113 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6 分

Answer "True" or "False" for each of the following statements; explanation is not needed.

(1) For equation of continuity
∂ρ∂t+∇⋅(ρv)=0\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0
(a) Equation of continuity cannot be used in an unsteady-state system.
(b) The above equation is derived from momentum balance.
(c) The density of the fluid is invariant.

(2) Regarding the equation of motion in terms of v\mathbf{v}
(a) Applicable to both incompressible and compressible fluids.
(b) Cannot be applied to non-Newtonian fluids.
(c) Cannot be applied to an unsteady-state problem.

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這一題的完整詳解

這題主要在考驗考生對連續性方程式（Continuity Equation）和運動方程式（Equation of Motion）基本形式及其適用範圍的理解。

  1. 連續性方程式：
    ∂ρ∂t+∇⋅(ρv)=0\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0
    此方程式代表質量的守恆，即在一個控制體積內，質量隨時間的變化率等於通過該體積表面的淨質量通量。
    (a) False。連續性方程式是描述質量守恆的普遍定律，無論系統是穩態（steady-state）還是非穩態（unsteady-state）都可以使用。在穩態系統中，∂ρ∂t=0\frac{\partial \rho}{\partial t} = 0。
    (b) False。此方程式是從質量守恆定律直接推導而來，而不是動量平衡。動量平衡則導出運動方程式。
    (c) False。此方程式適用於密度隨時間或空間變化的流體（可壓縮流體）。
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第 2 題7 分

A Newtonian and incompressible fluid is in a cylindrical container of radius R and at rest initially. There is a sudden motion applied to rotate the container at an angular speed of Ω\Omega. The cylindrical velocity profile along θ\theta direction in this system can be determined by the following partial differential equation:
ρ∂vθ∂t=μ[1r∂∂r(r∂vθ∂r)−vθr2]\rho \frac{\partial v_{\theta}}{\partial t} = \mu \left[ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial v_{\theta}}{\partial r} \right) - \frac{v_{\theta}}{r^2} \right]
(1) Use R and Ω\Omega as the characteristic length and velocity, respectively. Please express the above PDE in terms of three dimensionless variables r∗r^*, vθ∗v_{\theta}^*, t∗t^*, and vθ∗=vθ/ΩRv_{\theta}^* = v_{\theta}/\Omega R, r∗=r/Rr^* = r/R, t∗=tΩt^* = t\Omega. The answer should include Reynolds number, Re, which should also be clearly defined. (5 points)
(2) If you place the same fluid in two containers with different sizes: R and 2R. To maintain the dynamic similarity between the two systems, if the smaller one rotates at an angular velocity of Ω\Omega, what should the angular velocity of the larger container be? (2 points partial credits)

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這一題的完整詳解

核心觀念

本題考查偏微分方程的無因次化，以及雷諾數在動力相似中的應用。

選定特徵尺度：

  • 特徵長度：RR
  • 特徵速度：ΩR\Omega R
  • 特徵時間：1/Ω1/\Omega

定義無因次變數：

r∗=rR,vθ∗=vθΩR,t∗=tΩr^*=\frac{r}{R},\qquad v_\theta^*=\frac{v_\theta}{\Omega R},\qquad t^*=t\Omega

因此：

r=Rr∗,vθ=ΩRvθ∗,t=t∗Ωr=Rr^*,\qquad v_\theta=\Omega Rv_\theta^*,\qquad t=\frac{t^*}{\Omega}

雷諾數定義為慣性力與黏滯力的比值：

Re=ρULμ\mathrm{Re}=\frac{\rho UL}{\mu}

本題的特徵速度為 U=ΩRU=\Omega R，特徵長度為 L=RL=R，故

Re=ρΩR2μ\boxed{\mathrm{Re}=\frac{\rho\Omega R^2}{\mu}}

解題方法：PDE 無因次化

原方程式為

ρ∂vθ∂t=μ[1r∂∂r(r∂vθ∂r)−vθr2]\rho \frac{\partial v_{\theta}}{\partial t} = \mu \left[ \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial v_{\theta}}{\partial r} \right) -\frac{v_{\theta}}{r^2} \right]

左側時間微分項

由

vθ=ΩRvθ∗,t=t∗Ωv_\theta=\Omega Rv_\theta^*, \qquad t=\frac{t^*}{\Omega}

可得

∂vθ∂t=ΩR∂vθ∗∂t∗∂t∗∂t=Ω2R∂vθ∗∂t∗\frac{\partial v_\theta}{\partial t} = \Omega R \frac{\partial v_\theta^*}{\partial t^*} \frac{\partial t^*}{\partial t} = \Omega^2R \frac{\partial v_\theta^*}{\partial t^*}

因此左側為

ρ∂vθ∂t=ρΩ2R∂vθ∗∂t∗\rho\frac{\partial v_\theta}{\partial t} = \rho\Omega^2R \frac{\partial v_\theta^*}{\partial t^*}

右側第一項

首先，

∂vθ∂r=ΩR∂vθ∗∂r∗∂r∗∂r=Ω∂vθ∗∂r∗\frac{\partial v_\theta}{\partial r} = \Omega R \frac{\partial v_\theta^*}{\partial r^*} \frac{\partial r^*}{\partial r} = \Omega \frac{\partial v_\theta^*}{\partial r^*}

又因為 r=Rr∗r=Rr^*，所以

r∂vθ∂r=Rr∗Ω∂vθ∗∂r∗r\frac{\partial v_\theta}{\partial r} = Rr^*\Omega \frac{\partial v_\theta^*}{\partial r^*}

繼續微分：

∂∂r(r∂vθ∂r)=Ω∂∂r∗(r∗∂vθ∗∂r∗)\frac{\partial}{\partial r} \left( r\frac{\partial v_\theta}{\partial r} \right) = \Omega \frac{\partial}{\partial r^*} \left( r^* \frac{\partial v_\theta^*}{\partial r^*} \right)

因此

1r∂∂r(r∂vθ∂r)=ΩR1r∗∂∂r∗(r∗∂vθ∗∂r∗)\frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial v_\theta}{\partial r} \right) = \frac{\Omega}{R} \frac{1}{r^*} \frac{\partial}{\partial r^*} \left( r^* \frac{\partial v_\theta^*}{\partial r^*} \right)

右側第二項

vθr2=ΩRvθ∗R2r∗2=ΩRvθ∗r∗2\frac{v_\theta}{r^2} = \frac{\Omega Rv_\theta^*}{R^2r^{*2}} = \frac{\Omega}{R} \frac{v_\theta^*}{r^{*2}}

所以右側整體為

μΩR[1r∗∂∂r∗(r∗∂vθ∗∂r∗)−vθ∗r∗2]\mu\frac{\Omega}{R} \left[ \frac{1}{r^*} \frac{\partial}{\partial r^*} \left( r^* \frac{\partial v_\theta^*}{\partial r^*} \right) - \frac{v_\theta^*}{r^{*2}} \right]

代回原式：

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第 3 題17 分

A long cylindrical container with radius of R is rotating at an angular speed of Ω\Omega. The cylindrical velocity profile along θ\theta direction in this system can be determined by the following partial differential equation:
ρ∂vθ∂t=μ[1r∂∂r(r∂vθ∂r)−vθr2]\rho \frac{\partial v_{\theta}}{\partial t} = \mu \left[ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial v_{\theta}}{\partial r} \right) - \frac{v_{\theta}}{r^2} \right]
The fluid is assumed to be steady-state and laminar, the pressure of the fluid is a function of r and z; vr=0v_r = 0; vθ=0v_{\theta} = 0; vz=f(r)v_z = f(r); and the pressure at the surface is PatmP_{atm}.
🖼️【此處有附圖，請對照原卷】
(1) Use R and Ω\Omega as the characteristic length and velocity, respectively. Please express the above PDE in terms of three dimensionless variables r∗r^*, vθ∗v_{\theta}^*, t∗t^*, and vθ∗=vθ/ΩRv_{\theta}^* = v_{\theta}/\Omega R, r∗=r/Rr^* = r/R, t∗=tΩt^* = t\Omega. The answer should include Reynolds number, Re, which should also be clearly defined. (5 points)
(2) (10 points) Express the velocity profile of the height difference (Δh\Delta h) in the cylinder using the above given parameters.
(3) (2 points) Show the relation between the height difference of the fluid surface Δh\Delta h and angular velocity Ω\Omega by plotting Δh\Delta h (y-axis) vs Ω\Omega (x-axis).

🖼️ 本題含圖表，以下為原卷對應頁面：
原卷第 1 頁

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這一題的完整詳解

核心觀念

本題考查三個重點：

  1. 以 RR、ΩR\Omega R、1/Ω1/\Omega 進行無因次化，建立 Reynolds number。
  2. 穩態旋轉流的速度分布。依原卷圖示，速度條件為
    vr=vz=0,vθ=f(r).v_r=v_z=0,\qquad v_\theta=f(r).
  3. 徑向離心力與壓力梯度平衡，並利用自由液面上的 p=Patmp=P_{\mathrm{atm}} 求液面高度差。

(1) 無因次化與 Reynolds number

定義

r=Rr∗,vθ=ΩRvθ∗,t=t∗Ω.r=Rr^*,\qquad v_\theta=\Omega Rv_\theta^*,\qquad t=\frac{t^*}{\Omega}.

因此

∂vθ∂t=Ω2R∂vθ∗∂t∗,\frac{\partial v_\theta}{\partial t} =\Omega^2R\frac{\partial v_\theta^*}{\partial t^*},

且

1r∂∂r(r∂vθ∂r)−vθr2=ΩR[1r∗∂∂r∗(r∗∂vθ∗∂r∗)−vθ∗(r∗)2].\frac{1}{r}\frac{\partial}{\partial r} \left(r\frac{\partial v_\theta}{\partial r}\right) -\frac{v_\theta}{r^2} = \frac{\Omega}{R} \left[ \frac{1}{r^*}\frac{\partial}{\partial r^*} \left(r^*\frac{\partial v_\theta^*}{\partial r^*}\right) -\frac{v_\theta^*}{(r^*)^2} \right].

代入原 PDE：

ρΩ2R∂vθ∗∂t∗=μΩR[1r∗∂∂r∗(r∗∂vθ∗∂r∗)−vθ∗(r∗)2].\rho\Omega^2R \frac{\partial v_\theta^*}{\partial t^*} = \frac{\mu\Omega}{R} \left[ \frac{1}{r^*}\frac{\partial}{\partial r^*} \left(r^*\frac{\partial v_\theta^*}{\partial r^*}\right) -\frac{v_\theta^*}{(r^*)^2} \right].

定義 Reynolds number：

Re=ρ(ΩR)Rμ=ρΩR2μ\boxed{ \mathrm{Re} =\frac{\rho(\Omega R)R}{\mu} =\frac{\rho\Omega R^2}{\mu} }

故無因次 PDE 為

Re∂vθ∗∂t∗=1r∗∂∂r∗(r∗∂vθ∗∂r∗)−vθ∗(r∗)2\boxed{ \mathrm{Re}\frac{\partial v_\theta^*}{\partial t^*} = \frac{1}{r^*}\frac{\partial}{\partial r^*} \left(r^*\frac{\partial v_\theta^*}{\partial r^*}\right) -\frac{v_\theta^*}{(r^*)^2} }

亦可寫成

∂vθ∗∂t∗=1Re[1r∗∂∂r∗(r∗∂vθ∗∂r∗)−vθ∗(r∗)2]\boxed{ \frac{\partial v_\theta^*}{\partial t^*} = \frac{1}{\mathrm{Re}} \left[ \frac{1}{r^*}\frac{\partial}{\partial r^*} \left(r^*\frac{\partial v_\theta^*}{\partial r^*}\right) -\frac{v_\theta^*}{(r^*)^2} \right] }

圓柱壁面與中心的條件為

vθ∗(1)=1,vθ∗(0) 有限.v_\theta^*(1)=1, \qquad v_\theta^*(0)\ \text{有限}.

(2) 速度分布與液面高度差

穩態時

∂vθ∗∂t∗=0,\frac{\partial v_\theta^*}{\partial t^*}=0,

所以

1r∗ddr∗(r∗dvθ∗dr∗)−vθ∗(r∗)2=0.\frac{1}{r^*}\frac{d}{dr^*} \left(r^*\frac{dv_\theta^*}{dr^*}\right) -\frac{v_\theta^*}{(r^*)^2}=0.

其一般解為

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第 4 題15 分

To reduce the threat of predators, the sand grouse, a bird of Kenya, will lay its eggs in locations well removed from sources of contamination. To bring water to its chicks, the grouse will fly from its location, by submerging the lower part of its body, will drain water within its own body. The grouse will then return to its nest, and the chicks will imbibe water from the plumage. Of course, if the time of flight is too long, evaporative losses could cause a significant reduction in the water content of the plumage, and the chicks could succumb to dehydration. To gain a better understanding of convective transfer during flight, tunnel studies were performed with molded implants of the grouse. By heating the portion of the model that corresponds to the water-encapsulating plumage, an average convection heat transfer coefficient was determined. Results for different air speeds and model sizes were then used to develop an empirical correlation of the form, NuL=0.034ReL4/5Pr1/3Nu_L = 0.034 Re_L^{4/5} Pr^{1/3}, where the effective surface area is defined as Aeff=Ae3/2A_{eff} = A_{e}^{3/2}, and the characteristic length is defined as L=(Ae)1/2L = (A_e)^{1/2}.
Consider conditions for which a grouse has entrained 0.05 kg of water within plumage of Ae=0.04A_e = 0.04 m2^2 and is returning to its nest at a constant speed of V=30V = 30 m/s. The ambient air is stagnant and at a temperature and relative humidity of Tatm=37∘T_{atm} = 37^\circC and ϕatm=25%\phi_{atm} = 25\%, respectively. If, throughout the flight, the surface AsA_s is covered with a liquid water film at Ts=32∘T_s = 32^\circC, which is the maximum allowable distance of the nest from the water source, if the bird must return with at least 50% of its initial water supply? (Explicitly describe your assumption(s), if any, for solving the problem; the binary diffusion coefficient for water vapor in air DABD_{AB} is 0.26×10−40.26 \times 10^{-4} m2^2/s).

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這一題的完整詳解

核心觀念

本題考查高速飛行下水膜的對流質傳與蒸發損失，使用熱質傳遞類比：

NuL⟶ShL,Pr⟶ScNu_L \longrightarrow Sh_L,\qquad Pr \longrightarrow Sc

其中

ShL=kcLDAB,Sc=νDABSh_L=\frac{k_cL}{D_{AB}},\qquad Sc=\frac{\nu}{D_{AB}}

kck_c 為對流質傳遞係數，蒸發速率由

m˙A=kcAeff(ρA,s−ρA,∞)\dot m_A=k_cA_{\mathrm{eff}} \left(\rho_{A,s}-\rho_{A,\infty}\right)

求得。

解題方法與必要假設

採用下列假設：

  1. 飛行速度 V=30V=30 m/s 可視為羽毛相對於空氣的速度。
  2. 水膜表面維持 Ts=32∘CT_s=32^\circ\mathrm{C}，故表面水蒸氣達飽和。
  3. 空氣性質取膜溫約 34.5∘C34.5^\circ\mathrm{C}，視為定值： ρair≈1.15 kg/m3,μ≈1.90×10−5 Pa⋅s\rho_{\mathrm{air}}\approx1.15\ \mathrm{kg/m^3},\qquad \mu\approx1.90\times10^{-5}\ \mathrm{Pa\cdot s} 因此 ν=μρair≈1.65×10−5 m2/s\nu=\frac{\mu}{\rho_{\mathrm{air}}} \approx1.65\times10^{-5}\ \mathrm{m^2/s}
  4. 蒸發過程中水膜溫度及環境條件不變，因此蒸發速率為常數。
  5. 依題目指定，質傳有效面積採用 Aeff=Ae3/2A_{\mathrm{eff}}=A_e^{3/2}

由題目給定：

Ae=0.04 m2A_e=0.04\ \mathrm{m^2}

故

L=(Ae)1/2=(0.04)1/2=0.20 mL=(A_e)^{1/2}=(0.04)^{1/2}=0.20\ \mathrm{m}

以及

Aeff=(0.04)3/2=0.0080A_{\mathrm{eff}}=(0.04)^{3/2} =0.0080

第一步：計算 Reynolds 數與 Schmidt 數

ReL=VLν=(30)(0.20)1.65×10−5≈3.64×105Re_L=\frac{VL}{\nu} =\frac{(30)(0.20)}{1.65\times10^{-5}} \approx3.64\times10^5

水蒸氣在空氣中的二元擴散係數為

DAB=0.26×10−4=2.60×10−5 m2/sD_{AB}=0.26\times10^{-4} =2.60\times10^{-5}\ \mathrm{m^2/s}

因此

Sc=νDAB=1.65×10−52.60×10−5≈0.635Sc=\frac{\nu}{D_{AB}} =\frac{1.65\times10^{-5}}{2.60\times10^{-5}} \approx0.635

第二步：由關係式求 Sherwood 數

將題目給定的關係式作熱質傳遞類比：

ShL=0.034ReL4/5Sc1/3Sh_L=0.034Re_L^{4/5}Sc^{1/3}

代入數值：

ShL=0.034(3.64×105)4/5(0.635)1/3≈8.2×102Sh_L =0.034(3.64\times10^5)^{4/5}(0.635)^{1/3} \approx8.2\times10^2

所以質傳係數為

kc=ShLDABLk_c=\frac{Sh_LD_{AB}}{L} kc=(8.2×102)(2.60×10−5)0.20≈0.107 m/sk_c =\frac{(8.2\times10^2)(2.60\times10^{-5})}{0.20} \approx0.107\ \mathrm{m/s}

第三步：計算水蒸氣濃度差

32∘C32^\circ\mathrm{C} 時水的飽和蒸氣壓約為

PA,ssat=4.76 kPaP_{A,s}^{sat}=4.76\ \mathrm{kPa}

由於表面為液態水膜：

PA,s=4.76 kPaP_{A,s}=4.76\ \mathrm{kPa}

37∘C37^\circ\mathrm{C} 時水的飽和蒸氣壓約為

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第 5 題15 分

A laboratory apparatus to measure the diffusion coefficient of vapor-gas mixtures consists of a vertical, small-diameter column containing the liquid phase that evaporates into gas flowing over the mouth of the column. The gas flow rate is sufficient to maintain a negligible vapor concentration at the exit plane. The column is 150 mm high, and the pressure and temperature in the chamber are maintained at 0.25 atm and 320 K, respectively. Given the evaporation rate of water into air being 0.257 kg/m3^3 under the corresponding conditions, determine the value of the binary diffusion coefficient DABD_{AB} for the vapor-air mixture; (Explicitly describe your assumption(s), if any, for solving the problem; the vapor-air mixture may be treated as an ideal gas.)

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這一題的完整詳解

核心觀念

本題是以「Stefan 擴散管」測定水蒸氣 AA 在空氣 BB 中的二元擴散係數。水面蒸發產生水蒸氣，空氣則近似停滯，因此屬於：

  • 水蒸氣 AA 穿過停滯空氣 BB 的穩態擴散；
  • 一維、等溫、等壓；
  • 氣相可視為理想氣體；
  • 水蒸氣在液面達到氣液平衡；
  • 管口水蒸氣濃度可忽略，即 yA2≈0y_{A2}\approx 0。

對 AA 穿過停滯 BB 的 Stefan 擴散，通量方程為

NA=DABCLln⁡(1−yA21−yA1)N_A=\frac{D_{AB}C}{L} \ln\left(\frac{1-y_{A2}}{1-y_{A1}}\right)

其中：

  • NAN_A：水蒸氣質量通量；
  • DABD_{AB}：水蒸氣－空氣二元擴散係數；
  • CC：氣相總莫耳濃度；
  • LL：擴散距離；
  • yA1y_{A1}：液面處水蒸氣莫耳分率；
  • yA2y_{A2}：管口處水蒸氣莫耳分率。

解題方法

題目所給的「0.257 kg/m30.257\ \mathrm{kg/m^3}」為密度單位，無法直接作為蒸發通量使用。依 Stefan 擴散管實驗資料的通常表示方式，將其解讀為水的蒸發通量

NA=0.257 kg/(m2⋅h)N_A=0.257\ \mathrm{kg/(m^2\cdot h)}

這也是使本題具有唯一數值解的必要假設。

1. 液面處水蒸氣莫耳分率

在 T=320 KT=320\ \mathrm{K} 時，水的飽和蒸氣壓約為

PA∗=0.105 atmP_A^*=0.105\ \mathrm{atm}

總壓力為 P=0.25 atmP=0.25\ \mathrm{atm}，因此液面處

yA1=PA∗P=0.1050.25=0.420y_{A1}=\frac{P_A^*}{P} =\frac{0.105}{0.25} =0.420

管口處水蒸氣濃度可忽略：

yA2=0y_{A2}=0

故

ln⁡(1−yA21−yA1)=ln⁡(11−0.420)=ln⁡(1.724)=0.545\ln\left(\frac{1-y_{A2}}{1-y_{A1}}\right) = \ln\left(\frac{1}{1-0.420}\right) = \ln(1.724) = 0.545

2. 計算氣相總莫耳濃度

理想氣體關係式為

C=PRTC=\frac{P}{RT}

將壓力換算為 SI 單位：

P=0.25(101325)=25331 PaP=0.25(101325)=25331\ \mathrm{Pa}

因此

C=25331(8.314)(320)=9.52 mol/m3C=\frac{25331}{(8.314)(320)} =9.52\ \mathrm{mol/m^3}

3. 將蒸發通量換算為莫耳通量

水的分子量為

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第 6 題40 分

Due to the worsening environmental conditions attributable to plastic wastes, governments of advanced economies have been tightening their regulations on the manufacturing and distribution of plastic products, particularly those for single uses. Thus, the European Union, which boasts of the most stringent environmental regulations on the planet, revised their Waste Framework Directive and Packing Waster Directive in 2018 and, according to them, will gradually require single-use plastic products to be made of biodegradable plastics. One of the industries bearing the brunt of the new regulations is fast food restaurants and the food service providers specializing in food delivery. These industries have long been heavy users of disposable plastic products; an outstanding example being the disposable containers for hot drinks. Even though fast-food giants, such as MacDonald's, have shifted to the paper-based containers (which are not eco-friendly as well), many players in the industries still use the plastic-based containers for their hot drinks, under the consideration of cost and thermal performance. As a product designer of biodegradable plastics, you are tasked with the endeavor to develop a disposable coffee cup made of biodegradable plastic, poly lactic acid (PLA), with the aim of substituting it for its non-degradable plastic-and-paper-based counterparts currently used in the food industry. A key step for the task is to determine the wall thickness for your new PLA coffee cup. According to a consumer survey by a trade association of the food industry, people finish their coffee in 10 minutes and prefer the temperature of their coffee to be maintained within 10∘10^\circC from its initial temperature (−90∘-90^\circC) in this period. They decide to begin with determining the temperature distribution within coffee immediately after it is poured into a cup.
(a) (5 points) Given the cylindrical geometry immediately the most common design for a coffee cup, the temperature distribution within coffee inside the cup should also be described in the cylindrical coordinate. Where,
1r∂∂r(kr∂T∂r)+1r2∂∂θ(k∂T∂θ)+∂∂z(k∂T∂z)+q=ρcp∂T∂t\frac{1}{r} \frac{\partial}{\partial r} \left( kr \frac{\partial T}{\partial r} \right) + \frac{1}{r^2} \frac{\partial}{\partial \theta} \left( k \frac{\partial T}{\partial \theta} \right) + \frac{\partial}{\partial z} \left( k \frac{\partial T}{\partial z} \right) + q = \rho c_p \frac{\partial T}{\partial t}
where kk is the thermal conductivity, qq is the rate of heat generation/consumption from an internal source/drain, ρ\rho is the density and cpc_p is the specific heat of the coffee; describe the physical meanings of the entire equation and of each term in the equation.
(b) (3 points) Following (a), what assumptions must be valid such that the heat equation holds true?
(c) (3 points) Assuming the heat dissipation from the top and bottom ends of the cup being negligible for the moment and the thermal properties of coffee being constant, simplify the heat equation from (a).
(d) (3 points) You may make an extra assumption(s) when carrying out the simplification in (c). Why is the assumption reasonable for the considered problem?
(e) (11 points) With the effective thermal conductivity, density and specific heat of coffee being 2.8 W/m·∘^\circC, 1100 kg/m3^3 and 3400 J/kg·∘^\circC, respectively, for the water content of 70 wt%, exploit below the heat equation to determine the temperature distribution within coffee immediately after it is poured into a cup.
(f) (4 points) With the inner diameter of the cup being 6 cm, use the temperature distribution function you obtained in (e) to determine the temperature of the inner surface of the cup.
(g) (11 points) Assume that the heat dissipation from the top and bottom ends of the cup being negligible for the moment and the thermal properties of coffee being constant, simplify the heat equation from (a). (Hint: Ignore the surface area difference between the inner and outer surfaces of the cup. Due to the relatively small dimension of the wall thickness, the heat dissipation through the wall can be described in the context of the plane wall geometry. The consumers' preference depicted above should be taken into account in some rational manner.)

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這一題的完整詳解

核心觀念

本題屬於熱量傳送（Heat Transfer）中的經典題型，涵蓋**熱傳導控制方程式（Heat Conduction Equation）**的推導與化簡、座標幾何假設、邊界條件設定，以及非穩態／穩態導熱分析。核心觀念包含：

  1. 微觀能量平衡（Microscopic Energy Balance）：以能量守恆定律結合傅立葉熱傳導定律（Fourier's Law），建立圓柱座標系下的通用熱傳導方程式。
  2. 對稱性與維度簡化：辨析圓周方向對稱性（∂/∂θ=0\partial/\partial \theta = 0）、長度方向效應（∂/∂z=0\partial/\partial z = 0）、無內熱源（q=0q = 0）及無流體對流流動下的簡化。
  3. 初始狀態（t=0+t = 0^+）的熱分佈：熱咖啡剛倒入杯中（immediately after it is poured）時，內部液體維持均勻初溫，熱邊界層尚未向內發展。
  4. 平板近似（Plane Wall Approximation）與總熱阻/能量損耗：當杯壁厚度相較於半徑甚小時，壁面熱傳導可簡化為一維一維平板穩態導熱或集總熱容量法（Lumped Capacitance）評估降溫速率。

各子題詳解與推導

(a) 物理意義闡述

給定圓柱座標下的熱傳導方程式：

1r∂∂r(kr∂T∂r)+1r2∂∂θ(k∂T∂θ)+∂∂z(k∂T∂z)+q=ρcp∂T∂t\frac{1}{r} \frac{\partial}{\partial r} \left( kr \frac{\partial T}{\partial r} \right) + \frac{1}{r^2} \frac{\partial}{\partial \theta} \left( k \frac{\partial T}{\partial \theta} \right) + \frac{\partial}{\partial z} \left( k \frac{\partial T}{\partial z} \right) + q = \rho c_p \frac{\partial T}{\partial t}
  • 整條方程式的物理意義：
    基於能量守恆定律（一階熱力學定律）應用於微觀控制體積（Control Volume），即： (淨傳導熱流入速率)+(內部熱源生成速率)=(熱能蓄積速率)(\text{淨傳導熱流入速率}) + (\text{內部熱源生成速率}) = (\text{熱能蓄積速率})
  • 各項個別物理意義：
    1. 1r∂∂r(kr∂T∂r)\frac{1}{r} \frac{\partial}{\partial r} \left( kr \frac{\partial T}{\partial r} \right)：單位體積在徑向（rr 方向）因熱傳導所導致的淨熱流入速率。
    2. 1r2∂∂θ(k∂T∂θ)\frac{1}{r^2} \frac{\partial}{\partial \theta} \left( k \frac{\partial T}{\partial \theta} \right)：單位體積在角向／切向（θ\theta 方向）因熱傳導所導致的淨熱流入速率。
    3. ∂∂z(k∂T∂z)\frac{\partial}{\partial z} \left( k \frac{\partial T}{\partial z} \right)：單位體積在軸向（zz 方向）因熱傳導所導致的淨熱流入速率。
    4. qq：單位體積之內部熱源生成率（Generation）或消耗率（Sink，若為負值）。
    5. ρcp∂T∂t\rho c_p \frac{\partial T}{\partial t}：單位體積內熱焓／內能隨時間變化的蓄積速率（Thermal Energy Accumulation Rate）。

(b) 該方程式成立所需之假設

使上述熱傳導方程式嚴格成立的必備假設條件為：

  1. 無對流效應（No Bulk Motion / Convection）：流體處於靜止狀態或液體流動速度視為零（v=0\mathbf{v} = 0），熱量僅藉由微觀分子熱傳導傳遞。
  2. 符合傅立葉熱傳導定律（Fourier's Law of Heat Conduction）：熱通量向量正比於溫度梯度，即 q′′=−k∇T\mathbf{q}'' = -k \nabla T。
  3. 等方性介質（Isotropic Medium）：熱傳導係數 kk 在各方向上相同（無方向性差異）。
  4. 可忽略黏滯耗散（Viscous Dissipation Negligible）與膨脹功（Work done by pressure/compression is negligible）：不可壓縮流體（ρ≈const\rho \approx \text{const}）且無剪切發熱。

(c) 忽略杯頂與底部的熱散失且熱物性為常數時之化簡

根據題意：

  1. 熱物性為常數：kk、ρ\rho、cpc_p 均為常數，熱擴散率定義為 α=kρcp\alpha = \frac{k}{\rho c_p}。
  2. 忽略頂部與底部散熱：軸向熱通量可忽略，即溫度在 zz 方向無梯度，∂T∂z=0\frac{\partial T}{\partial z} = 0。
  3. 無內部熱源：咖啡內部無化學反應或相變化發熱，q=0q = 0。
  4. 圓周對稱性：杯子為旋轉對稱柱體，周圍散熱沿角向均勻，∂T∂θ=0\frac{\partial T}{\partial \theta} = 0。

代入方程式化簡為：

kr∂∂r(r∂T∂r)=ρcp∂T∂t\frac{k}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) = \rho c_p \frac{\partial T}{\partial t}

或寫為：

1r∂∂r(r∂T∂r)=1α∂T∂t\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) = \frac{1}{\alpha} \frac{\partial T}{\partial t}

展開微分項亦可表示為：

∂2T∂r2+1r∂T∂r=1α∂T∂t\frac{\partial^2 T}{\partial r^2} + \frac{1}{r}\frac{\partial T}{\partial r} = \frac{1}{\alpha}\frac{\partial T}{\partial t}

(d) (c) 中所做額外假設及其合理性

在 (c) 的化簡過程中，額外引入的關鍵假設為：

  1. 軸對稱假設（Axial/Angular Symmetry，∂T∂θ=0\frac{\partial T}{\partial \theta} = 0）：
    • 合理性：咖啡杯幾何形狀為圓柱對稱結構，杯內液體呈旋轉對稱分佈；外圍空氣的自然對流環境均勻包覆杯壁，周向無顯著溫差或不對稱冷卻效應。
  2. 無內熱源（q=0q = 0）：
    • 合理性：熱咖啡注入杯中後，僅進行單純的降溫冷卻過程，不存在核反應、微波加熱或強烈化學放熱／吸熱反應。

(e) 咖啡剛倒入杯中瞬間（Immediately after pouring）的溫度分佈

  • 物理機制分析：
    題目探討的時間點為咖啡剛倒入杯中的初始瞬間（t=0+t = 0^+）。在倒入杯子前，咖啡液經過混合注入，整杯液體內部處於均勻之初始溫度 T0T_0（題幹給定 initial temperature 為 90∘90^\circC）。
    熱傳導需要傳遞時間，熱擴散滲透深度（Thermal penetration depth）正比於 αt\sqrt{\alpha t}。當 t→0+t \to 0^+ 時，滲透深度 δt→0\delta_t \to 0。
  • 數學解與溫度分佈：
    在控制方程式： ∂2T∂r2+1r∂T∂r=1α∂T∂t\frac{\partial^2 T}{\partial r^2} + \frac{1}{r}\frac{\partial T}{\partial r} = \frac{1}{\alpha}\frac{\partial T}{\partial t} 於初始條件： t=0,0≤r<R:T(r,0)=T0=90∘Ct = 0,\quad 0 \le r < R: \quad T(r, 0) = T_0 = 90^\circ\text{C} 因此，在咖啡剛注入杯內瞬間（t=0+t = 0^+），除了與冷杯壁接觸的無限薄界面外，咖啡內部的溫度分佈函數為一均勻常數分佈： T(r)=90∘C(0≤r<R)T(r) = 90^\circ\text{C} \quad (0 \le r < R)

(f) 杯內表面（Inner Surface）在接觸瞬間的溫度

  • 物理模型：
    當高溫咖啡（介質 1，初溫 T1,0=90∘CT_{1,0} = 90^\circ\text{C}）剛接觸室溫下之杯壁（介質 2，PLA 杯壁，初溫設為室溫 T2,0≈20∘C∼25∘CT_{2,0} \approx 20^\circ\text{C} \sim 25^\circ\text{C}）時，由於時間極短（t→0+t \to 0^+），此接觸過程可視為**兩半無限固體接觸（Contact of two semi-infinite media）**的瞬態熱傳導問題。
  • 界面接觸溫度（Interface Temperature）公式：
    由界面熱通量與溫度連續性，瞬時接觸界面溫度 TsT_s 為定值，決定於兩者的熱蓄積係數（Thermal Effusivity）e=kρcpe = \sqrt{k \rho c_p}： Ts=e1T1,0+e2T2,0e1+e2T_s = \frac{e_1 T_{1,0} + e_2 T_{2,0}}{e_1 + e_2}
  • 參數代入：
    • 咖啡（介質 1）： e1=k1ρ1cp1=2.8×1100×3400=1.0472×107≈3236 W⋅s1/2/(m2⋅∘C)e_1 = \sqrt{k_1 \rho_1 c_{p1}} = \sqrt{2.8 \times 1100 \times 3400} = \sqrt{1.0472 \times 10^7} \approx 3236 \text{ W}\cdot\text{s}^{1/2}/(\text{m}^2\cdot^\circ\text{C})
    • PLA 杯壁（介質 2）：
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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112 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