流线正交 两流线间流量 q ′ = Δ ψ q'=\Delta\psi q ′ = Δ ψ 二维流量，单位为 m 2 / s \mathrm{m^2/s} m 2 /s 流网测速 u ≈ Δ ψ / Δ n u\approx\Delta\psi/\Delta n u ≈ Δ ψ /Δ n 相邻流线间距越小，速度越大
典型基本流：
uniform flow: ϕ = U ( x cos α + y sin α ) , ψ = U ( y cos α − x sin α ) , source: ϕ = Q 2 π ln r , ψ = Q 2 π θ , free vortex: ϕ = Γ 2 π θ , ψ = − Γ 2 π ln r . \begin{aligned}
&\text{uniform flow: } \phi=U(x\cos\alpha+y\sin\alpha),\quad
\psi=U(y\cos\alpha-x\sin\alpha),\\
&\text{source: } \phi=\frac{Q}{2\pi}\ln r,\quad
\psi=\frac{Q}{2\pi}\theta,\\
&\text{free vortex: } \phi=\frac{\Gamma}{2\pi}\theta,\quad
\psi=-\frac{\Gamma}{2\pi}\ln r.
\end{aligned} uniform flow: ϕ = U ( x cos α + y sin α ) , ψ = U ( y cos α − x sin α ) , source: ϕ = 2 π Q ln r , ψ = 2 π Q θ , free vortex: ϕ = 2 π Γ θ , ψ = − 2 π Γ ln r . A5-1 判断是否为势流# 【教材例题改编】
中文题目： 二维速度场 u = 2 x + y u=2x+y u = 2 x + y v = x − 2 y v=x-2y v = x − 2 y
English version: Determine whether the two-dimensional velocity field u = 2 x + y u=2x+y u = 2 x + y v = x − 2 y v=x-2y v = x − 2 y
答案与讲解 二维无旋条件为
∂ v ∂ x = ∂ u ∂ y . \frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}. ∂ x ∂ v = ∂ y ∂ u . 这里
∂ v ∂ x = 1 , ∂ u ∂ y = 1. \frac{\partial v}{\partial x}=1,
\qquad
\frac{\partial u}{\partial y}=1. ∂ x ∂ v = 1 , ∂ y ∂ u = 1. 故
ω z = 0 , the flow is irrotational and a velocity potential exists . \boxed{\omega_z=0,\ \text{the flow is irrotational and a velocity potential exists}.} ω z = 0 , the flow is irrotational and a velocity potential exists . 同时 ∂ u / ∂ x + ∂ v / ∂ y = 2 − 2 = 0 \partial u/\partial x+\partial v/\partial y=2-2=0 ∂ u / ∂ x + ∂ v / ∂ y = 2 − 2 = 0
A5-2 由速度求势函数# 【教材例题改编】
中文题目： 对 A5-1 的速度场，求速度势 ϕ ( x , y ) \phi(x,y) ϕ ( x , y )
English version: For the velocity field in A5-1, determine the velocity potential ϕ ( x , y ) \phi(x,y) ϕ ( x , y )
答案与讲解 由 u = ϕ x = 2 x + y u=\phi_x=2x+y u = ϕ x = 2 x + y x x x
ϕ = x 2 + x y + f ( y ) . \phi=x^2+xy+f(y). ϕ = x 2 + x y + f ( y ) . 再利用 v = ϕ y = x + f ′ ( y ) = x − 2 y v=\phi_y=x+f'(y)=x-2y v = ϕ y = x + f ′ ( y ) = x − 2 y
f ′ ( y ) = − 2 y ⇒ f ( y ) = − y 2 + C . f'(y)=-2y\Rightarrow f(y)=-y^2+C. f ′ ( y ) = − 2 y ⇒ f ( y ) = − y 2 + C . ϕ = x 2 + x y − y 2 + C \boxed{\phi=x^2+xy-y^2+C} ϕ = x 2 + x y − y 2 + C 只对一个变量积分时，积分“常数”应写成另一个变量的函数。
A5-3 由速度求流函数# 【教材例题改编】
中文题目： 对 A5-1 的速度场，求流函数 ψ ( x , y ) \psi(x,y) ψ ( x , y )
English version: For the velocity field in A5-1, determine the stream function ψ ( x , y ) \psi(x,y) ψ ( x , y )
答案与讲解 由
u = ∂ ψ ∂ y = 2 x + y , u=\frac{\partial\psi}{\partial y}=2x+y, u = ∂ y ∂ ψ = 2 x + y , 对 y y y
ψ = 2 x y + 1 2 y 2 + f ( x ) . \psi=2xy+\frac12y^2+f(x). ψ = 2 x y + 2 1 y 2 + f ( x ) . 再由
v = − ∂ ψ ∂ x = x − 2 y , v=-\frac{\partial\psi}{\partial x}=x-2y, v = − ∂ x ∂ ψ = x − 2 y , 得到
− ( 2 y + f ′ ( x ) ) = x − 2 y ⇒ f ′ ( x ) = − x . -(2y+f'(x))=x-2y\Rightarrow f'(x)=-x. − ( 2 y + f ′ ( x )) = x − 2 y ⇒ f ′ ( x ) = − x . 所以
ψ = 2 x y + 1 2 y 2 − 1 2 x 2 + C . \boxed{\psi=2xy+\frac12y^2-\frac12x^2+C}. ψ = 2 x y + 2 1 y 2 − 2 1 x 2 + C . A5-4 任意方向均匀流# 【21–22 真题】
中文题目： 平面均匀流速度大小为 U U U x x x α \alpha α
English version: A two-dimensional uniform flow of speed U U U α \alpha α x x x
答案与讲解 速度分量
u = U cos α , v = U sin α . u=U\cos\alpha,\qquad v=U\sin\alpha. u = U cos α , v = U sin α . 积分得
ϕ = U ( x cos α + y sin α ) + C 1 , \boxed{\phi=U(x\cos\alpha+y\sin\alpha)+C_1}, ϕ = U ( x cos α + y sin α ) + C 1 , ψ = U ( y cos α − x sin α ) + C 2 . \boxed{\psi=U(y\cos\alpha-x\sin\alpha)+C_2}. ψ = U ( y cos α − x sin α ) + C 2 . ϕ = \phi= ϕ = ψ = \psi= ψ =
A5-5 极坐标中的点源# 【教材例题改编】
中文题目： 平面点源强度为 Q = 0.020 m 2 / s Q=0.020\ \mathrm{m^2/s} Q = 0.020 m 2 /s r = 0.10 m r=0.10\ \mathrm m r = 0.10 m ϕ \phi ϕ ψ \psi ψ
English version: A two-dimensional source has strength Q = 0.020 m 2 / s Q=0.020\ \mathrm{m^2/s} Q = 0.020 m 2 /s r = 0.10 m r=0.10\ \mathrm m r = 0.10 m
答案与讲解 圆周上单位宽流量守恒：
2 π r u r = Q . 2\pi r u_r=Q. 2 π r u r = Q . 所以
u r = Q 2 π r = 0.020 2 π × 0.10 = 0.0318 m / s . u_r=\frac{Q}{2\pi r}
=\frac{0.020}{2\pi\times0.10}=0.0318\ \mathrm{m/s}. u r = 2 π r Q = 2 π × 0.10 0.020 = 0.0318 m/s . ϕ = Q 2 π ln r + C , ψ = Q 2 π θ + C . \boxed{\phi=\frac{Q}{2\pi}\ln r+C},
\qquad
\boxed{\psi=\frac{Q}{2\pi}\theta+C}. ϕ = 2 π Q ln r + C , ψ = 2 π Q θ + C . A5-6 两条流线之间的流量与平均流速# 【24–25 真题】
中文题目： 两条相邻流线的流函数差为 Δ ψ \Delta\psi Δ ψ Δ n \Delta n Δ n
English version: The difference in stream-function values between two adjacent streamlines is Δ ψ \Delta\psi Δ ψ Δ n \Delta n Δ n
答案与讲解 二维不可压缩流中：
q ′ = Δ ψ . \boxed{q'=\Delta\psi}. q ′ = Δ ψ . 单位宽度过流面积近似为 Δ n × 1 \Delta n\times1 Δ n × 1
u ˉ ≈ Δ ψ Δ n . \boxed{\bar u\approx\frac{\Delta\psi}{\Delta n}}. u ˉ ≈ Δ n Δ ψ . 流线越密集，Δ n \Delta n Δ n
A5-7 均匀流与点源叠加的驻点# 【教材例题改编】
中文题目： 沿 + x +x + x U U U Q Q Q
English version: A uniform flow of speed U U U x x x Q Q Q
答案与讲解 极坐标速度：
u r = U cos θ + Q 2 π r , u θ = − U sin θ . u_r=U\cos\theta+\frac{Q}{2\pi r},
\qquad
u_\theta=-U\sin\theta. u r = U cos θ + 2 π r Q , u θ = − U sin θ . 驻点要求两分量均为零。由 u θ = 0 u_\theta=0 u θ = 0 θ = 0 \theta=0 θ = 0 π \pi π θ = π \theta=\pi θ = π
− U + Q 2 π r s = 0. -U+\frac{Q}{2\pi r_s}=0. − U + 2 π r s Q = 0. 因此
r s = Q 2 π U , θ = π , \boxed{r_s=\frac{Q}{2\pi U},\quad \theta=\pi}, r s = 2 π U Q , θ = π , 即
x s = − Q 2 π U , y s = 0 . \boxed{x_s=-\frac{Q}{2\pi U},\quad y_s=0}. x s = − 2 π U Q , y s = 0 . A5-8 流网性质# 【23–24 真题改编】
中文题目： 简述二维不可压缩势流流网的三个主要性质，并说明如何由流网判断高速区。
English version: State three principal properties of a flow net for a two-dimensional incompressible potential flow, and explain how a high-speed region can be identified from the flow net.
答案与讲解
流线 ψ = \psi= ψ = ϕ = \phi= ϕ =
两族曲线组成近似曲线正方形网格；
相邻流线之间的流量 Δ ψ \Delta\psi Δ ψ
由
u ≈ Δ ψ Δ n , u\approx\frac{\Delta\psi}{\Delta n}, u ≈ Δ n Δ ψ , 可知流线或等势线越密集，速度越大。
A6 总流连续方程、伯努利方程与水头线# A6.1 重要公式、符号与关系#
内容 公式 说明 总流连续方程 Q = v A Q=vA Q = v A 不可压缩恒定总流，各断面相同 实际流体伯努利方程 z 1 + p 1 ρ g + α 1 v 1 2 2 g + H p = z 2 + p 2 ρ g + α 2 v 2 2 2 g + H t + h w z_1+\dfrac{p_1}{\rho g}+\alpha_1\dfrac{v_1^2}{2g}+H_p=z_2+\dfrac{p_2}{\rho g}+\alpha_2\dfrac{v_2^2}{2g}+H_t+h_w z 1 + ρ g p 1 + α 1 2 g v 1 2 + H p = z 2 + ρ g p 2 + α 2 2 g v 2 2 + H t + h w H p H_p H p H t H_t H t 测压管水头 H p ( i e z ) = z + p / ( ρ g ) H_p^{(iez)}=z+p/(\rho g) H p ( i ez ) = z + p / ( ρ g ) hydraulic/piezometric head 总水头 H = z + p / ( ρ g ) + α v 2 / ( 2 g ) H=z+p/(\rho g)+\alpha v^2/(2g) H = z + p / ( ρ g ) + α v 2 / ( 2 g ) total head 水头损失 h w = H 1 − H 2 h_w=H_1-H_2 h w = H 1 − H 2 实际流动沿程总水头下降 毕托管 v = c 2 g Δ h v=c\sqrt{2g\Delta h} v = c 2 g Δ h c c c 文丘里流量计 连续方程 + 伯努利方程 + 差压计 先把差压计读数换成测压管水头差 泵轴功率 P s h a f t = ρ g Q H p / η P_{shaft}=\rho gQH_p/\eta P s ha f t = ρ g Q H p / η η \eta η
三选一列：
选基准面；
选两个渐变流/均匀流断面；
选断面代表点；
列伯努利方程，并用连续方程消元。
水头线：
EGL − HGL = α v 2 2 g . \text{EGL}-\text{HGL}=\alpha\frac{v^2}{2g}. EGL − HGL = α 2 g v 2 . 实际流动 EGL 沿流向下降；HGL 可以上升、下降或水平。
A6-1 水箱自由出流# 【教材例题改编】
中文题目： 大水箱通过小孔自由出流，孔中心位于自由液面下 H = 4.0 m H=4.0\ \mathrm m H = 4.0 m
English version: A large tank discharges freely through a small orifice located 4.0 m 4.0\ \mathrm m 4.0 m
答案与讲解 自由面和出口均为大气压：
H = v 2 2 g . H=\frac{v^2}{2g}. H = 2 g v 2 . v = 2 g H = 2 × 9.81 × 4 = 8.86 m / s \boxed{v=\sqrt{2gH}=\sqrt{2\times9.81\times4}=8.86\ \mathrm{m/s}} v = 2 g H = 2 × 9.81 × 4 = 8.86 m/s 这是 Torricelli 定律。
A6-2 突然扩大后测压管水头为何升高# 【21–22 真题】
中文题目： 水平管道突然扩大，A 2 > A 1 A_2>A_1 A 2 > A 1
English version: A horizontal pipe undergoes a sudden expansion, with A 2 > A 1 A_2>A_1 A 2 > A 1
答案与讲解 连续方程给出 v 1 > v 2 v_1>v_2 v 1 > v 2
H p 1 + v 1 2 2 g = H p 2 + v 2 2 2 g + h j . H_{p1}+\frac{v_1^2}{2g}
=H_{p2}+\frac{v_2^2}{2g}+h_j. H p 1 + 2 g v 1 2 = H p 2 + 2 g v 2 2 + h j . 突然扩大损失
h j = ( v 1 − v 2 ) 2 2 g . h_j=\frac{(v_1-v_2)^2}{2g}. h j = 2 g ( v 1 − v 2 ) 2 . 所以
H p 2 − H p 1 = v 1 2 − v 2 2 − ( v 1 − v 2 ) 2 2 g = v 2 ( v 1 − v 2 ) g > 0. H_{p2}-H_{p1}
=\frac{v_1^2-v_2^2-(v_1-v_2)^2}{2g}
=\frac{v_2(v_1-v_2)}{g}>0. H p 2 − H p 1 = 2 g v 1 2 − v 2 2 − ( v 1 − v 2 ) 2 = g v 2 ( v 1 − v 2 ) > 0. H p 2 > H p 1 \boxed{H_{p2}>H_{p1}} H p 2 > H p 1 动能的一部分转化为压强能，另一部分耗散。
A6-3 虹吸管顶部负压# 【课件例题改编】
中文题目： 等径虹吸管出口比上游水箱自由液面低 3.0 m 3.0\ \mathrm m 3.0 m 2.0 m 2.0\ \mathrm m 2.0 m 0.6 v 2 / ( 2 g ) 0.6v^2/(2g) 0.6 v 2 / ( 2 g ) 0.5 v 2 / ( 2 g ) 0.5v^2/(2g) 0.5 v 2 / ( 2 g ) α = 1 \alpha=1 α = 1
English version: In a constant-diameter siphon, the outlet is 3.0 m 3.0\ \mathrm m 3.0 m 2.0 m 2.0\ \mathrm m 2.0 m 0.6 v 2 / ( 2 g ) 0.6v^2/(2g) 0.6 v 2 / ( 2 g ) 0.5 v 2 / ( 2 g ) 0.5v^2/(2g) 0.5 v 2 / ( 2 g ) α = 1 \alpha=1 α = 1
答案与讲解 自由面至出口：
3.0 = ( 1 + 0.6 + 0.5 ) v 2 2 g . 3.0=\left(1+0.6+0.5\right)\frac{v^2}{2g}. 3.0 = ( 1 + 0.6 + 0.5 ) 2 g v 2 . v 2 2 g = 1.4286 m , v = 5.29 m / s . \frac{v^2}{2g}=1.4286\ \mathrm m,
\qquad
\boxed{v=5.29\ \mathrm{m/s}}. 2 g v 2 = 1.4286 m , v = 5.29 m/s . 自由面至顶部：
0 = 2.0 + p 2 ρ g + v 2 2 g + 0.6 v 2 2 g . 0=2.0+\frac{p_2}{\rho g}+\frac{v^2}{2g}+0.6\frac{v^2}{2g}. 0 = 2.0 + ρ g p 2 + 2 g v 2 + 0.6 2 g v 2 . p 2 ρ g = − 2.0 − 1.6 ( 1.4286 ) = − 4.29 m . \frac{p_2}{\rho g}=-2.0-1.6(1.4286)=-4.29\ \mathrm m. ρ g p 2 = − 2.0 − 1.6 ( 1.4286 ) = − 4.29 m . p 2 / ( ρ g ) = − 4.29 m H 2 O \boxed{p_2/(\rho g)=-4.29\ \mathrm{mH_2O}} p 2 / ( ρ g ) = − 4.29 m H 2 O A6-4 毕托管测速# 【教材例题改编】
中文题目： 毕托管读得动压水头差 Δ h = 0.25 m \Delta h=0.25\ \mathrm m Δ h = 0.25 m c = 0.98 c=0.98 c = 0.98
English version: A Pitot tube measures a dynamic-head difference of 0.25 m 0.25\ \mathrm m 0.25 m 0.98 0.98 0.98
答案与讲解 v = c 2 g Δ h = 0.98 2 × 9.81 × 0.25 . v=c\sqrt{2g\Delta h}
=0.98\sqrt{2\times9.81\times0.25}. v = c 2 g Δ h = 0.98 2 × 9.81 × 0.25 . v = 2.17 m / s \boxed{v=2.17\ \mathrm{m/s}} v = 2.17 m/s A6-5 铅直文丘里流量计# 【24–25 真题】
中文题目： 铅直放置的文丘里管测量水流量，d 1 = 0.40 m d_1=0.40\ \mathrm m d 1 = 0.40 m d 1 / d 2 = 2 d_1/d_2=2 d 1 / d 2 = 2 Δ h = 30 m m \Delta h=30\ \mathrm{mm} Δ h = 30 mm 0.2 v 2 2 / ( 2 g ) 0.2v_2^2/(2g) 0.2 v 2 2 / ( 2 g ) α 1 = 1.0 \alpha_1=1.0 α 1 = 1.0 α 2 = 1.2 \alpha_2=1.2 α 2 = 1.2 13.6 13.6 13.6
English version: A vertical Venturi meter measures water flow. The inlet diameter is 0.40 m 0.40\ \mathrm m 0.40 m d 1 / d 2 = 2 d_1/d_2=2 d 1 / d 2 = 2 30 m m 30\ \mathrm{mm} 30 mm 0.2 v 2 2 / ( 2 g ) 0.2v_2^2/(2g) 0.2 v 2 2 / ( 2 g ) α 1 = 1.0 \alpha_1=1.0 α 1 = 1.0 α 2 = 1.2 \alpha_2=1.2 α 2 = 1.2 13.6 13.6 13.6
答案与讲解 差压计给出测压管水头差：
H p 1 − H p 2 = ( 13.6 − 1 ) ( 0.030 ) = 0.378 m . H_{p1}-H_{p2}=(13.6-1)(0.030)=0.378\ \mathrm m. H p 1 − H p 2 = ( 13.6 − 1 ) ( 0.030 ) = 0.378 m . 连续方程：
A 1 A 2 = ( d 1 d 2 ) 2 = 4 , v 2 = 4 v 1 . \frac{A_1}{A_2}=\left(\frac{d_1}{d_2}\right)^2=4,
\qquad v_2=4v_1. A 2 A 1 = ( d 2 d 1 ) 2 = 4 , v 2 = 4 v 1 . 伯努利方程：
0.378 = ( α 2 + 0.2 ) v 2 2 2 g − α 1 v 1 2 2 g . 0.378
=\left(\alpha_2+0.2\right)\frac{v_2^2}{2g}
-\alpha_1\frac{v_1^2}{2g}. 0.378 = ( α 2 + 0.2 ) 2 g v 2 2 − α 1 2 g v 1 2 . 代入 v 1 = v 2 / 4 v_1=v_2/4 v 1 = v 2 /4
0.378 = ( 1.4 − 1 16 ) v 2 2 2 g . 0.378=\left(1.4-\frac1{16}\right)\frac{v_2^2}{2g}. 0.378 = ( 1.4 − 16 1 ) 2 g v 2 2 . 解得
v 2 = 2.355 m / s . v_2=2.355\ \mathrm{m/s}. v 2 = 2.355 m/s . Q = A 2 v 2 = π ( 0.20 ) 2 4 × 2.355 = 7.40 × 10 − 2 m 3 / s . Q=A_2v_2=\frac{\pi(0.20)^2}{4}\times2.355
=7.40\times10^{-2}\ \mathrm{m^3/s}. Q = A 2 v 2 = 4 π ( 0.20 ) 2 × 2.355 = 7.40 × 1 0 − 2 m 3 /s . Q ≈ 0.0740 m 3 / s \boxed{Q\approx0.0740\ \mathrm{m^3/s}} Q ≈ 0.0740 m 3 /s A6-6 总水头线与测压管水头线# 【24–25 真题改编】
中文题目： 对一段实际等径管流，说明总水头线 EGL 和测压管水头线 HGL 的相对位置与沿程变化；若管径突然缩小，两线间距如何变化？
English version: For real flow in a constant-diameter pipe, describe the relative positions and streamwise variations of the energy grade line (EGL) and hydraulic grade line (HGL). What happens to their separation when the pipe diameter suddenly decreases?
答案与讲解 EGL = z + p ρ g + α v 2 2 g , HGL = z + p ρ g . \text{EGL}=z+\frac p{\rho g}+\alpha\frac{v^2}{2g},
\qquad
\text{HGL}=z+\frac p{\rho g}. EGL = z + ρ g p + α 2 g v 2 , HGL = z + ρ g p . 因此 EGL 始终位于 HGL 上方，间距为 α v 2 / ( 2 g ) \alpha v^2/(2g) α v 2 / ( 2 g )
突然缩小时 v v v
A6-7 自由射流末端的水头线# 【教材概念题改编】
中文题目： 管道末端向大气自由出流。说明末端断面 HGL 和 EGL 分别位于何处。
English version: A pipe discharges freely into the atmosphere. State the locations of the HGL and EGL at the outlet section.
答案与讲解 出口处 p g = 0 p_g=0 p g = 0
HGL = z o u t , \text{HGL}=z_{out}, HGL = z o u t , 即 HGL 通过出口断面的中心（若以中心为代表点）。EGL 比 HGL 高一个速度水头：
EGL = z o u t + α v 2 2 g . \text{EGL}=z_{out}+\alpha\frac{v^2}{2g}. EGL = z o u t + α 2 g v 2 . 若出口进入大水池并最终速度降为零，则出口速度水头通常作为出口局部损失耗散。
A6-8 泵的轴功率# 【教材习题改编】
中文题目： 水泵输水流量 Q = 0.050 m 3 / s Q=0.050\ \mathrm{m^3/s} Q = 0.050 m 3 /s H p = 20 m H_p=20\ \mathrm m H p = 20 m η = 0.80 \eta=0.80 η = 0.80
English version: A pump delivers water at Q = 0.050 m 3 / s Q=0.050\ \mathrm{m^3/s} Q = 0.050 m 3 /s 20 m 20\ \mathrm m 20 m 0.80 0.80 0.80
答案与讲解 水流获得的水力功率：
P h = ρ g Q H p . P_h=\rho gQH_p. P h = ρ g Q H p . 轴功率应除以效率：
P s h a f t = ρ g Q H p η = 1000 × 9.81 × 0.050 × 20 0.80 = 12.26 k W . P_{shaft}=\frac{\rho gQH_p}{\eta}
=\frac{1000\times9.81\times0.050\times20}{0.80}
=12.26\ \mathrm{kW}. P s ha f t = η ρ g Q H p = 0.80 1000 × 9.81 × 0.050 × 20 = 12.26 kW . P s h a f t = 12.3 k W \boxed{P_{shaft}=12.3\ \mathrm{kW}} P s ha f t = 12.3 kW A7 动量方程与水流作用力# A7.1 重要公式、符号与关系#
内容 公式 说明 恒定总流动量方程 ∑ F = ρ Q ( β 2 v 2 − β 1 v 1 ) \sum\mathbf F=\rho Q(\beta_2\mathbf v_2-\beta_1\mathbf v_1) ∑ F = ρQ ( β 2 v 2 − β 1 v 1 ) 对每个坐标方向分别列式 多进多出 ∑ F = ∑ o u t ρ Q β v − ∑ i n ρ Q β v \sum\mathbf F=\sum_{out}\rho Q\beta\mathbf v-\sum_{in}\rho Q\beta\mathbf v ∑ F = ∑ o u t ρQβ v − ∑ in ρQβ v 速度必须带方向 压力力 F p = − p A n o u t \mathbf F_p=-pA\mathbf n_{out} F p = − p A n o u t 用表压可自动消去大气压 重力 G = ρ g V C V \mathbf G=\rho gV_{CV} G = ρ g V C V 需要时计入控制体内流体重量 固定平板射流 F = ρ Q v = ρ A v 2 F=\rho Qv=\rho Av^2 F = ρQ v = ρ A v 2 射流法向速度降为零 180° 转弯射流 F = 2 ρ Q v = 2 ρ A v 2 F=2\rho Qv=2\rho Av^2 F = 2 ρQ v = 2 ρ A v 2 速度改变量为 2 v 2v 2 v 运动平板 F = ρ A ( v − u ) 2 F=\rho A(v-u)^2 F = ρ A ( v − u ) 2 A ( v − u ) A(v-u) A ( v − u ) 动量修正系数 β = ∫ A u 2 d A v 2 A \beta=\dfrac{\int_Au^2\,\mathrm dA}{v^2A} β = v 2 A ∫ A u 2 d A 层流圆管 β = 4 / 3 \beta=4/3 β = 4/3
**控制体受力清单：**压力力 + 固体边界力 + 重力。求“水对管壁的力”时，动量方程先求“管壁对水的力”，最后反号。
A7-1 180° 导叶允许的最大射流速度# 【24–25 真题】
中文题目： 水平圆射流经导叶转弯 180 ∘ 180^\circ 18 0 ∘ d d d F F F
English version: A horizontal circular water jet is turned through 180 ∘ 180^\circ 18 0 ∘ d d d F F F
答案与讲解 取来流方向为正。入口速度 + v +v + v − v -v − v
ρ Q ( − v − v ) = − 2 ρ Q v . \rho Q(-v-v)=-2\rho Qv. ρQ ( − v − v ) = − 2 ρQ v . 导叶所受反力大小
F = 2 ρ Q v = 2 ρ A v 2 , A = π d 2 4 . F=2\rho Qv=2\rho Av^2,
\qquad A=\frac{\pi d^2}{4}. F = 2 ρQ v = 2 ρ A v 2 , A = 4 π d 2 . 所以
v max = 2 F ρ π d 2 . \boxed{v_{\max}=\sqrt{\frac{2F}{\rho\pi d^2}}}. v m a x = ρ π d 2 2 F . A7-2 射流冲击固定平板# 【教材例题改编】
中文题目： 直径 d = 50 m m d=50\ \mathrm{mm} d = 50 mm v = 10 m / s v=10\ \mathrm{m/s} v = 10 m/s
English version: A water jet of diameter 50 m m 50\ \mathrm{mm} 50 mm 10 m / s 10\ \mathrm{m/s} 10 m/s
答案与讲解 冲击后法向速度变为零：
F = ρ Q v = ρ A v 2 . F=\rho Qv=\rho Av^2. F = ρQ v = ρ A v 2 . A = π ( 0.05 ) 2 4 = 1.9635 × 10 − 3 m 2 . A=\frac{\pi(0.05)^2}{4}=1.9635\times10^{-3}\ \mathrm{m^2}. A = 4 π ( 0.05 ) 2 = 1.9635 × 1 0 − 3 m 2 . F = 1000 × 1.9635 × 10 − 3 × 10 2 = 196.3 N . F=1000\times1.9635\times10^{-3}\times10^2
=196.3\ \mathrm N. F = 1000 × 1.9635 × 1 0 − 3 × 1 0 2 = 196.3 N . F = 196 N \boxed{F=196\ \mathrm N} F = 196 N A7-3 射流冲击同向运动平板# 【教材例题改编】
中文题目： 直径 d = 40 m m d=40\ \mathrm{mm} d = 40 mm v = 8 m / s v=8\ \mathrm{m/s} v = 8 m/s u = 2 m / s u=2\ \mathrm{m/s} u = 2 m/s
English version: A water jet of diameter 40 m m 40\ \mathrm{mm} 40 mm 8 m / s 8\ \mathrm{m/s} 8 m/s 2 m / s 2\ \mathrm{m/s} 2 m/s
答案与讲解 单位时间真正撞上平板的水量按相对速度计算：
m ˙ = ρ A ( v − u ) . \dot m=\rho A(v-u). m ˙ = ρ A ( v − u ) . 水的绝对速度从 v v v u u u
F = m ˙ ( v − u ) = ρ A ( v − u ) 2 . F=\dot m(v-u)=\rho A(v-u)^2. F = m ˙ ( v − u ) = ρ A ( v − u ) 2 . F = 1000 π ( 0.04 ) 2 4 ( 8 − 2 ) 2 = 45.2 N . F=1000\frac{\pi(0.04)^2}{4}(8-2)^2=45.2\ \mathrm N. F = 1000 4 π ( 0.04 ) 2 ( 8 − 2 ) 2 = 45.2 N . F = 45.2 N \boxed{F=45.2\ \mathrm N} F = 45.2 N A7-4 水平 90° 弯管的作用力# 【教材习题改编】
中文题目： 水以 Q = 0.030 m 3 / s Q=0.030\ \mathrm{m^3/s} Q = 0.030 m 3 /s 90 ∘ 90^\circ 9 0 ∘ d = 0.20 m d=0.20\ \mathrm m d = 0.20 m + x +x + x + y +y + y 150 k P a 150\ \mathrm{kPa} 150 kPa 100 k P a 100\ \mathrm{kPa} 100 kPa β = 1 \beta=1 β = 1
English version: Water flows at 0.030 m 3 / s 0.030\ \mathrm{m^3/s} 0.030 m 3 /s 90 ∘ 90^\circ 9 0 ∘ 0.20 m 0.20\ \mathrm m 0.20 m + x +x + x + y +y + y 150 k P a 150\ \mathrm{kPa} 150 kPa 100 k P a 100\ \mathrm{kPa} 100 kPa β = 1 \beta=1 β = 1
答案与讲解 A = 0.031416 m 2 , v = Q / A = 0.955 m / s . A=0.031416\ \mathrm{m^2},\qquad v=Q/A=0.955\ \mathrm{m/s}. A = 0.031416 m 2 , v = Q / A = 0.955 m/s . 设管壁对水的力为 ( R x , R y ) (R_x,R_y) ( R x , R y )
R x + p 1 A = ρ Q ( 0 − v ) , R_x+p_1A=\rho Q(0-v), R x + p 1 A = ρQ ( 0 − v ) , R y − p 2 A = ρ Q ( v − 0 ) . R_y-p_2A=\rho Q(v-0). R y − p 2 A = ρQ ( v − 0 ) . 得到
R x = − 4.741 k N , R y = 3.170 k N . R_x=-4.741\ \mathrm{kN},\qquad R_y=3.170\ \mathrm{kN}. R x = − 4.741 kN , R y = 3.170 kN . 水对弯管的力反向：
F x = + 4.741 k N , F y = − 3.170 k N . \boxed{F_x=+4.741\ \mathrm{kN},\quad F_y=-3.170\ \mathrm{kN}}. F x = + 4.741 kN , F y = − 3.170 kN . 合力
F = 5.70 k N . \boxed{F=5.70\ \mathrm{kN}}. F = 5.70 kN . A7-5 水平喷嘴的轴向反力# 【教材习题改编】
中文题目： 水平喷嘴入口直径 d 1 = 0.10 m d_1=0.10\ \mathrm m d 1 = 0.10 m d 2 = 0.050 m d_2=0.050\ \mathrm m d 2 = 0.050 m Q = 0.020 m 3 / s Q=0.020\ \mathrm{m^3/s} Q = 0.020 m 3 /s
English version: A horizontal nozzle has inlet diameter 0.10 m 0.10\ \mathrm m 0.10 m 0.050 m 0.050\ \mathrm m 0.050 m 0.020 m 3 / s 0.020\ \mathrm{m^3/s} 0.020 m 3 /s
答案与讲解 v 1 = 2.546 m / s , v 2 = 10.186 m / s . v_1=2.546\ \mathrm{m/s},\qquad v_2=10.186\ \mathrm{m/s}. v 1 = 2.546 m/s , v 2 = 10.186 m/s . 由伯努利方程：
p 1 = 1 2 ρ ( v 2 2 − v 1 2 ) = 48.6 k P a . p_1=\frac12\rho(v_2^2-v_1^2)=48.6\ \mathrm{kPa}. p 1 = 2 1 ρ ( v 2 2 − v 1 2 ) = 48.6 kPa . 对水列动量方程：
R + p 1 A 1 = ρ Q ( v 2 − v 1 ) . R+p_1A_1=\rho Q(v_2-v_1). R + p 1 A 1 = ρQ ( v 2 − v 1 ) . R = 152.8 − 381.9 = − 229.1 N . R=152.8-381.9=-229.1\ \mathrm N. R = 152.8 − 381.9 = − 229.1 N . R R R
229 N downstream . \boxed{229\ \mathrm N\ \text{downstream}}. 229 N downstream . A7-6 明渠闸门所受水平力# 【教材习题改编】
中文题目： 水平矩形明渠宽 b = 2.0 m b=2.0\ \mathrm m b = 2.0 m h 1 = 1.2 m h_1=1.2\ \mathrm m h 1 = 1.2 m h 2 = 0.4 m h_2=0.4\ \mathrm m h 2 = 0.4 m Q = 2.0 m 3 / s Q=2.0\ \mathrm{m^3/s} Q = 2.0 m 3 /s
English version: A horizontal rectangular channel is 2.0 m 2.0\ \mathrm m 2.0 m 1.2 m 1.2\ \mathrm m 1.2 m 0.4 m 0.4\ \mathrm m 0.4 m 2.0 m 3 / s 2.0\ \mathrm{m^3/s} 2.0 m 3 /s
答案与讲解 v 1 = Q b h 1 = 0.833 m / s , v 2 = Q b h 2 = 2.50 m / s . v_1=\frac{Q}{bh_1}=0.833\ \mathrm{m/s},
\qquad
v_2=\frac{Q}{bh_2}=2.50\ \mathrm{m/s}. v 1 = b h 1 Q = 0.833 m/s , v 2 = b h 2 Q = 2.50 m/s . 两断面静水压力合力：
P i = 1 2 ρ g b h i 2 . P_i=\frac12\rho gbh_i^2. P i = 2 1 ρ g b h i 2 . P 1 = 14.13 k N , P 2 = 1.57 k N . P_1=14.13\ \mathrm{kN},\qquad P_2=1.57\ \mathrm{kN}. P 1 = 14.13 kN , P 2 = 1.57 kN . 设闸门对水的力为 R R R
P 1 − P 2 + R = ρ Q ( v 2 − v 1 ) = 3.333 k N . P_1-P_2+R=\rho Q(v_2-v_1)=3.333\ \mathrm{kN}. P 1 − P 2 + R = ρQ ( v 2 − v 1 ) = 3.333 kN . R = − 9.22 k N . R=-9.22\ \mathrm{kN}. R = − 9.22 kN . 因此水对闸门的力向下游：
F = 9.22 k N . \boxed{F=9.22\ \mathrm{kN}}. F = 9.22 kN . A7-7 为什么压力力用表压即可# 【23–24 真题改编】
中文题目： 用控制体动量方程计算管件受力时，为什么开口断面上的压力力通常可以直接采用表压？
English version: In a control-volume momentum analysis of a pipe fitting, why can gauge pressure usually be used directly for the pressure forces on open control surfaces?
答案与讲解 若使用绝对压强，控制体外部大气压还会作用于管件和控制面未显式画出的外表面。均匀大气压对封闭控制面的合力为零：
∮ p a n d A = 0. \oint p_a\mathbf n\,\mathrm dA=0. ∮ p a n d A = 0. 因此可以把所有压力统一减去 p a p_a p a p g = 0 p_g=0 p g = 0
A7-8 动量与动能修正系数# 【教材例题改编】
中文题目： 圆管充分发展层流速度分布为 u = 2 v ( 1 − r 2 / R 2 ) u=2v(1-r^2/R^2) u = 2 v ( 1 − r 2 / R 2 ) β \beta β α \alpha α
English version: The fully developed laminar velocity profile in a circular pipe is u = 2 v ( 1 − r 2 / R 2 ) u=2v(1-r^2/R^2) u = 2 v ( 1 − r 2 / R 2 ) v v v β \beta β α \alpha α
答案与讲解 定义：
β = ∫ A u 2 d A v 2 A , α = ∫ A u 3 d A v 3 A . \beta=\frac{\int_Au^2\,\mathrm dA}{v^2A},
\qquad
\alpha=\frac{\int_Au^3\,\mathrm dA}{v^3A}. β = v 2 A ∫ A u 2 d A , α = v 3 A ∫ A u 3 d A . 令 s = r / R s=r/R s = r / R d A = 2 π R 2 s d s \mathrm dA=2\pi R^2s\,\mathrm ds d A = 2 π R 2 s d s
β = 4 3 , α = 2 . \boxed{\beta=\frac43},
\qquad
\boxed{\alpha=2}. β = 3 4 , α = 2 . 层流速度分布很不均匀，因此 α , β \alpha,\beta α , β
A8 流动阻力与能量损失# A8.1 后两章速成路线（一）：阻力题先做什么# 看到管流或明渠阻力题，固定按以下顺序：
几何量 → v → R e → 判流态 → λ / ζ → h w → 伯努利方程 \boxed{\text{几何量}\to v\to Re\to\text{判流态}\to\lambda/\zeta\to h_w\to\text{伯努利方程}} 几何量 → v → R e → 判流态 → λ / ζ → h w → 伯努利方程
算面积 A A A χ \chi χ R = A / χ R=A/\chi R = A / χ R = d / 4 R=d/4 R = d /4
算平均流速 v = Q / A v=Q/A v = Q / A
圆管用 R e = v d / ν Re=vd/\nu R e = v d / ν R e R = v R / ν Re_R=vR/\nu R e R = v R / ν
层流圆管直接用 λ = 64 / R e \lambda=64/Re λ = 64/ R e
沿程损失与局部损失相加：h w = h f + ∑ h j h_w=h_f+\sum h_j h w = h f + ∑ h j
若未知流量出现在 v v v R e Re R e λ \lambda λ
A8.2 重要公式、符号与关系#
内容 公式 说明 Darcy–Weisbach h f = λ l d v 2 2 g h_f=\lambda\dfrac{l}{d}\dfrac{v^2}{2g} h f = λ d l 2 g v 2 非圆管以 d h = 4 R d_h=4R d h = 4 R d d d 局部损失 h j = ζ v 2 2 g h_j=\zeta\dfrac{v^2}{2g} h j = ζ 2 g v 2 必须确认 ζ \zeta ζ 均匀流基本式 τ 0 = ρ g R J \tau_0=\rho gRJ τ 0 = ρ g R J J = h f / l J=h_f/l J = h f / l 摩阻流速 v ∗ = τ 0 / ρ v_*=\sqrt{\tau_0/\rho} v ∗ = τ 0 / ρ friction velocity 圆管壁面切应力 τ 0 = λ ρ v 2 / 8 \tau_0=\lambda\rho v^2/8 τ 0 = λ ρ v 2 /8 与 Darcy 摩阻系数配套 雷诺数 R e = v d / ν Re=vd/\nu R e = v d / ν 圆管：约 R e < 2300 Re<2300 R e < 2300 层流速度分布 u ( r ) = ρ g J 4 μ ( R 2 − r 2 ) = 2 v ( 1 − r 2 / R 2 ) u(r)=\dfrac{\rho gJ}{4\mu}(R^2-r^2)=2v(1-r^2/R^2) u ( r ) = 4 μ ρ g J ( R 2 − r 2 ) = 2 v ( 1 − r 2 / R 2 ) 抛物线分布，u m a x = 2 v u_{max}=2v u ma x = 2 v Hagen–Poiseuille Q = π d 4 128 μ Δ p l Q=\dfrac{\pi d^4}{128\mu}\dfrac{\Delta p}{l} Q = 128 μ π d 4 l Δ p Q ∝ d 4 Q\propto d^4 Q ∝ d 4 层流摩阻系数 λ = 64 / R e \lambda=64/Re λ = 64/ R e 与相对粗糙度无关 光滑湍流近似 λ ≈ 0.3164 R e − 0.25 \lambda\approx0.3164Re^{-0.25} λ ≈ 0.3164 R e − 0.25 Blasius，适用于一定 Re 范围 完全粗糙区 λ = f ( Δ / d ) \lambda=f(\Delta/d) λ = f ( Δ/ d ) 与 Re 近似无关 Stokes 阻力 F D = 3 π μ d u F_D=3\pi\mu du F D = 3 π μ d u 要求 R e d < 1 Re_d<1 R e d < 1 Stokes 沉速 u s = ( ρ s − ρ ) g d 2 18 μ u_s=\dfrac{(\rho_s-\rho)gd^2}{18\mu} u s = 18 μ ( ρ s − ρ ) g d 2 求完后必须回代检查 Re
A8-1 圆管流态与最大层流速度# 【教材例题 6-1 改编】
中文题目： 水在直径 d = 0.10 m d=0.10\ \mathrm m d = 0.10 m v = 1.0 m / s v=1.0\ \mathrm{m/s} v = 1.0 m/s ν = 1.31 × 10 − 6 m 2 / s \nu=1.31\times10^{-6}\ \mathrm{m^2/s} ν = 1.31 × 1 0 − 6 m 2 /s 2300 2300 2300
English version: Water flows in a circular pipe of diameter 0.10 m 0.10\ \mathrm m 0.10 m 1.0 m / s 1.0\ \mathrm{m/s} 1.0 m/s 1.31 × 10 − 6 m 2 / s 1.31\times10^{-6}\ \mathrm{m^2/s} 1.31 × 1 0 − 6 m 2 /s 2300 2300 2300
答案与讲解 R e = v d ν = 1.0 × 0.10 1.31 × 10 − 6 = 7.63 × 10 4 > 2300. Re=\frac{vd}{\nu}
=\frac{1.0\times0.10}{1.31\times10^{-6}}
=7.63\times10^4>2300. R e = ν v d = 1.31 × 1 0 − 6 1.0 × 0.10 = 7.63 × 1 0 4 > 2300. 因此为湍流。最大层流速度：
v m a x = R e c ν d = 2300 × 1.31 × 10 − 6 0.10 = 0.0301 m / s . v_{max}=\frac{Re_c\nu}{d}
=\frac{2300\times1.31\times10^{-6}}{0.10}
=0.0301\ \mathrm{m/s}. v ma x = d R e c ν = 0.10 2300 × 1.31 × 1 0 − 6 = 0.0301 m/s . turbulent ; v m a x , l a m = 0.030 m / s \boxed{\text{turbulent};\quad v_{max,lam}=0.030\ \mathrm{m/s}} turbulent ; v ma x , l am = 0.030 m/s A8-2 明渠/平流池流态判断# 【教材例题 6-2 改编】
中文题目： 矩形平流池宽 b = 7 m b=7\ \mathrm m b = 7 m h = 1.5 m h=1.5\ \mathrm m h = 1.5 m Q = 0.010 m 3 / s Q=0.010\ \mathrm{m^3/s} Q = 0.010 m 3 /s ν = 1.31 × 10 − 6 m 2 / s \nu=1.31\times10^{-6}\ \mathrm{m^2/s} ν = 1.31 × 1 0 − 6 m 2 /s R e R = v R / ν Re_R=vR/\nu R e R = v R / ν 575 575 575
English version: A rectangular settling basin is 7 m 7\ \mathrm m 7 m 1.5 m 1.5\ \mathrm m 1.5 m 0.010 m 3 / s 0.010\ \mathrm{m^3/s} 0.010 m 3 /s ν = 1.31 × 10 − 6 m 2 / s \nu=1.31\times10^{-6}\ \mathrm{m^2/s} ν = 1.31 × 1 0 − 6 m 2 /s R e R = v R / ν Re_R=vR/\nu R e R = v R / ν 575 575 575
答案与讲解 A = b h = 10.5 m 2 , χ = b + 2 h = 10 m , A=bh=10.5\ \mathrm{m^2},
\qquad
\chi=b+2h=10\ \mathrm m, A = bh = 10.5 m 2 , χ = b + 2 h = 10 m , R = A χ = 1.05 m , v = Q A = 9.52 × 10 − 4 m / s . R=\frac{A}{\chi}=1.05\ \mathrm m,
\qquad
v=\frac{Q}{A}=9.52\times10^{-4}\ \mathrm{m/s}. R = χ A = 1.05 m , v = A Q = 9.52 × 1 0 − 4 m/s . R e R = v R ν = 9.52 × 10 − 4 × 1.05 1.31 × 10 − 6 = 763 > 575. Re_R=\frac{vR}{\nu}
=\frac{9.52\times10^{-4}\times1.05}{1.31\times10^{-6}}
=763>575. R e R = ν v R = 1.31 × 1 0 − 6 9.52 × 1 0 − 4 × 1.05 = 763 > 575. turbulent according to the open-channel criterion \boxed{\text{turbulent according to the open-channel criterion}} turbulent according to the open-channel criterion A8-3 圆管层流速度分布# 【教材推导题改编】
中文题目： 写出圆管充分发展层流的速度分布，并说明最大速度、平均速度和壁面剪应力之间的关系。
English version: Write the velocity distribution for fully developed laminar flow in a circular pipe, and state the relations among maximum velocity, mean velocity, and wall shear stress.
答案与讲解 由剪应力线性分布与牛顿内摩擦定律积分：
u ( r ) = − d p / d x 4 μ ( R 2 − r 2 ) = u m a x ( 1 − r 2 R 2 ) . u(r)=\frac{-\mathrm dp/\mathrm dx}{4\mu}(R^2-r^2)
=u_{max}\left(1-\frac{r^2}{R^2}\right). u ( r ) = 4 μ − d p / d x ( R 2 − r 2 ) = u ma x ( 1 − R 2 r 2 ) . 截面平均后：
u m a x = 2 v , u ( r ) = 2 v ( 1 − r 2 R 2 ) . \boxed{u_{max}=2v},
\qquad
\boxed{u(r)=2v\left(1-\frac{r^2}{R^2}\right)}. u ma x = 2 v , u ( r ) = 2 v ( 1 − R 2 r 2 ) . 壁面剪应力
τ 0 = − R 2 d p d x \boxed{\tau_0=-\frac R2\frac{\mathrm dp}{\mathrm dx}} τ 0 = − 2 R d x d p 且从管轴到管壁按 r r r
A8-4 管径变化对层流流量的影响# 【24–25 真题】
中文题目： 圆管层流中，管长和两端压强差保持不变，管径增大一倍。流量变为原来的多少倍？
English version: In laminar flow through a circular pipe, the pipe length and pressure difference are unchanged, while the diameter is doubled. By what factor does the discharge change?
答案与讲解 Hagen–Poiseuille 定律：
Q = π d 4 128 μ Δ p l . Q=\frac{\pi d^4}{128\mu}\frac{\Delta p}{l}. Q = 128 μ π d 4 l Δ p . 因此
Q 2 Q 1 = ( d 2 d 1 ) 4 = 2 4 = 16. \frac{Q_2}{Q_1}=\left(\frac{d_2}{d_1}\right)^4=2^4=16. Q 1 Q 2 = ( d 1 d 2 ) 4 = 2 4 = 16. Q 2 = 16 Q 1 \boxed{Q_2=16Q_1} Q 2 = 16 Q 1 所以“增大一倍后流量变成 2 倍”的判断是错误的。
A8-5 层流摩阻系数、沿程损失与壁面剪应力# 【教材习题改编】
中文题目： 水在 d = 0.050 m d=0.050\ \mathrm m d = 0.050 m l = 10 m l=10\ \mathrm m l = 10 m Q = 5.0 × 10 − 5 m 3 / s Q=5.0\times10^{-5}\ \mathrm{m^3/s} Q = 5.0 × 1 0 − 5 m 3 /s ν = 1.0 × 10 − 6 m 2 / s \nu=1.0\times10^{-6}\ \mathrm{m^2/s} ν = 1.0 × 1 0 − 6 m 2 /s R e Re R e λ \lambda λ h f h_f h f τ 0 \tau_0 τ 0
English version: Water flows through a circular pipe of diameter 0.050 m 0.050\ \mathrm m 0.050 m 10 m 10\ \mathrm m 10 m Q = 5.0 × 10 − 5 m 3 / s Q=5.0\times10^{-5}\ \mathrm{m^3/s} Q = 5.0 × 1 0 − 5 m 3 /s ν = 1.0 × 10 − 6 m 2 / s \nu=1.0\times10^{-6}\ \mathrm{m^2/s} ν = 1.0 × 1 0 − 6 m 2 /s R e Re R e λ \lambda λ h f h_f h f τ 0 \tau_0 τ 0
答案与讲解 v = Q A = 0.02546 m / s , R e = v d ν = 1273 < 2300. v=\frac Q A=0.02546\ \mathrm{m/s},
\qquad
Re=\frac{vd}{\nu}=1273<2300. v = A Q = 0.02546 m/s , R e = ν v d = 1273 < 2300. 层流：
λ = 64 R e = 0.05027. \lambda=\frac{64}{Re}=0.05027. λ = R e 64 = 0.05027. h f = λ l d v 2 2 g = 3.32 × 10 − 4 m . h_f=\lambda\frac ld\frac{v^2}{2g}
=3.32\times10^{-4}\ \mathrm m. h f = λ d l 2 g v 2 = 3.32 × 1 0 − 4 m . τ 0 = λ 8 ρ v 2 = 4.07 × 10 − 3 P a . \tau_0=\frac{\lambda}{8}\rho v^2
=4.07\times10^{-3}\ \mathrm{Pa}. τ 0 = 8 λ ρ v 2 = 4.07 × 1 0 − 3 Pa . R e = 1273 , λ = 0.0503 , h f = 3.32 × 10 − 4 m , τ 0 = 4.07 × 10 − 3 P a \boxed{Re=1273,\ \lambda=0.0503,\ h_f=3.32\times10^{-4}\ \mathrm m,\ \tau_0=4.07\times10^{-3}\ \mathrm{Pa}} R e = 1273 , λ = 0.0503 , h f = 3.32 × 1 0 − 4 m , τ 0 = 4.07 × 1 0 − 3 Pa A8-6 已知摩阻系数求沿程损失# 【教材习题改编】
中文题目： 直径 d = 0.30 m d=0.30\ \mathrm m d = 0.30 m l = 1000 m l=1000\ \mathrm m l = 1000 m Q = 0.050 m 3 / s Q=0.050\ \mathrm{m^3/s} Q = 0.050 m 3 /s λ = 0.020 \lambda=0.020 λ = 0.020
English version: A pipe of diameter 0.30 m 0.30\ \mathrm m 0.30 m 1000 m 1000\ \mathrm m 1000 m 0.050 m 3 / s 0.050\ \mathrm{m^3/s} 0.050 m 3 /s 0.020 0.020 0.020
答案与讲解 v = Q π d 2 / 4 = 0.707 m / s . v=\frac{Q}{\pi d^2/4}=0.707\ \mathrm{m/s}. v = π d 2 /4 Q = 0.707 m/s . h f = 0.020 1000 0.30 0.707 2 2 × 9.81 = 1.70 m . h_f=0.020\frac{1000}{0.30}\frac{0.707^2}{2\times9.81}
=1.70\ \mathrm m. h f = 0.020 0.30 1000 2 × 9.81 0.70 7 2 = 1.70 m . h f = 1.70 m \boxed{h_f=1.70\ \mathrm m} h f = 1.70 m A8-7 非圆形过流断面的水力半径# 【教材习题改编】
中文题目： 矩形明渠宽 b = 2.5 m b=2.5\ \mathrm m b = 2.5 m h = 0.30 m h=0.30\ \mathrm m h = 0.30 m Q = 0.010 m 3 / s Q=0.010\ \mathrm{m^3/s} Q = 0.010 m 3 /s ν = 1.01 × 10 − 6 m 2 / s \nu=1.01\times10^{-6}\ \mathrm{m^2/s} ν = 1.01 × 1 0 − 6 m 2 /s R e R = v R / ν Re_R=vR/\nu R e R = v R / ν
English version: A rectangular open channel is 2.5 m 2.5\ \mathrm m 2.5 m 0.30 m 0.30\ \mathrm m 0.30 m 0.010 m 3 / s 0.010\ \mathrm{m^3/s} 0.010 m 3 /s ν = 1.01 × 10 − 6 m 2 / s \nu=1.01\times10^{-6}\ \mathrm{m^2/s} ν = 1.01 × 1 0 − 6 m 2 /s R e R = v R / ν Re_R=vR/\nu R e R = v R / ν
答案与讲解 A = b h = 0.75 m 2 , χ = b + 2 h = 3.10 m . A=bh=0.75\ \mathrm{m^2},
\qquad
\chi=b+2h=3.10\ \mathrm m. A = bh = 0.75 m 2 , χ = b + 2 h = 3.10 m . R = A χ = 0.2419 m , d h = 4 R = 0.9677 m . R=\frac{A}{\chi}=0.2419\ \mathrm m,
\qquad
d_h=4R=0.9677\ \mathrm m. R = χ A = 0.2419 m , d h = 4 R = 0.9677 m . v = Q A = 0.01333 m / s , v=\frac Q A=0.01333\ \mathrm{m/s}, v = A Q = 0.01333 m/s , R e R = v R ν = 3.19 × 10 3 . Re_R=\frac{vR}{\nu}=3.19\times10^3. R e R = ν v R = 3.19 × 1 0 3 . R = 0.242 m , d h = 0.968 m , R e R = 3.19 × 10 3 \boxed{R=0.242\ \mathrm m,\quad d_h=0.968\ \mathrm m,\quad Re_R=3.19\times10^3} R = 0.242 m , d h = 0.968 m , R e R = 3.19 × 1 0 3 A8-8 光滑区、过渡粗糙区与完全粗糙区# 【24–25 真题改编】
中文题目： 说明湍流光滑区、过渡粗糙区和完全粗糙区中摩阻系数 λ \lambda λ λ \lambda λ R e Re R e
English version: State the parameter dependence of the friction factor λ \lambda λ λ \lambda λ R e Re R e
答案与讲解
光滑区：λ = f ( R e ) \lambda=f(Re) λ = f ( R e )
过渡粗糙区：λ = f ( R e , Δ / d ) \lambda=f(Re,\Delta/d) λ = f ( R e , Δ/ d )
完全粗糙区：λ = f ( Δ / d ) \lambda=f(\Delta/d) λ = f ( Δ/ d )
完全粗糙区中，粗糙峰穿透黏性底层，阻力主要由粗糙元引起的压差阻力决定。增大 R e Re R e λ \lambda λ R e Re R e
A8-9 黏性底层厚度随流量的变化# 【24–25 真题】
中文题目： 在管径、水温和沿程阻力系数均不变时，流量增大，黏性底层厚度如何变化？
English version: For fixed pipe diameter, water temperature, and friction factor, how does the viscous sublayer thickness change as the discharge increases?
答案与讲解 黏性底层厚度近似满足
δ v ∝ ν v ∗ , v ∗ = τ 0 ρ = v λ 8 . \delta_v\propto\frac{\nu}{v_*},
\qquad
v_*=\sqrt{\frac{\tau_0}{\rho}}
=v\sqrt{\frac\lambda8}. δ v ∝ v ∗ ν , v ∗ = ρ τ 0 = v 8 λ . 当 ν , λ \nu,\lambda ν , λ v v v v ∗ v_* v ∗
δ v decreases . \boxed{\delta_v\ \text{decreases}.} δ v decreases . A8-10 局部水头损失# 【教材习题改编】
中文题目： 直径 d = 0.10 m d=0.10\ \mathrm m d = 0.10 m Q = 0.010 m 3 / s Q=0.010\ \mathrm{m^3/s} Q = 0.010 m 3 /s 0.5 0.5 0.5 0.9 0.9 0.9 2.0 2.0 2.0
English version: A 0.10 m 0.10\ \mathrm m 0.10 m 0.010 m 3 / s 0.010\ \mathrm{m^3/s} 0.010 m 3 /s 0.5 0.5 0.5 0.9 0.9 0.9 2.0 2.0 2.0
答案与讲解 v = Q A = 1.273 m / s , ∑ ζ = 3.4. v=\frac{Q}{A}=1.273\ \mathrm{m/s},
\qquad
\sum\zeta=3.4. v = A Q = 1.273 m/s , ∑ ζ = 3.4. h j = ∑ ζ v 2 2 g = 3.4 1.273 2 2 × 9.81 = 0.281 m . h_j=\sum\zeta\frac{v^2}{2g}
=3.4\frac{1.273^2}{2\times9.81}
=0.281\ \mathrm m. h j = ∑ ζ 2 g v 2 = 3.4 2 × 9.81 1.27 3 2 = 0.281 m . h j = 0.281 m \boxed{h_j=0.281\ \mathrm m} h j = 0.281 m 不同管径处的局部构件不能盲目共用同一个速度水头。
A8-11 水库经管道自由出流# 【教材综合题改编】
中文题目： 大水库经直径 d = 0.15 m d=0.15\ \mathrm m d = 0.15 m l = 200 m l=200\ \mathrm m l = 200 m H = 15 m H=15\ \mathrm m H = 15 m λ = 0.025 \lambda=0.025 λ = 0.025 ∑ ζ = 3.0 \sum\zeta=3.0 ∑ ζ = 3.0
English version: A large reservoir discharges freely through a pipe of diameter 0.15 m 0.15\ \mathrm m 0.15 m 200 m 200\ \mathrm m 200 m 15 m 15\ \mathrm m 15 m 0.025 0.025 0.025 3.0 3.0 3.0
答案与讲解 自由面至自由射流出口：
H = ( 1 + λ l d + ∑ ζ ) v 2 2 g . H=\left(1+\lambda\frac ld+\sum\zeta\right)\frac{v^2}{2g}. H = ( 1 + λ d l + ∑ ζ ) 2 g v 2 . 系数
1 + 0.025 200 0.15 + 3 = 37.333. 1+0.025\frac{200}{0.15}+3=37.333. 1 + 0.025 0.15 200 + 3 = 37.333. v = 2 g H 37.333 = 2.807 m / s . v=\sqrt{\frac{2gH}{37.333}}=2.807\ \mathrm{m/s}. v = 37.333 2 g H = 2.807 m/s . Q = π d 2 4 v = 0.0496 m 3 / s . Q=\frac{\pi d^2}{4}v=0.0496\ \mathrm{m^3/s}. Q = 4 π d 2 v = 0.0496 m 3 /s . Q = 4.96 × 10 − 2 m 3 / s \boxed{Q=4.96\times10^{-2}\ \mathrm{m^3/s}} Q = 4.96 × 1 0 − 2 m 3 /s A8-12 Stokes 沉降速度与适用性检查# 【教材例题改编】
中文题目： 直径 d = 0.050 m m d=0.050\ \mathrm{mm} d = 0.050 mm ρ s = 2650 k g / m 3 \rho_s=2650\ \mathrm{kg/m^3} ρ s = 2650 kg/ m 3 ρ = 1000 k g / m 3 \rho=1000\ \mathrm{kg/m^3} ρ = 1000 kg/ m 3 μ = 1.0 × 10 − 3 P a ⋅ s \mu=1.0\times10^{-3}\ \mathrm{Pa\cdot s} μ = 1.0 × 1 0 − 3 Pa ⋅ s R e d < 1 Re_d<1 R e d < 1
English version: A spherical particle of diameter 0.050 m m 0.050\ \mathrm{mm} 0.050 mm 2650 k g / m 3 2650\ \mathrm{kg/m^3} 2650 kg/ m 3 ρ = 1000 k g / m 3 \rho=1000\ \mathrm{kg/m^3} ρ = 1000 kg/ m 3 μ = 1.0 × 10 − 3 P a ⋅ s \mu=1.0\times10^{-3}\ \mathrm{Pa\cdot s} μ = 1.0 × 1 0 − 3 Pa ⋅ s R e d < 1 Re_d<1 R e d < 1
答案与讲解 终速时，有效重力等于 Stokes 阻力：
u s = ( ρ s − ρ ) g d 2 18 μ . u_s=\frac{(\rho_s-\rho)gd^2}{18\mu}. u s = 18 μ ( ρ s − ρ ) g d 2 . d = 5.0 × 10 − 5 m d=5.0\times10^{-5}\ \mathrm m d = 5.0 × 1 0 − 5 m
u s = ( 2650 − 1000 ) × 9.81 × ( 5.0 × 10 − 5 ) 2 18 × 10 − 3 = 2.25 × 10 − 3 m / s . u_s=\frac{(2650-1000)\times9.81\times(5.0\times10^{-5})^2}{18\times10^{-3}}
=2.25\times10^{-3}\ \mathrm{m/s}. u s = 18 × 1 0 − 3 ( 2650 − 1000 ) × 9.81 × ( 5.0 × 1 0 − 5 ) 2 = 2.25 × 1 0 − 3 m/s . R e d = ρ u s d μ = 0.112 < 1. Re_d=\frac{\rho u_sd}{\mu}
=0.112<1. R e d = μ ρ u s d = 0.112 < 1. u s = 2.25 × 10 − 3 m / s , R e d = 0.112 \boxed{u_s=2.25\times10^{-3}\ \mathrm{m/s},\quad Re_d=0.112} u s = 2.25 × 1 0 − 3 m/s , R e d = 0.112 适用性满足。若算出 R e d ≫ 1 Re_d\gg1 R e d ≫ 1
A8-13 边界层分离与绕流阻力# 【教材思考题改编】
中文题目： 说明平板边界层在顺压梯度与逆压梯度下的变化，并解释逆压梯度为什么可能导致边界层分离。
English version: Describe the behavior of a boundary layer under favorable and adverse pressure gradients, and explain why an adverse pressure gradient may cause boundary-layer separation.
答案与讲解
顺压梯度：沿流向压强降低，外流加速，近壁流体获得动量，边界层较薄，不易分离；
逆压梯度：沿流向压强升高，外流减速，低动量近壁流体受到更强阻滞。
壁面切应力
τ w = μ ( ∂ u ∂ y ) y = 0 . \tau_w=\mu\left(\frac{\partial u}{\partial y}\right)_{y=0}. τ w = μ ( ∂ y ∂ u ) y = 0 . 当逆压梯度足够强时，壁面速度梯度逐渐减小；分离点满足
τ w = 0 . \boxed{\tau_w=0}. τ w = 0 . 分离后近壁出现回流与尾涡，压差阻力显著增大。流线型物体通过延缓分离来降低压差阻力。
A9 量纲分析与相似理论# A9.1 后两章速成路线（二）：量纲与模型题先做什么# 量纲分析题# 列变量 → 写量纲 → 选基本变量 → 配成 π 项 → 写函数关系 \boxed{\text{列变量}\to\text{写量纲}\to\text{选基本变量}\to\text{配成}\ \pi\text{项}\to\text{写函数关系}} 列变量 → 写量纲 → 选基本变量 → 配成 π 项 → 写函数关系
Rayleigh 法适合变量较少、可以合理假设幂乘积的情况；
Buckingham π \pi π n n n m m m n − m n-m n − m
基本变量必须量纲独立，并共同覆盖 M , L , T M,L,T M , L , T
相似模型题#
统一规定比尺：本节均采用
λ x = x p x m \boxed{\lambda_x=\frac{x_p}{x_m}} λ x = x m x p 其中下标 p p p m m m
A9.2 重要公式、准数与比尺关系#
内容 表达式 物理意义 Reynolds number R e = v L ν = ρ v L μ Re=\dfrac{vL}{\nu}=\dfrac{\rho vL}{\mu} R e = ν vL = μ ρ vL 惯性力/黏性力 Froude number F r = v g L Fr=\dfrac{v}{\sqrt{gL}} F r = gL v 惯性力/重力的平方根形式；也常用 F r 2 = v 2 / ( g L ) Fr^2=v^2/(gL) F r 2 = v 2 / ( gL ) Euler number E u = Δ p ρ v 2 Eu=\dfrac{\Delta p}{\rho v^2} E u = ρ v 2 Δ p 压力/惯性力 Weber number W e = ρ v 2 L σ We=\dfrac{\rho v^2L}{\sigma} W e = σ ρ v 2 L 惯性力/表面张力 Strouhal number S t = L v t St=\dfrac{L}{vt} St = v t L 当地惯性/迁移惯性 几何比尺 λ A = λ L 2 \lambda_A=\lambda_L^2 λ A = λ L 2 λ V = λ L 3 \lambda_V=\lambda_L^3 λ V = λ L 3 几何相似 Froude 相似 λ v = λ L 1 / 2 \lambda_v=\lambda_L^{1/2} λ v = λ L 1/2 λ t = λ L 1 / 2 \lambda_t=\lambda_L^{1/2} λ t = λ L 1/2 λ Q = λ L 5 / 2 \lambda_Q=\lambda_L^{5/2} λ Q = λ L 5/2 同一重力场 Froude 力/功率比尺 λ F = λ ρ λ L 3 \lambda_F=\lambda_\rho\lambda_L^3 λ F = λ ρ λ L 3 λ P = λ ρ λ L 7 / 2 \lambda_P=\lambda_\rho\lambda_L^{7/2} λ P = λ ρ λ L 7/2 P = F v P=Fv P = F v Reynolds 相似 λ v = λ ν / λ L \lambda_v=\lambda_\nu/\lambda_L λ v = λ ν / λ L 同种流体时 λ v = 1 / λ L \lambda_v=1/\lambda_L λ v = 1/ λ L Euler 相似 λ Δ p = λ ρ λ v 2 \lambda_{\Delta p}=\lambda_\rho\lambda_v^2 λ Δ p = λ ρ λ v 2 压差换算
A9-1 常用物理量的量纲# 【23–24 真题改编】
中文题目： 写出 μ \mu μ ν \nu ν σ \sigma σ p p p P P P M M M L L L T T T
English version: State the M M M L L L T T T μ \mu μ ν \nu ν σ \sigma σ p p p P P P
答案与讲解 [ μ ] = M L − 1 T − 1 , \boxed{[\mu]=ML^{-1}T^{-1}}, [ μ ] = M L − 1 T − 1 , [ ν ] = L 2 T − 1 , \boxed{[\nu]=L^2T^{-1}}, [ ν ] = L 2 T − 1 , 表面张力是单位长度上的力：
[ σ ] = M T − 2 , \boxed{[\sigma]=MT^{-2}}, [ σ ] = M T − 2 , [ p ] = M L − 1 T − 2 , \boxed{[p]=ML^{-1}T^{-2}}, [ p ] = M L − 1 T − 2 , [ P ] = M L 2 T − 3 . \boxed{[P]=ML^2T^{-3}}. [ P ] = M L 2 T − 3 . A9-2 Rayleigh 法求绕流阻力结构# 【教材例题改编】
中文题目： 假设物体绕流阻力 F D F_D F D ρ \rho ρ U U U D D D
English version: Assume that the drag force F D F_D F D ρ \rho ρ U U U D D D
答案与讲解 设
F D = k ρ a U b D c . F_D=k\rho^aU^bD^c. F D = k ρ a U b D c . 代入量纲：
M L T − 2 = ( M L − 3 ) a ( L T − 1 ) b L c . MLT^{-2}=(ML^{-3})^a(LT^{-1})^bL^c. M L T − 2 = ( M L − 3 ) a ( L T − 1 ) b L c . 比较指数：
a = 1 , b = 2 , − 3 a + b + c = 1 ⇒ c = 2. a=1,\qquad b=2,\qquad -3a+b+c=1\Rightarrow c=2. a = 1 , b = 2 , − 3 a + b + c = 1 ⇒ c = 2. 所以
F D = k ρ U 2 D 2 . \boxed{F_D=k\rho U^2D^2}. F D = k ρ U 2 D 2 . 通常写成
F D = C D ρ U 2 2 A , F_D=C_D\frac{\rho U^2}{2}A, F D = C D 2 ρ U 2 A , 其中无量纲系数 C D C_D C D
A9-3 Rayleigh 法推导毕托管测速关系# 【教材例题 7-1 / 课件例题】
中文题目： 假设毕托管测得的流速 u u u Δ p \Delta p Δ p ρ \rho ρ g g g u u u
English version: Assume that the velocity measured by a Pitot tube depends only on the pressure difference Δ p \Delta p Δ p ρ \rho ρ g g g u u u
答案与讲解 设
u = k ( Δ p ) a ρ b g c . u=k(\Delta p)^a\rho^bg^c. u = k ( Δ p ) a ρ b g c . 量纲方程：
L T − 1 = ( M L − 1 T − 2 ) a ( M L − 3 ) b ( L T − 2 ) c . LT^{-1}=(ML^{-1}T^{-2})^a(ML^{-3})^b(LT^{-2})^c. L T − 1 = ( M L − 1 T − 2 ) a ( M L − 3 ) b ( L T − 2 ) c . 比较 M , L , T M,L,T M , L , T
a = 1 2 , b = − 1 2 , c = 0. a=\frac12,\qquad b=-\frac12,\qquad c=0. a = 2 1 , b = − 2 1 , c = 0. 所以
u = k Δ p ρ . \boxed{u=k\sqrt{\frac{\Delta p}{\rho}}}. u = k ρ Δ p . 若 Δ p = ρ g Δ h \Delta p=\rho g\Delta h Δ p = ρ g Δ h
u = c 2 g Δ h , \boxed{u=c\sqrt{2g\Delta h}}, u = c 2 g Δ h , 无量纲常数由理论或标定确定。
A9-4 Buckingham π \pi π # 【教材例题 7-3】
中文题目： 水平圆管压降 Δ p \Delta p Δ p d d d l l l v v v ρ \rho ρ μ \mu μ Δ \Delta Δ d , v , ρ d,v,\rho d , v , ρ
English version: The pressure drop Δ p \Delta p Δ p d d d l l l v v v ρ \rho ρ μ \mu μ Δ \Delta Δ d d d v v v ρ \rho ρ
答案与讲解 共有 n = 7 n=7 n = 7 m = 3 m=3 m = 3 4 4 4 π \pi π
π 1 = Δ p ρ v 2 , π 2 = l d , π 3 = μ ρ v d = 1 R e , π 4 = Δ d . \pi_1=\frac{\Delta p}{\rho v^2},
\qquad
\pi_2=\frac{l}{d},
\qquad
\pi_3=\frac{\mu}{\rho vd}=\frac1{Re},
\qquad
\pi_4=\frac{\Delta}{d}. π 1 = ρ v 2 Δ p , π 2 = d l , π 3 = ρ v d μ = R e 1 , π 4 = d Δ . 故
Δ p ρ v 2 = f ( l d , R e , Δ d ) . \boxed{\frac{\Delta p}{\rho v^2}
=f\left(\frac ld,Re,\frac\Delta d\right)}. ρ v 2 Δ p = f ( d l , R e , d Δ ) . 把压降换成水头损失，可写成
h f = λ ( R e , Δ d ) l d v 2 2 g . \boxed{h_f=\lambda\left(Re,\frac\Delta d\right)\frac ld\frac{v^2}{2g}}. h f = λ ( R e , d Δ ) d l 2 g v 2 . A9-5 相似准数的物理意义# 【24–25、23–24 真题改编】
中文题目： 分别说明 R e Re R e F r Fr F r E u Eu E u W e We W e
English version: State the force ratios represented by R e Re R e F r Fr F r E u Eu E u W e We W e
答案与讲解
准数 力比 典型问题 R e Re R e 惯性力/黏性力 管流、边界层、低速绕流 F r 2 Fr^2 F r 2 惯性力/重力 明渠、溢流坝、船舶兴波 E u Eu E u 压力/惯性力 压降、泵与风机、压力相似 W e We W e 惯性力/表面张力 液滴、射流破碎、毛细现象
考试填空“Re 表征 ______ 力与 ______ 力之比”应答：
inertia force to viscous force . \boxed{\text{inertia force to viscous force}.} inertia force to viscous force . A9-6 Froude 相似的全套比尺# 【24–25 真题】
中文题目： 原型与模型满足重力相似，采用比尺 λ L = L p / L m \lambda_L=L_p/L_m λ L = L p / L m
English version: A prototype and its model satisfy Froude similarity. Define λ L = L p / L m \lambda_L=L_p/L_m λ L = L p / L m
答案与讲解 Froude 相似：
v p g L p = v m g L m . \frac{v_p}{\sqrt{gL_p}}=\frac{v_m}{\sqrt{gL_m}}. g L p v p = g L m v m . 所以
λ v = λ L 1 / 2 . \boxed{\lambda_v=\lambda_L^{1/2}}. λ v = λ L 1/2 . 由 t ∼ L / v t\sim L/v t ∼ L / v
λ t = λ L 1 / 2 . \boxed{\lambda_t=\lambda_L^{1/2}}. λ t = λ L 1/2 . 由 Q ∼ v L 2 Q\sim vL^2 Q ∼ v L 2
λ Q = λ L 5 / 2 . \boxed{\lambda_Q=\lambda_L^{5/2}}. λ Q = λ L 5/2 . 由重力/惯性力尺度 F ∼ ρ g L 3 F\sim\rho gL^3 F ∼ ρ g L 3
λ F = λ L 3 . \boxed{\lambda_F=\lambda_L^3}. λ F = λ L 3 . 由 P = F v P=Fv P = F v
λ P = λ L 7 / 2 . \boxed{\lambda_P=\lambda_L^{7/2}}. λ P = λ L 7/2 . A9-7 溢流坝模型流量# 【21–22 真题】
中文题目： 某溢流坝原型特征高度为 12 m 12\ \mathrm m 12 m 60 m 3 / s 60\ \mathrm{m^3/s} 60 m 3 /s 0.48 m 0.48\ \mathrm m 0.48 m
English version: A prototype spillway has a characteristic height of 12 m 12\ \mathrm m 12 m 60 m 3 / s 60\ \mathrm{m^3/s} 60 m 3 /s 0.48 m 0.48\ \mathrm m 0.48 m
答案与讲解 λ L = 12 0.48 = 25. \lambda_L=\frac{12}{0.48}=25. λ L = 0.48 12 = 25. Froude 相似下
λ Q = λ L 5 / 2 = 25 5 / 2 = 3125. \lambda_Q=\lambda_L^{5/2}=25^{5/2}=3125. λ Q = λ L 5/2 = 2 5 5/2 = 3125. Q m = Q p λ Q = 60 3125 = 0.0192 m 3 / s . Q_m=\frac{Q_p}{\lambda_Q}
=\frac{60}{3125}=0.0192\ \mathrm{m^3/s}. Q m = λ Q Q p = 3125 60 = 0.0192 m 3 /s . Q m = 1.92 × 10 − 2 m 3 / s \boxed{Q_m=1.92\times10^{-2}\ \mathrm{m^3/s}} Q m = 1.92 × 1 0 − 2 m 3 /s A9-8 Reynolds 相似下模型速度# 【教材概念题改编】
中文题目： 管流模型与原型使用同一种流体，长度比尺 λ L = L p / L m = 20 \lambda_L=L_p/L_m=20 λ L = L p / L m = 20 v p = 2.0 m / s v_p=2.0\ \mathrm{m/s} v p = 2.0 m/s
English version: A pipe-flow model and prototype use the same fluid. The length scale is λ L = L p / L m = 20 \lambda_L=L_p/L_m=20 λ L = L p / L m = 20 2.0 m / s 2.0\ \mathrm{m/s} 2.0 m/s
答案与讲解 同种流体 λ ν = 1 \lambda_\nu=1 λ ν = 1
R e p = R e m Re_p=Re_m R e p = R e m 得到
λ v = v p v m = λ ν λ L = 1 20 . \lambda_v=\frac{v_p}{v_m}=\frac{\lambda_\nu}{\lambda_L}=\frac1{20}. λ v = v m v p = λ L λ ν = 20 1 . 所以
v m = 20 v p = 40 m / s . v_m=20v_p=40\ \mathrm{m/s}. v m = 20 v p = 40 m/s . v m = 40 m / s \boxed{v_m=40\ \mathrm{m/s}} v m = 40 m/s 小模型反而要求更高速度，往往导致试验困难，这正是黏性相似模型的尺度效应来源之一。
A9-9 船模兴波阻力、速度与功率换算# 【教材习题 7-9 改编】
中文题目： 长度比尺 λ L = 50 \lambda_L=50 λ L = 50 v m = 1.0 m / s v_m=1.0\ \mathrm{m/s} v m = 1.0 m/s F m = 0.020 N F_m=0.020\ \mathrm N F m = 0.020 N
English version: A ship model has a length scale λ L = 50 \lambda_L=50 λ L = 50 1.0 m / s 1.0\ \mathrm{m/s} 1.0 m/s 0.020 N 0.020\ \mathrm N 0.020 N
答案与讲解 速度比尺：
λ v = 50 = 7.071 , v p = 7.071 m / s . \lambda_v=\sqrt{50}=7.071,
\qquad
v_p=7.071\ \mathrm{m/s}. λ v = 50 = 7.071 , v p = 7.071 m/s . 力比尺：
λ F = 50 3 = 125000 , F p = 0.020 × 125000 = 2500 N . \lambda_F=50^3=125000,
\qquad
F_p=0.020\times125000=2500\ \mathrm N. λ F = 5 0 3 = 125000 , F p = 0.020 × 125000 = 2500 N . 功率：
P p = F p v p = 2500 × 7.071 = 17.7 k W . P_p=F_pv_p=2500\times7.071=17.7\ \mathrm{kW}. P p = F p v p = 2500 × 7.071 = 17.7 kW . v p = 7.07 m / s , F p = 2.50 k N , P p = 17.7 k W \boxed{v_p=7.07\ \mathrm{m/s},\quad F_p=2.50\ \mathrm{kN},\quad P_p=17.7\ \mathrm{kW}} v p = 7.07 m/s , F p = 2.50 kN , P p = 17.7 kW 这里换算的是兴波阻力；实际总阻力还包含摩擦阻力，需要分项修正。
A9-10 矩形薄壁堰流量的结构形式# 【教材习题 7-4 改编】
中文题目： 矩形薄壁堰的单位时间过流量 Q Q Q b b b H H H g g g
English version: The discharge Q Q Q b b b H H H g g g
答案与讲解 设
Q = k b a H c g d . Q=kb^aH^cg^d. Q = k b a H c g d . 量纲：
L 3 T − 1 = L a L c ( L T − 2 ) d . L^3T^{-1}=L^aL^c(LT^{-2})^d. L 3 T − 1 = L a L c ( L T − 2 ) d . 时间指数：
− 2 d = − 1 ⇒ d = 1 2 . -2d=-1\Rightarrow d=\frac12. − 2 d = − 1 ⇒ d = 2 1 . 长度指数：
a + c + d = 3. a+c+d=3. a + c + d = 3. 仅靠量纲分析，a a a c c c a = 1 a=1 a = 1
c = 3 2 . c=\frac32. c = 2 3 . 因此
Q = C d b g H 3 / 2 . \boxed{Q=C_db\sqrt g\,H^{3/2}}. Q = C d b g H 3/2 . 这道题提醒：量纲分析有时需要额外物理信息，不能凭量纲唯一确定所有指数。
A9-11 为什么同种流体难以同时满足 Re 与 Fr 相似# 【教材思考题改编】
中文题目： 原型和模型使用同种流体、处于同一重力场。证明除 λ L = 1 \lambda_L=1 λ L = 1
English version: A model and its prototype use the same fluid in the same gravitational field. Prove that exact Reynolds and Froude similarity cannot be satisfied simultaneously unless λ L = 1 \lambda_L=1 λ L = 1
答案与讲解 同种流体下 Reynolds 相似要求
λ v = 1 λ L . \lambda_v=\frac1{\lambda_L}. λ v = λ L 1 . Froude 相似要求
λ v = λ L 1 / 2 . \lambda_v=\lambda_L^{1/2}. λ v = λ L 1/2 . 二者同时成立：
λ L − 1 = λ L 1 / 2 ⇒ λ L 3 / 2 = 1 ⇒ λ L = 1 . \lambda_L^{-1}=\lambda_L^{1/2}
\Rightarrow \lambda_L^{3/2}=1
\Rightarrow \boxed{\lambda_L=1}. λ L − 1 = λ L 1/2 ⇒ λ L 3/2 = 1 ⇒ λ L = 1 . 因此真正缩尺模型只能根据主导力选择主要相似准则，或通过改变模型流体黏度来缓解冲突。
A9-12 溢流坝模型数据换算# 【教材习题 7-12 改编】
中文题目： 溢流坝模型长度比尺为 λ L = 60 \lambda_L=60 λ L = 60 Q p = 500 m 3 / s Q_p=500\ \mathrm{m^3/s} Q p = 500 m 3 /s H m = 0.060 m H_m=0.060\ \mathrm m H m = 0.060 m
English version: A spillway model has a length scale λ L = 60 \lambda_L=60 λ L = 60 500 m 3 / s 500\ \mathrm{m^3/s} 500 m 3 /s 0.060 m 0.060\ \mathrm m 0.060 m
答案与讲解 λ Q = λ L 5 / 2 = 60 5 / 2 = 2.7885 × 10 4 . \lambda_Q=\lambda_L^{5/2}=60^{5/2}=2.7885\times10^4. λ Q = λ L 5/2 = 6 0 5/2 = 2.7885 × 1 0 4 . Q m = Q p λ Q = 500 2.7885 × 10 4 = 1.793 × 10 − 2 m 3 / s . Q_m=\frac{Q_p}{\lambda_Q}
=\frac{500}{2.7885\times10^4}
=1.793\times10^{-2}\ \mathrm{m^3/s}. Q m = λ Q Q p = 2.7885 × 1 0 4 500 = 1.793 × 1 0 − 2 m 3 /s . 水头属于长度量：
H p = λ L H m = 60 × 0.060 = 3.60 m . H_p=\lambda_LH_m=60\times0.060=3.60\ \mathrm m. H p = λ L H m = 60 × 0.060 = 3.60 m . Q m = 0.0179 m 3 / s , H p = 3.60 m \boxed{Q_m=0.0179\ \mathrm{m^3/s},\qquad H_p=3.60\ \mathrm m} Q m = 0.0179 m 3 /s , H p = 3.60 m 