112 年 國立中央大學化學工程與材料工程學系碩士班《單元操作與輸送現象》

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第 1 題15 分

  1. (15%) The McCabe-Thiele method is a chemical engineering technique for the analysis of binary distillation.
    (a) (7%) What is/are the major assumption(s) made in McCabe-Thiele method? At what conditions, can we make such assumption(s)?
    (b) (8%) Please draw a typical McCabe-Thiele diagram for the distillation of a binary feed and identify the equilibrium line, two operating lines, q-line, feed tray, and the compositions of feed, bottoms and distillate in the diagram.

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這一題的完整詳解

核心觀念

McCabe–Thiele 法把整座二成分蒸餾塔的質量平衡與相平衡畫在同一張 xx–yy 圖上，用「階梯」逐板求理論板數。能用直線當操作線，靠的是恆定莫耳溢流（constant molar overflow, CMO）這個假設。

(a) 主要假設與成立條件

  1. 恆定莫耳溢流（最核心）：在精餾段內液體流率 LL、蒸氣流率 VV 各自維持定值，提餾段內 Lˉ\bar L、Vˉ\bar V 也各自維持定值；只有在進料板（或側流處）才會改變。
    這表示每冷凝 1 mol 蒸氣，恰好汽化 1 mol 液體。
  2. 每一板都是理想平衡級：離開某一板的汽、液兩相互成平衡（實際板數再以板效率修正）。
  3. 塔為絕熱、壓力固定：沒有熱損失，整座塔的平衡曲線可視為同一條。
  4. 二成分系統、穩態操作。

要讓恆定莫耳溢流成立，需要下列條件：

  • 兩成分的莫耳汽化熱相近（這是最重要的條件），例如苯–甲苯、正庚烷–正辛烷這類化學性質相似的物系；
  • 顯熱變化與混合熱可忽略：塔頂到塔底的溫差不大、溶液接近理想溶液；
  • 塔身保溫良好，熱損失可忽略。

注意：McCabe–Thiele 法不需要相對揮發度為定值——平衡線可以直接用實驗數據畫，任何形狀都行；相對揮發度恆定只是用來寫出平衡線解析式的額外簡化。若兩成分汽化熱差很多（例如乙醇–水在嚴格計算時），就要改用焓–濃度圖（Ponchon–Savarit 法）。

(b) 典型 McCabe–Thiele 圖的畫法與各元素

座標：橫軸為液相中易揮發成分的莫耳分率 xx，縱軸為汽相莫耳分率 yy，範圍皆為 0∼10\sim1。圖上依序畫出：

  1. 對角線 y=xy=x：作為定位用的參考線。
  2. 平衡線（equilibrium line）：y∗=f(x)y^*=f(x)，位於對角線上方的曲線，兩端通過 (0,0)(0,0) 與 (1,1)(1,1)。
  3. 三個組成（皆標在對角線上）：
    • 塔底產物 xBx_B：靠近左下角（xx 小）；
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第 2 題20 分

  1. (20%) A droplet of liquid A, of radius r1r_1, is suspended in a stream of gas B. We postulate that there is a spherical stagnant gas film of radius r2r_2 surrounding the droplet. The concentration of A in the gas stream is xA1x_{A1} at r=r1r = r_1 and xA2x_{A2} at the outer edge of the film, r=r2r = r_2. Please find the molar flux NAr1N_{Ar1} at r=r1r = r_1.

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這一題的完整詳解

核心觀念

這是「A 經過停滯的 B 擴散」在球座標下的版本（BSL Problem 18.B 系列）。氣體 B 不溶於液滴，所以 NBr=0N_{Br}=0；穩態、無反應時，通過每一個同心球面的總莫耳流率相同。

假設

  • 擬穩態：液滴半徑變化很慢，可視為穩態；
  • 等溫、等壓的理想氣體，總濃度 cc 與擴散係數 DABD_{AB} 為定值；
  • B 在液體中不溶、不反應：NBr=0N_{Br}=0；
  • 只有徑向擴散。

1. 殼層質量平衡

在 rr 到 r+Δrr+\Delta r 的球殼上，穩態下

ddr(r2NAr)=0 ⟹ r2NAr=const=r12NAr1.\frac{d}{dr}\left(r^2N_{Ar}\right)=0 \ \Longrightarrow\ r^2N_{Ar}=\text{const}=r_1^2N_{Ar1}.

2. 通量式（Fick 定律＋整體流動）

NAr=−cDABdxAdr+xA(NAr+NBr),NBr=0N_{Ar}=-cD_{AB}\frac{dx_A}{dr}+x_A\left(N_{Ar}+N_{Br}\right),\qquad N_{Br}=0 ⟹ NAr=−cDAB1−xAdxAdr.\Longrightarrow\ N_{Ar}=-\frac{cD_{AB}}{1-x_A}\frac{dx_A}{dr}.

3. 合併並積分

r12NAr11r2=−cDAB1−xAdxAdrr_1^2N_{Ar1}\frac{1}{r^2}=-\frac{cD_{AB}}{1-x_A}\frac{dx_A}{dr}

分離變數，由 r1r_1（xA=xA1x_A=x_{A1}）積分到 r2r_2（xA=xA2x_A=x_{A2}）：

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第 3 題10 分

  1. (10%) Answer the following problems
    (a) (4%, 2% for each problem) Regarding the Navier-Stokes equation:
ρDvDt=ρ(∂v∂t+v⋅∇v)=−∇P+μ∇2v+ρg\rho\frac{D\mathbf{v}}{Dt}=\rho\left(\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right)=-\nabla P+\mu\nabla^{2}\mathbf{v}+\rho\mathbf{g}

(1) Please list the assumptions made by the Navier-Stokes equation.
(2) Please write down the simplified equation when Reynold's number is very small. (Assume it is steady-state).
(b) (3%, 1% for each problem) Regarding the Hagen-Poiseuille equation, which following statements are true?
(1) Applicable to flows that are driven only by gravity.
(2) Applicable to both Newtonian and non-Newtonian fluids.
(3) Applicable to flows with Re < 0.1.
(c) (3%, 1% for each problem) For a fluid that is Newtonian but compressible, and its flow is un-steady-state with a Reynold's number of about 10. Answer whether the following equation (Yes or No) is applicable to the fluid and the flow. No explanation is required.
(1) Equation of continuity
(2) Equation of motion in terms of viscous force
(3) Stokes (creeping) flow equation

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這一題的完整詳解

(a) Navier–Stokes equation

(1) 題目所寫形式的假設

ρDvDt=−∇P+μ∇2v+ρg\rho\frac{D\mathbf{v}}{Dt}=-\nabla P+\mu\nabla^2\mathbf{v}+\rho\mathbf{g}

這個形式是由一般的運動方程式（in terms of τ\boldsymbol{\tau}）代入牛頓黏度定律後化簡而來，所以需要：

  • 牛頓流體（應力與應變率成線性）；
  • 不可壓縮：ρ\rho 為定值，∇⋅v=0\nabla\cdot\mathbf{v}=0（這樣 (23μ−κ)(∇⋅v)(\tfrac23\mu-\kappa)(\nabla\cdot\mathbf{v}) 項才會消失）；
  • 黏度 μ\mu 為定值（等溫或黏度不隨溫度變化，才能把 μ\mu 提到微分外面）；
  • 連體假設（continuum），重力為唯一體積力。

(2) Reynolds number 很小、穩態時

Re≪1Re\ll1 表示慣性力遠小於黏性力，左邊 ρ v⋅∇v\rho\,\mathbf{v}\cdot\nabla\mathbf{v} 可忽略；穩態再去掉 ∂v/∂t\partial\mathbf{v}/\partial t，得 Stokes（creeping flow）方程式：

0=−∇P+μ∇2v+ρg或∇P=μ∇2v,  P=P+ρgh.\mathbf{0}=-\nabla P+\mu\nabla^2\mathbf{v}+\rho\mathbf{g} \qquad\text{或}\qquad \nabla\mathcal{P}=\mu\nabla^2\mathbf{v},\ \ \mathcal{P}=P+\rho g h .

(b) Hagen–Poiseuille equation

Q=π(P0−PL)R48μLQ=\frac{\pi(\mathcal{P}_0-\mathcal{P}_L)R^4}{8\mu L}

推導前提：穩態、層流、完全發展、不可壓縮的牛頓流體在長直圓管中流動、管壁無滑移。

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第 4 題20 分

  1. (20%) An incompressible (density ρ\rho) and non-Newtonian liquid flows in an inclined circular pipe with a radius of RR and length of LL. The pipe is tilted with an angle of 30 degree. A schematic illustration of the system can be seen below (with the definition of system coordinates). The flow is laminar, steady-state, and it is fully developed with no edge and entrance/exit effects. PoP_o and PLP_L represent the pressures on the two ends of the tube. In this system, we have: vz=vz(r)v_z = v_z(r), vr=0v_r = 0, and vθ=0v_\theta = 0; τθz=τzz=0\tau_{\theta z} = \tau_{zz} = 0.

Please use the coordinates defined in the figure for derivation.

🖼️【此處有附圖：與水平夾 30° 的傾斜圓管，長度 LL；下端壓力 PoP_o、上端壓力 PLP_L，zz 軸沿管軸由 PoP_o 端指向 PLP_L 端（向上），rr 為徑向座標；重力 gg 鉛直向下】

For a Bingham fluid, the shear stress of the fluid can be described as the following

dVzdr=0if ∣τrz∣<τo,where τo=ρgR8\frac{dV_z}{dr}=0 \quad \text{if } |\tau_{rz}|<\tau_o,\qquad \text{where } \tau_o=\frac{\rho g R}{8} τrz=±τo−μdVzdr=0;if ∣τrz∣≥τo,{τrz=τo−μdVzdrif τrz>0τrz=−τo−μdVzdrif τrz<0\tau_{rz}=\pm\tau_o-\mu\frac{dV_z}{dr}=0;\quad \text{if } |\tau_{rz}|\ge\tau_o,\quad \begin{cases} \tau_{rz}=\tau_o-\mu\dfrac{dV_z}{dr} & \text{if } \tau_{rz}>0\\[4pt] \tau_{rz}=-\tau_o-\mu\dfrac{dV_z}{dr} & \text{if } \tau_{rz}<0 \end{cases}

(a) (10%) Should we expect fluid (Bingham fluid) to flow down under the condition of Po=PLP_o=P_L? (No pressure drop along z-direction.) Please provide detailed to support your answer.
(b) (10%) Follow the previous question. If a pressure drop along the positive z direction (Po>PLP_o>P_L) is provided, with a sufficiently large pressure drop we should be able to stop the fluid to flow downward and push the fluid upward. Please determine the critical pressure drop that allows the fluid to flow along the positive z direction (Please present the critical pressure drop Po−PLP_o-P_L in terms of ρ\rho, gg, RR, LL, and/or μ\mu).

Newton's law of viscosity（cylindrical coordinates (r,θ,z)(r,\theta,z)）

τrr=−μ[2∂vr∂r]+(23μ−κ)(∇⋅v)τθθ=−μ[2(1r∂vθ∂θ+vrr)]+(23μ−κ)(∇⋅v)τzz=−μ[2∂vz∂z]+(23μ−κ)(∇⋅v)τrθ=τθr=−μ[r∂∂r(vθr)+1r∂vr∂θ]τθz=τzθ=−μ[1r∂vz∂θ+∂vθ∂z]τzr=τrz=−μ[∂vz∂r+∂vr∂z]\begin{aligned} \tau_{rr}&=-\mu\left[2\frac{\partial v_r}{\partial r}\right]+\left(\tfrac{2}{3}\mu-\kappa\right)(\nabla\cdot\mathbf{v})\\ \tau_{\theta\theta}&=-\mu\left[2\left(\frac{1}{r}\frac{\partial v_\theta}{\partial \theta}+\frac{v_r}{r}\right)\right]+\left(\tfrac{2}{3}\mu-\kappa\right)(\nabla\cdot\mathbf{v})\\ \tau_{zz}&=-\mu\left[2\frac{\partial v_z}{\partial z}\right]+\left(\tfrac{2}{3}\mu-\kappa\right)(\nabla\cdot\mathbf{v})\\ \tau_{r\theta}=\tau_{\theta r}&=-\mu\left[r\frac{\partial}{\partial r}\left(\frac{v_\theta}{r}\right)+\frac{1}{r}\frac{\partial v_r}{\partial \theta}\right]\\ \tau_{\theta z}=\tau_{z\theta}&=-\mu\left[\frac{1}{r}\frac{\partial v_z}{\partial \theta}+\frac{\partial v_\theta}{\partial z}\right]\\ \tau_{zr}=\tau_{rz}&=-\mu\left[\frac{\partial v_z}{\partial r}+\frac{\partial v_r}{\partial z}\right] \end{aligned}

in which

(∇⋅v)=1r∂∂r(rvr)+1r∂vθ∂θ+∂vz∂z(\nabla\cdot\mathbf{v})=\frac{1}{r}\frac{\partial}{\partial r}(r v_r)+\frac{1}{r}\frac{\partial v_\theta}{\partial \theta}+\frac{\partial v_z}{\partial z}

The equation of continuity [∂ρ/∂t+(∇⋅ρv)=0]\left[\partial\rho/\partial t+(\nabla\cdot\rho\mathbf{v})=0\right]

Cartesian (x,y,z):∂ρ∂t+∂∂x(ρvx)+∂∂y(ρvy)+∂∂z(ρvz)=0\text{Cartesian }(x,y,z):\quad \frac{\partial\rho}{\partial t}+\frac{\partial}{\partial x}(\rho v_x)+\frac{\partial}{\partial y}(\rho v_y)+\frac{\partial}{\partial z}(\rho v_z)=0 Cylindrical (r,θ,z):∂ρ∂t+1r∂∂r(ρrvr)+1r∂∂θ(ρvθ)+∂∂z(ρvz)=0\text{Cylindrical }(r,\theta,z):\quad \frac{\partial\rho}{\partial t}+\frac{1}{r}\frac{\partial}{\partial r}(\rho r v_r)+\frac{1}{r}\frac{\partial}{\partial \theta}(\rho v_\theta)+\frac{\partial}{\partial z}(\rho v_z)=0

The equation of motion in terms of τ\tau [ρDv/Dt=−∇p−[∇⋅τ]+ρg]\left[\rho D\mathbf{v}/Dt=-\nabla p-[\nabla\cdot\boldsymbol{\tau}]+\rho\mathbf{g}\right]（cylindrical coordinates (r,θ,z)(r,\theta,z)）

ρ(∂vr∂t+vr∂vr∂r+vθr∂vr∂θ+vz∂vr∂z−vθ2r)=−∂p∂r−[1r∂∂r(rτrr)+1r∂∂θτθr+∂∂zτzr−τθθr]+ρgrρ(∂vθ∂t+vr∂vθ∂r+vθr∂vθ∂θ+vz∂vθ∂z+vrvθr)=−1r∂p∂θ−[1r2∂∂r(r2τrθ)+1r∂∂θτθθ+∂∂zτzθ+τθr−τrθr]+ρgθρ(∂vz∂t+vr∂vz∂r+vθr∂vz∂θ+vz∂vz∂z)=−∂p∂z−[1r∂∂r(rτrz)+1r∂∂θτθz+∂∂zτzz]+ρgz\begin{aligned} &\rho\left(\frac{\partial v_r}{\partial t}+v_r\frac{\partial v_r}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_r}{\partial \theta}+v_z\frac{\partial v_r}{\partial z}-\frac{v_\theta^2}{r}\right)=-\frac{\partial p}{\partial r}-\left[\frac{1}{r}\frac{\partial}{\partial r}(r\tau_{rr})+\frac{1}{r}\frac{\partial}{\partial \theta}\tau_{\theta r}+\frac{\partial}{\partial z}\tau_{zr}-\frac{\tau_{\theta\theta}}{r}\right]+\rho g_r\\ &\rho\left(\frac{\partial v_\theta}{\partial t}+v_r\frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_\theta}{\partial \theta}+v_z\frac{\partial v_\theta}{\partial z}+\frac{v_r v_\theta}{r}\right)=-\frac{1}{r}\frac{\partial p}{\partial \theta}-\left[\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\tau_{r\theta})+\frac{1}{r}\frac{\partial}{\partial \theta}\tau_{\theta\theta}+\frac{\partial}{\partial z}\tau_{z\theta}+\frac{\tau_{\theta r}-\tau_{r\theta}}{r}\right]+\rho g_\theta\\ &\rho\left(\frac{\partial v_z}{\partial t}+v_r\frac{\partial v_z}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_z}{\partial \theta}+v_z\frac{\partial v_z}{\partial z}\right)=-\frac{\partial p}{\partial z}-\left[\frac{1}{r}\frac{\partial}{\partial r}(r\tau_{rz})+\frac{1}{r}\frac{\partial}{\partial \theta}\tau_{\theta z}+\frac{\partial}{\partial z}\tau_{zz}\right]+\rho g_z \end{aligned}
🖼️ 本題含圖表，以下為原卷對應頁面：
原卷第 1 頁原卷第 2 頁原卷第 3 頁

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座標與已知

依附圖：zz 軸沿管軸、由下端 PoP_o 指向上端 PLP_L，管軸與水平夾 30∘30^\circ，重力鉛直向下，所以重力在 zz 方向的分量為

gz=−gsin⁡30∘=−g2.g_z=-g\sin30^\circ=-\frac{g}{2}.

題目給定 vz=vz(r)v_z=v_z(r)、vr=vθ=0v_r=v_\theta=0、τθz=τzz=0\tau_{\theta z}=\tau_{zz}=0，Bingham 流體的屈服應力 τo=ρgR8\tau_o=\dfrac{\rho gR}{8}。

（原卷的本構式寫成「τrz=±τo−μ dVz/dr=0\tau_{rz}=\pm\tau_o-\mu\,dV_z/dr=0」，最後的「=0=0」是筆誤；實際使用的是後面分段寫出的 τrz=±τo−μ dVz/dr\tau_{rz}=\pm\tau_o-\mu\,dV_z/dr。）

1. 由 zz 方向運動方程式求剪應力分布（與流體種類無關）

穩態、完全發展，左邊慣性項全為零；τθz=τzz=0\tau_{\theta z}=\tau_{zz}=0：

0=−∂p∂z−1rddr(rτrz)+ρgz.0=-\frac{\partial p}{\partial z}-\frac{1}{r}\frac{d}{dr}\left(r\tau_{rz}\right)+\rho g_z .

完全發展時壓力沿管長線性變化，−∂p∂z=Po−PLL-\dfrac{\partial p}{\partial z}=\dfrac{P_o-P_L}{L}。令淨驅動力

G≡Po−PLL−ρg2(單位體積、指向 +z 的淨推力),G\equiv\frac{P_o-P_L}{L}-\frac{\rho g}{2}\qquad(\text{單位體積、指向 }+z\text{ 的淨推力}),

則 ddr(rτrz)=Gr\dfrac{d}{dr}(r\tau_{rz})=Gr，積分並要求 r=0r=0 處應力有限：

τrz=G r2,∣τrz∣max⁡=∣τw∣=∣G∣R2 (在管壁).\tau_{rz}=\frac{G\,r}{2},\qquad |\tau_{rz}|_{\max}=|\tau_w|=\frac{|G|R}{2}\ (\text{在管壁}).

流動判準：Bingham 流體只有在 ∣τrz∣≥τo|\tau_{rz}|\ge\tau_o 的區域才會產生剪切。只要管壁應力 ∣τw∣>τo|\tau_w|>\tau_o，靠近管壁就有剪切層，流體會流動；若 ∣τw∣≤τo|\tau_w|\le\tau_o，整個截面都低於屈服應力，流體像固體一樣不動。

(a) Po=PLP_o=P_L 時會不會往下流？

沒有壓力差時 G=−ρg2G=-\dfrac{\rho g}{2}，

τrz=−ρg r4,∣τw∣=ρgR4=2τo>τo.\tau_{rz}=-\frac{\rho g\,r}{4},\qquad |\tau_w|=\frac{\rho gR}{4}=2\tau_o>\tau_o .

管壁應力是屈服應力的兩倍，所以會往下流（−z-z 方向）。

  • 柱塞區（plug）：∣τrz∣<τo⇒ρgr4<ρgR8⇒r<ro=R2|\tau_{rz}|<\tau_o\Rightarrow \dfrac{\rho g r}{4}<\dfrac{\rho gR}{8}\Rightarrow r<r_o=\dfrac{R}{2}，
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第 5 題35 分

  1. (35%) To satisfy the consumer market's demand for wearable devices which are simultaneously powerful, multifunctional and lightweight, electronics manufacturers have been gearing up in their R&D efforts to miniaturize their products while maintaining or even beefing up the computing powers of their devices. One key to winning this miniaturization arms race is on the ability to shrink the sizes of the chips by packaging even more transistors into ever smaller chip dies. In addition to the technical challenges associated with the electronics designs and materials development, packing unprecedented numbers of transistors into increasingly small spaces is also faced with a daunting problem: dissipation of heat generated from the transistors packed in a tiny space, a make-or-break issue for any sophisticated electronics design and industrial project aiming to miniaturize an electronic device. As an engineer, you are tasked with resolving the heat-dissipation issue for the microchip of a new sporting wearable device currently in the R&D phase of your company. You plan to make a heat sink which can dissipate heat generated from the microchip into flowing water when users of the device are swimming (or flowing air when the users are walking or running, situations that are not considered here). As the first step, you decide to construct a model system with a geometry identical to that of the real system but with a larger dimension which allows you to test with ease your heat sink designs even before the microchip comes into being. Undoubtedly, the model system, consisting of the heat sink mock-up and a flowing fluid, must reflect the heat transfer condition of the real system. And you recognize that this demands a thorough consideration from the perspective of dimensionless parameters.

(1) (2%) To ensure the similarity in heat transfer between the model and real systems, what dimensionless parameters (groups) have to be identical across the two systems?
(2) (2%) Explain the physical meanings of the dimensionless parameters in (1)?
(3) (2%) Explain the physical reasons behind the fact that the dimensionless parameters in (1) must be identical across the two systems if their heat transfer conditions are to be similar.

When the dimensionless parameters in (1) are identical across the two systems, another dimensionless parameter may give the same functional form for the two systems.

(4) (2%) Identify this dimensionless parameter and describe the information from this dimensionless parameter, which is relevant to heat transfer?
(5) (4%) Based on the definition of the dimensionless temperature T∗≡T−TsT∞−TsT^*\equiv\dfrac{T-T_s}{T_\infty-T_s}, where T∞T_\infty is the free-streaming fluid and TsT_s is the surface temperature of the heat sink, as well as the boundary layer equations, conceptually prove that the dimensionless parameter in (4) indeed has the same functional form for the model and real systems when the dimensionless parameters in (1) are identical across the two systems.

You are informed that the dimension of the chip die is set to 1 mm across (so will the heat sink). And it is known that the water flow (at 17°C) experienced by a swimmer is of the velocity 0.8 m/s on average. You employ air at 27°C, instead of water at 17°C, as the flowing fluid in your model system.

(6) (3%) With the model system in the dimension of 10 cm, in what velocity should you operate the water flow to make the heat transfer conditions similar between the model and real systems?
(7) (2%) Is air at 27°C a right choice for your model system? Why?

Under the assumption that the heat-transfer similarity can be achieved with your choice of the conditions for the model system, you supply a fixed heat flux of 200 W/m2200\ \text{W/m}^2 to the heat sink mock-up, which is transferred to air, and measure the surface temperature of the mock-up. In the steady-state condition, the surface temperature is measured to be 55°C.

(8) (5%) What is the convection heat transfer coefficient for the real system in operation?
(9) (3%) How will the convection heat transfer coefficient evolve along the direction parallel to the surface of the heat sink?

You decide to use sapphire as the material for the heat sink in the real system, for the sake of the aesthetic appeal to consumers. A single (one-time) operation of the microchip brings the surface temperature of the heat sink to 65°C, which is surely uncomfortable to the users.

(10) (5%) Given that both of the two faces of the heat sink are in contact with flowing water, how long does it take for the surface temperature of the heat sink to drop to a more comfortable temperature of 40°C?
(11) (5%) How much heat is transferred during the period?

Air（M=28.97\mathcal{M}=28.97 kg/kmol）

TT (K)ρ\rho (kg/m³)cpc_p (kJ/kg·K)μ⋅107\mu\cdot10^7 (N·s/m²)ν⋅106\nu\cdot10^6 (m²/s)k⋅103k\cdot10^3 (W/m·K)α⋅106\alpha\cdot10^6 (m²/s)PrPr
1003.55621.03271.12.009.342.540.786
1502.33641.012103.44.42613.85.840.758
2001.74581.007132.57.59018.110.30.737
2501.39471.006159.611.4422.315.90.720
3001.16141.007184.615.8926.322.50.707
3500.99501.009208.220.9230.029.90.700
4000.87111.014230.126.4133.838.30.690
4500.77401.021250.732.3937.347.20.686
5000.69641.030270.138.7940.756.70.684
5500.63291.040288.445.5743.966.70.683

Aluminum oxide, sapphire

Melting point (K)ρ\rho (kg/m³)，300 Kcpc_p (J/kg·K)，300 Kkk (W/m·K)，300 Kα⋅106\alpha\cdot10^6 (m²/s)，300 K
232339707654615.1

Properties at various temperatures (K)，kk (W/m·K) / cpc_p (J/kg·K)：100 K：450 / —；200 K：82 / —；400 K：32.4 / 940；600 K：18.9 / 1110；800 K：13.0 / 1180；1000 K：10.5 / 1225

Table A.6 Thermophysical Properties of Saturated Water

TT (K)pp (bars)vf⋅103v_f\cdot10^3 (m³/kg)vgv_g (m³/kg)hfgh_{fg} (kJ/kg)cp,fc_{p,f} (kJ/kg·K)cp,gc_{p,g} (kJ/kg·K)μf⋅106\mu_f\cdot10^6 (N·s/m²)μg⋅106\mu_g\cdot10^6 (N·s/m²)kf⋅103k_f\cdot10^3 (W/m·K)kg⋅103k_g\cdot10^3 (W/m·K)PrfPr_fPrgPr_gσf⋅103\sigma_f\cdot10^3 (N/m)βf⋅106\beta_f\cdot10^6 (K⁻¹)
273.150.006111.000206.325024.2171.85417508.0256918.212.990.81575.5−68.05
2750.006971.000181.724974.2111.85516528.0957418.312.220.81775.3−32.74
2800.009901.000130.424854.1981.85814228.2958218.610.260.82574.846.04
2850.013871.00099.424734.1891.86112258.4959018.98.810.83374.3114.1
2900.019171.00169.724614.1841.86410808.6959819.37.560.84173.7174.0
2950.026171.00251.9424494.1811.8689598.8960619.56.620.84972.7227.5
3000.035311.00339.1324384.1791.8728559.0961319.65.830.85771.7276.1
3050.047121.00529.7424264.1781.8777699.2962020.15.200.86570.9320.6
3100.062211.00722.9324144.1781.8826959.4962820.44.620.87370.0361.9
3150.081321.00917.8224024.1791.8886319.6963420.74.160.88369.2400.4

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這一題的完整詳解

核心觀念

這題是「用放大模型做對流熱傳實驗」的相似律問題：只要幾何相似、而且無因次群相同，模型與實體的無因次邊界層方程式與邊界條件就完全一樣，量到的 Nusselt number 可以直接搬回實體系統。後半段再用實體的 hh 計算藍寶石散熱片的暫態冷卻。

(1) 必須相同的無因次群

強制對流（幾何相似）時：

ReL=VLν,Pr=να=cpμk.Re_L=\frac{VL}{\nu},\qquad Pr=\frac{\nu}{\alpha}=\frac{c_p\mu}{k}.

(2) 物理意義

  • ReLRe_L：慣性力與黏性力之比；決定速度邊界層的厚度與流況（層流／紊流）。
  • PrPr：動量擴散係數與熱擴散係數之比；決定速度邊界層與熱邊界層的相對厚度（層流時 δ/δt≈Pr1/3\delta/\delta_t\approx Pr^{1/3}）。

(3) 為什麼必須相同

把連續、動量、能量邊界層方程式無因次化後，方程式裡只剩下 ReLRe_L 與 PrPr 兩個參數（壓力梯度 dp∗/dx∗dp^*/dx^* 由幾何決定，幾何相似即相同）；無因次邊界條件（壁面 u∗=0u^*=0、T∗=0T^*=0，遠處 u∗=1u^*=1、T∗=1T^*=1）也完全相同。方程式＋邊界條件都相同，無因次解 u∗(x∗,y∗)u^*(x^*,y^*)、T∗(x∗,y∗)T^*(x^*,y^*) 就相同，兩系統的熱傳行為才會相似。只要其中一個群不同，方程式就不同，解也就不同。

(4) 另一個無因次參數：Nusselt number

Nu=hLkf=∂T∗∂y∗∣y∗=0Nu=\frac{hL}{k_f}=\left.\frac{\partial T^*}{\partial y^*}\right|_{y^*=0}

它是壁面處的無因次溫度梯度，代表「對流熱傳相對於同一層流體純傳導」的增強倍數。量到 NuNu 就能得到對流熱傳係數 h=Nu kf/Lh=Nu\,k_f/L——這正是熱傳設計需要的資訊。

(5) 概念證明：NuNu 有相同的函數形式

令 x∗=x/Lx^*=x/L、y∗=y/Ly^*=y/L、u∗=u/Vu^*=u/V、v∗=v/Vv^*=v/V、p∗=p/(ρV2)p^*=p/(\rho V^2)、T∗=T−TsT∞−TsT^*=\dfrac{T-T_s}{T_\infty-T_s}，邊界層方程式成為

u∗∂u∗∂x∗+v∗∂u∗∂y∗=−dp∗dx∗+1ReL∂2u∗∂y∗2,u∗∂T∗∂x∗+v∗∂T∗∂y∗=1ReL Pr∂2T∗∂y∗2.\begin{aligned} &u^*\frac{\partial u^*}{\partial x^*}+v^*\frac{\partial u^*}{\partial y^*}=-\frac{dp^*}{dx^*}+\frac{1}{Re_L}\frac{\partial^2u^*}{\partial y^{*2}},\\ &u^*\frac{\partial T^*}{\partial x^*}+v^*\frac{\partial T^*}{\partial y^*}=\frac{1}{Re_L\,Pr}\frac{\partial^2T^*}{\partial y^{*2}}. \end{aligned}

因此解的函數形式為

u∗=f ⁣(x∗,y∗,ReL,dp∗dx∗),T∗=g ⁣(x∗,y∗,ReL,Pr,dp∗dx∗).u^*=f\!\left(x^*,y^*,Re_L,\tfrac{dp^*}{dx^*}\right),\qquad T^*=g\!\left(x^*,y^*,Re_L,Pr,\tfrac{dp^*}{dx^*}\right).

T∗T^* 對 y∗y^* 微分、取壁面值，y∗y^* 這個變數就消失了：

Nu=∂T∗∂y∗∣y∗=0=ϕ ⁣(x∗,ReL,Pr),Nu‾=Φ ⁣(ReL,Pr).Nu=\left.\frac{\partial T^*}{\partial y^*}\right|_{y^*=0}=\phi\!\left(x^*,Re_L,Pr\right),\qquad \overline{Nu}=\Phi\!\left(Re_L,Pr\right).

dp∗/dx∗dp^*/dx^* 只與幾何有關，幾何相似時兩系統相同。所以 ReLRe_L、PrPr 相同時，兩系統的 Nu‾\overline{Nu} 是同一個函數、同一個數值。

物性（查題附表）

  • 實體：水 17∘17^\circC ≈290\approx290 K：μf=1080×10−6\mu_f=1080\times10^{-6} N·s/m²、vf=1.001×10−3v_f=1.001\times10^{-3} m³/kg（ρ≈999\rho\approx999 kg/m³）、kf=0.598k_f=0.598 W/m·K、Pr=7.56Pr=7.56。
νw=μfvf=1080×10−6×1.001×10−3=1.081×10−6 m2/s\nu_w=\mu_f v_f=1080\times10^{-6}\times1.001\times10^{-3}=1.081\times10^{-6}\ \text{m}^2/\text{s}
  • 模型：空氣 27∘27^\circC =300=300 K：νa=15.89×10−6\nu_a=15.89\times10^{-6} m²/s、ka=0.0263k_a=0.0263 W/m·K、Pr=0.707Pr=0.707。

(6) 模型的操作速度

實體 Lr=1L_r=1 mm、Vr=0.8V_r=0.8 m/s：

Rer=0.8×0.0011.081×10−6=740(層流).Re_r=\frac{0.8\times0.001}{1.081\times10^{-6}}=740\quad(\text{層流}).

模型 Lm=0.1L_m=0.1 m，要求 Rem=RerRe_m=Re_r：

Vm=RerνaLm=740×15.89×10−60.1≈0.118 m/s.V_m=Re_r\frac{\nu_a}{L_m}=740\times\frac{15.89\times10^{-6}}{0.1}\approx0.118\ \text{m/s}.

（題目寫「operate the water flow」，但前文說模型用的是空氣，這裡算的是模型中空氣的流速。）

(7) 用 27∘27^\circC 的空氣恰當嗎？——不恰當

🔒

後續完整解題步驟與【答案】

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其他考古題

中央大學《單元操作與輸送現象》其他年度

115 年114 年113 年111 年108 年

其他學校的化工與材料考古題

臺灣大學106清華大學10成功大學64中正大學14中山大學6高雄大學22宜蘭大學6聯合大學6臺南大學1