115 年 國立成功大學化學工程學系碩士班甲組《單元操作與輸送現象》

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第 1 題

Dimethyl sulfoxide in a storage tank is transported to a reactor through a cast-iron pipe (e = 0.00006 m) with a constant flow rate of 11.8 kg/s. The absolute pressure at the entrance of the pipe at the storage tank is 159080 Pa, and the pressure at the end of the pipe at the reactor is 101300 Pa. The location of the storage tank is 5.0 m above the reactor, and the length of the pipe is 120 m. The density and viscosity of dimethyl sulfoxide are 1100 kg/m³ and 0.002 Pa-s, respectively. According to the chart below, what are the Fanning friction factor (8%) and the required diameter of the pipe (8%) for this application, respectively?
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這一題的完整詳解

核心觀念

本題用伯努力方程式求管路可用的摩擦損失水頭,再以 Fanning 摩擦因子形式的 Darcy–Weisbach 式,配合圖表中的雷諾數與相對粗糙度,迭代求管徑。

假設管路兩端管徑相同、忽略局部損失,且題目給定的壓力與高度分別代表管路入口、出口狀態。兩端平均流速相同,因此伯努力方程式中的速度水頭相消。

解題方法

先將質量流率換算為體積流率:

Q=m˙ρ=11.81100=0.01073 m3/sQ=\frac{\dot m}{\rho} =\frac{11.8}{1100} =0.01073\ \mathrm{m^3/s}

伯努力方程式給出的摩擦損失水頭為:

hf=p1−p2ρg+(z1−z2)h_f=\frac{p_1-p_2}{\rho g}+(z_1-z_2)

儲槽比反應器高 5.0 m5.0\ \mathrm{m},所以高度差會增加可供克服管路摩擦的水頭:

hf=159080−1013001100(9.81)+5.0=10.35 mh_f=\frac{159080-101300}{1100(9.81)}+5.0 =10.35\ \mathrm{m}

注意高度項是加上 5.0 m5.0\ \mathrm{m},因為流體由較高處流向較低處。

使用 Fanning 摩擦因子 fFf_F 時,Darcy–Weisbach 式為:

hf=4fFLDV22gh_f=4f_F\frac{L}{D}\frac{V^2}{2g}

而管內平均流速與雷諾數為:

V=4QπD2,Re=ρVDμ=4m˙πμDV=\frac{4Q}{\pi D^2}, \qquad Re=\frac{\rho VD}{\mu} =\frac{4\dot m}{\pi\mu D}
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第 2 題

Boundary-layer theory is usually used to estimate the shear force between the surface and the fluid flowing on it. For turbulent boundary layer, since no analytical solution can be found, the von Kármán approximation shown below should be used to estimate the shear force.
τ0ρ=ddx∫0δ(v∞−vx)vxdy+v∞ddx∫0δ(v∞−vx)dy\frac{\tau_0}{\rho} = \frac{d}{dx} \int_0^\delta (v_\infty - v_x) v_x dy + v_\infty \frac{d}{dx} \int_0^\delta (v_\infty - v_x) dy
In addition, for turbulent boundary layer, the following Blasius's correlation is usually used to express the surface shear stress.
τ0=0.0225ρv∞2(νv∞δ)1/4\tau_0 = 0.0225 \rho v_\infty^2 \left( \frac{\nu}{v_\infty \delta} \right)^{1/4}
For a fluid flowing on the x direction with a constant speed of v∞v_\infty, the boundary layer is formed when it flows through a flat plate. The front edge of the flat plate is defined as the leading edge, and the entire boundary layer is assumed as turbulent boundary layer.
(a) Express the velocity profile (vxv_x) in the boundary layer by using the power-law velocity profile with n=10 (4%).
(b) With the velocity profile from (a), derive the expression of δ\delta as the function of x, μ\mu, ρ\rho and v∞v_\infty (10%).

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這一題的完整詳解

核心觀念

本題使用兩項關係:

  1. 湍流邊界層的冪次速度分布

    vxv∞=(yδ)1/n\frac{v_x}{v_\infty}=\left(\frac{y}{\delta}\right)^{1/n}

    題目給定 n=10n=10。

  2. von Kármán 動量積分式

τ0ρ=ddx∫0δ(v∞−vx)vx dy+v∞ddx∫0δ(v∞−vx) dy\frac{\tau_0}{\rho} = \frac{d}{dx}\int_0^\delta (v_\infty-v_x)v_x\,dy + v_\infty\frac{d}{dx}\int_0^\delta (v_\infty-v_x)\,dy

因為平板外流速度 v∞v_\infty 為常數,邊界層厚度 δ\delta 是 xx 的函數。


(a) 速度分布

代入 n=10n=10,邊界層內的速度分布為

vx=v∞(yδ)1/10\boxed{ v_x=v_\infty\left(\frac{y}{\delta}\right)^{1/10} }

其中:

  • y=0y=0 時,vx=0v_x=0,符合無滑移邊界條件;
  • y=δy=\delta 時,vx=v∞v_x=v_\infty,符合邊界層外緣條件。

(b) 邊界層厚度 δ\delta 與 xx 的關係

第一步:計算第一項積分

令

η=yδ\eta=\frac{y}{\delta}

因此 y=δηy=\delta\eta、dy=δ dηdy=\delta\,d\eta,且

vxv∞=η1/n\frac{v_x}{v_\infty}=\eta^{1/n}

則

∫0δ(v∞−vx)vx dy=v∞2δ∫01(η1/n−η2/n)dη\int_0^\delta (v_\infty-v_x)v_x\,dy = v_\infty^2\delta \int_0^1 \left(\eta^{1/n}-\eta^{2/n}\right)d\eta

利用

∫01ηa dη=1a+1\int_0^1\eta^a\,d\eta=\frac{1}{a+1}

可得

∫01η1/ndη=nn+1\int_0^1\eta^{1/n}d\eta=\frac{n}{n+1}

以及

∫01η2/ndη=nn+2\int_0^1\eta^{2/n}d\eta=\frac{n}{n+2}

因此

∫0δ(v∞−vx)vx dy=v∞2δ(nn+1−nn+2)\int_0^\delta (v_\infty-v_x)v_x\,dy = v_\infty^2\delta \left( \frac{n}{n+1}-\frac{n}{n+2} \right)

整理為

∫0δ(v∞−vx)vx dy=v∞2δn(n+1)(n+2)\int_0^\delta (v_\infty-v_x)v_x\,dy = v_\infty^2\delta \frac{n}{(n+1)(n+2)}

第二步:計算第二項積分

∫0δ(v∞−vx) dy=v∞δ∫01(1−η1/n)dη\int_0^\delta(v_\infty-v_x)\,dy = v_\infty\delta \int_0^1(1-\eta^{1/n})d\eta

所以

∫0δ(v∞−vx) dy=v∞δ(1−nn+1)\int_0^\delta(v_\infty-v_x)\,dy = v_\infty\delta \left( 1-\frac{n}{n+1} \right)

即

∫0δ(v∞−vx) dy=v∞δn+1\int_0^\delta(v_\infty-v_x)\,dy = \frac{v_\infty\delta}{n+1}

第三步:代入 von Kármán 動量積分式

由於 v∞v_\infty 為常數,且 nn 為常數:

ddx∫0δ(v∞−vx)vx dy=v∞2n(n+1)(n+2)dδdx\frac{d}{dx} \int_0^\delta(v_\infty-v_x)v_x\,dy = v_\infty^2 \frac{n}{(n+1)(n+2)} \frac{d\delta}{dx} v∞ddx∫0δ(v∞−vx) dy=v∞21n+1dδdxv_\infty\frac{d}{dx} \int_0^\delta(v_\infty-v_x)\,dy = v_\infty^2 \frac{1}{n+1} \frac{d\delta}{dx}

相加得

τ0ρ=v∞2dδdx[n(n+1)(n+2)+1n+1]\frac{\tau_0}{\rho} = v_\infty^2\frac{d\delta}{dx} \left[ \frac{n}{(n+1)(n+2)} + \frac{1}{n+1} \right]

整理括號內的係數:

n(n+1)(n+2)+1n+1=n+n+2(n+1)(n+2)=2n+2\frac{n}{(n+1)(n+2)} + \frac{1}{n+1} = \frac{n+n+2}{(n+1)(n+2)} = \frac{2}{n+2}

因此

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第 3 題

Consider the liquid film as illustrated in the following figure. At the position of z = 0, the concentration of species A is CA0C_{A0}. A surface reaction occurs at z = L. The surface reaction is assumed to be first-order in species A with a surface reaction rate constant, k. You may assume that the transport of products of the reaction is ignored.
(a) What is the concentration of A at z = L? State all reasonable assumptions as necessary (11%).
(b) What is the molar flux of A at z = L (3%)?
(c) If consumption of A by the reaction is considerably faster than diffusion of A through the liquid film, nondimensionalize the concentration of A at z = L (Hint: the concentration of A at z = L is not zero.) (3%).
(d) If the reaction is considerably slower than diffusion of A, simplify the concentration of A at z = L (3%).

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這一題的完整詳解

核心觀念

本題考察「液膜內穩態分子擴散+表面一階反應」的串聯阻力概念。

合理假設如下:

  • 液膜內為穩態、一維擴散,僅沿 zz 方向傳質。
  • 液膜內沒有均相反應,且擴散係數 DAD_A 為常數。
  • z=0z=0 處濃度固定為 CA0C_{A0}。
  • z=Lz=L 處發生不可逆一階表面反應,反應消耗 species A。
  • 液膜內無對流,且產物傳輸忽略不計。
  • 表面反應速率等於 A 通過液膜到達表面的莫耳通量。

解題方法

液膜內沒有反應,因此穩態物料衡算為

d2CAdz2=0\frac{d^2 C_A}{dz^2}=0

積分後濃度分布為線性形式:

CA=a+bzC_A=a+bz

由 z=0z=0 時 CA=CA0C_A=C_{A0},可寫成

CA=CA0+bzC_A=C_{A0}+bz

由 Fick 第一定律,沿 +z+z 方向的 A 莫耳通量為

NA=−DAdCAdz=−DAbN_A=-D_A\frac{dC_A}{dz}=-D_A b

因此濃度分布也可寫為

CA=CA0−NADAzC_A=C_{A0}-\frac{N_A}{D_A}z

(a) z=Lz=L 處的 A 濃度

在液膜表面 z=Lz=L,表面反應消耗 A,其反應通量為

NA=kCALN_A=kC_{AL}

其中 CALC_{AL} 表示 z=Lz=L 處的 A 濃度。

由液膜內的濃度分布,在 z=Lz=L 處有

CAL=CA0−NALDAC_{AL}=C_{A0}-\frac{N_A L}{D_A}

代入表面反應邊界條件 NA=kCALN_A=kC_{AL}:

CAL=CA0−kCALLDAC_{AL}=C_{A0}-\frac{kC_{AL}L}{D_A}

整理得

CAL(1+kLDA)=CA0C_{AL}\left(1+\frac{kL}{D_A}\right)=C_{A0}

因此

CAL=CA01+kLDA\boxed{ C_{AL}=\frac{C_{A0}}{1+\dfrac{kL}{D_A}} }

亦可寫成

CAL=DACA0DA+kL\boxed{ C_{AL}=\frac{D_A C_{A0}}{D_A+kL} }

其中 LDA\dfrac{L}{D_A} 是液膜擴散阻力,1k\dfrac{1}{k} 是表面反應阻力。


(b) z=Lz=L 處的 A 莫耳通量

由表面一階反應的邊界條件:

NA=kCALN_A=kC_{AL}

代入 (a) 的結果:

NA=kCA01+kLDAN_A = \frac{kC_{A0}}{1+\dfrac{kL}{D_A}}

因此

NA=kCA01+kLDA\boxed{ N_A=\frac{kC_{A0}}{1+\dfrac{kL}{D_A}} }

等價形式為

NA=DAkCA0DA+kL\boxed{ N_A=\frac{D_AkC_{A0}}{D_A+kL} }

通量方向為由 z=0z=0 指向 z=Lz=L,即 A 從液膜內擴散至反應表面。


(c) 反應遠快於液膜擴散

反應遠快於擴散表示

k≫DALk\gg \frac{D_A}{L}

定義液膜傳質的無因次濃度與質傳 Biot 數:

cL=CALCA0c_L=\frac{C_{AL}}{C_{A0}} Bim=kLDA\mathrm{Bi}_m=\frac{kL}{D_A}
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第 4 題

A shell-and-tube heat exchanger with one shell pass and two tube passes (1-2 configuration) is used to heat a cold process stream (Stream C) by a hot utility stream (Stream H). The cold stream flows through the tubes, while the hot stream flows on the shell side. The heat transfer process involves a series of thermal resistances: internal convection, wall conduction, and external convection.
(Multiple-selection questions. Full credit requires selecting all and only the correct options. Half credit is given for at most one incorrect or missing option; two or more errors receive zero credit.)
(a) To characterize the heat transfer efficiency, we examine the tube-side local Nusselt number (NuxNux). The figure below illustrates the variation of NuxNux as a function of the Graetz number, defined as Gz=Pe⋅(D/x)Gz = Pe \cdot (D/x), for laminar flow. Specifically, the figure compares cases of constant wall temperature and constant wall heat flux. Regarding the trend shown in the figure where NuxNux generally increases with Pe(D/x)Pe(D/x) and reaches a constant value at lower Pe(D/x)Pe(D/x), which of the following statements correctly describe the underlying physical mechanisms (12%)?
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A. The increase in NuxNux at large Pe(D/x)Pe(D/x) is primarily because the average fluid velocity is higher in the region where Pe(D/x)Pe(D/x) is large.
B. The decrease of NuxNux in the region of lower Pe(D/x)Pe(D/x) is primarily due to the enhancement of the fluid's thermal conductivity (k) along the axial direction.
C. The variation of NuxNux is fundamentally governed by the development and thickness of the thermal boundary layer.
D. At very small values of Pe(D/x)Pe(D/x), NuxNux reaches a constant value because the dimensionless temperature profile becomes invariant with respect to the axial position.
E. The sharp increase in NuxNux at high Pe(D/x)Pe(D/x) indicates a transition from laminar flow to turbulent flow.
F. For fluids with a very high Prandtl number (Pr=ν/α≫1Pr = \nu/\alpha \gg 1), the thermal entry length is significantly shorter than the hydrodynamic entry length, causing NuxNux to reach its constant value more rapidly.
G. Under the constant wall temperature condition, the wall-to-bulk temperature difference decreases along the flow direction as the fluid temperature approaches the wall temperature, resulting in a reduced local wall temperature gradient despite the continued growth of the thermal boundary layer.
H. The Nusselt number is higher for a constant heat flux wall because the imposed uniform heat flux maintains a smaller wall-to-bulk temperature difference, and according to q′′=h(Tw−Tb)q'' = h(T_w - T_b), this necessarily leads to a larger heat transfer coefficient.

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這一題的完整詳解

此題考驗對流散熱中, Nusselt 數 (NuNu) 與 Graetz 數 (GzGz)、Peclet 數 (PePe)、Prandtl 數 (PrPr) 等無因次參數的理解,以及熱邊界層發展的物理機制。

首先,我們理解圖中顯示的趨勢:NuxNu_x 隨著 Pe(D/x)Pe(D/x) 的增加而增加,並在較低的 Pe(D/x)Pe(D/x) 值下達到一個常數值。這裡的 Pe(D/x)Pe(D/x) 是 Peclet 數的一種形式,與 Graetz 數 GzGz 相關。一般而言,Gz=Pe⋅(D/x)=Re⋅Pr⋅(D/x)Gz = Pe \cdot (D/x) = Re \cdot Pr \cdot (D/x)。圖中是以 Pe(D/x)Pe(D/x) 為橫座標,這與 GzGz 的定義略有不同,但它們都反映了軸向的發展。

我們逐一分析選項:

A. The increase in NuxNux at large Pe(D/x)Pe(D/x) is primarily because the average fluid velocity is higher in the region where Pe(D/x)Pe(D/x) is large.
Pe(D/x)=Re⋅Pr⋅(D/x)Pe(D/x) = Re \cdot Pr \cdot (D/x)。在固定管徑 DD 和流體性質(Re,PrRe, Pr)下,Pe(D/x)Pe(D/x) 實際上與 1/x1/x 成正比。因此,較大的 Pe(D/x)Pe(D/x) 對應於較小的 xx 值,即靠近入口處。在入口區域,流體速度是恆定的,但邊界層正在發展。NuxNux 的增加是由於熱邊界層的發展,而不是流速的增加。此選項不正確。

B. The decrease of NuxNux in the region of lower Pe(D/x)Pe(D/x) is primarily due to the enhancement of the fluid's thermal conductivity (k) along the axial direction.
Pe(D/x)Pe(D/x) 較大時對應於較小的 xx (靠近入口),此時 NuxNux 較高。Pe(D/x)Pe(D/x) 較小時對應於較大的 xx (遠離入口),此時 NuxNux 趨於常數。圖顯示 NuxNux 在較高 Pe(D/x)Pe(D/x) 時 (靠近入口) 較高,然後下降並趨於常數。選項 B 描述的是 NuxNux 的下降,並將其歸因於熱導率的增強,這是不正確的。熱導率 kk 是流體性質,通常不隨軸向變化。NuxNux 的下降是因為熱邊界層發展趨於成熟。此選項不正確。

C. The variation of NuxNux is fundamentally governed by the development and thickness of the thermal boundary layer.
Nusselt 數 NuxNu_x 本質上是衡量表面傳熱係數 hxh_x 的無因次量。Nux=hxD/kNu_x = h_x D / k。傳熱係數 hxh_x 直接與熱邊界層的發展和厚度相關。當熱邊界層尚未發展完全時,傳熱效率較高,NuxNu_x 較大。隨著熱邊界層的發展並趨於成熟,傳熱效率降低,NuxNu_x 趨於常數。因此,此選項是正確的。

D. At very small values of Pe(D/x)Pe(D/x), NuxNux reaches a constant value because the dimensionless temperature profile becomes invariant with respect to the axial position.
這裡的「very small values of Pe(D/x)Pe(D/x)」對應於圖中的右側區域,即較大的 xx 值(遠離入口)。在這些區域,NuxNux 確實趨於一個常數值。這是因為熱邊界層已經充分發展,並且其厚度不再隨軸向顯著變化,從而導致無因次溫度分佈(例如 Tw−T(x,y)Tw−Tb\frac{T_w - T(x,y)}{T_w - T_b})在軸向上趨於穩定。此選項是正確的。

E. The sharp increase in NuxNux at high Pe(D/x)Pe(D/x) indicates a transition from laminar flow to turbulent flow.
圖中顯示的曲線是針對層流 (laminar flow) 的。

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第 4 題

(b) For heat exchangers that deviate from ideal counterflow, the logarithmic-mean temperature difference (LMTD) must be corrected using a correction factor F. The figure below shows the LMTD correction factor F for a 1-2 shell-and-tube heat exchanger, plotted as a function of Y and z, where subscripts t and s denote tube-side and shell-side streams, respectively. Which of the following statements correctly describe the physical meaning and trends of the correction factor shown in the figure (13%)?
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A. At a constant z value, increasing Y leads to a reduction in the actual heat transfer rate because a lower correction factor implies a smaller overall heat-transfer driving force between the hot and cold streams.
B. A higher z value indicates that the temperature change of the hot stream is larger than that of the cold stream from inlet to outlet, implying that more heat is transferred to the hot stream than the cold stream receives.
C. At a fixed Y value, the correction factor decreases with increasing z because transferring more heat to the hot stream enhances shell-side induced vortices, making the heat-transfer process less ideal.
D. When Y increases, the correction factor decreases more dramatically at higher z values than at lower z values because a smaller heat-capacity coefficient (mCp) of the hot stream causes a smaller change in Y for a given increase in heat-transfer rate.
E. For this 1-2 shell-and-tube heat exchanger, the correction factor is always less than unity because this configuration inherently contains regions of both counterflow and parallel flow within the exchanger.
F. For the same correction factor value, a smaller Y implies that the hot-stream temperature can be changed more easily.
G. The decrease in the correction factor is a direct consequence of a larger fraction of the maximum possible heat being transferred within the exchanger.
H. Increasing the inlet temperature of the hot stream reduces the value of Y, thereby increasing the correction factor.

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這一題的完整詳解

核心觀念

1–2 shell-and-tube heat exchanger 的實際對數平均溫差為

ΔTlm,actual=FΔTlm,counterflow,\Delta T_{\mathrm{lm,actual}} =F\Delta T_{\mathrm{lm,counterflow}},

因此熱傳量為

Q=UAFΔTlm,counterflow.Q=UA F\Delta T_{\mathrm{lm,counterflow}}.

其中 FF 是相對於理想逆流熱交換器的修正因子,1–2 管殼式熱交換器因同時含有逆流與並流區域,故通常有 F<1F<1。

由能量平衡:

Q=CH(TH1−TH2)=CC(TC2−TC1),Q=C_H(T_{H1}-T_{H2}) =C_C(T_{C2}-T_{C1}),

其中 C=m˙CpC=\dot m C_p。因此

z=TH1−TH2TC2−TC1=CCCH.z=\frac{T_{H1}-T_{H2}}{T_{C2}-T_{C1}} =\frac{C_C}{C_H}.

而

Y=Tt,out−Tt,inTs,in−Tt,inY=\frac{T_{t,\mathrm{out}}-T_{t,\mathrm{in}}} {T_{s,\mathrm{in}}-T_{t,\mathrm{in}}}

代表 tube-side 流體的無因次溫升;在固定 zz 下,YY 越大表示熱傳量及熱回收程度越高。圖中顯示:

  • 固定 zz 時,YY 增加,FF 下降。
  • 固定 YY 時,zz 增加,FF 下降。
  • 固定 FF 時,較小的 YY 對應較大的 zz。

選項分析

A:錯誤。
雖然固定 zz 時,YY 增加會使 FF 降低,但這不代表實際熱傳量降低。由

Q=CCY(TH1−TC1),Q=C_CY(T_{H1}-T_{C1}),

在其他條件固定時,YY 增加反而表示 QQ 增加。較低的 FF 只表示相同端溫條件下,相較於理想逆流的有效平均溫差較小。

B:錯誤。
較大的 zz 表示熱流體的溫降相對於冷流體的溫升較大,也就是熱流體的熱容量率較小:

z=CCCH.z=\frac{C_C}{C_H}.

但熱流體放出的熱量必然等於冷流體吸收的熱量:

QH=QC.Q_H=Q_C.

不能因此判定傳給熱流體的熱量多於冷流體接收的熱量。

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第 4 題

(b) For heat exchangers that deviate from ideal counterflow, the logarithmic-mean temperature difference (LMTD) must be corrected using a correction factor F. The figure below shows the LMTD correction factor F for a 1-2 shell-and-tube heat exchanger, plotted as a function of Y and z, where subscripts t and s denote tube-side and shell-side streams, respectively. Which of the following statements correctly describe the physical meaning and trends of the correction factor shown in the figure (13%)?
🖼️【此處有附圖,請對照原卷】
A. At a constant z value, increasing Y leads to a reduction in the actual heat transfer rate because a lower correction factor implies a smaller overall heat-transfer driving force between the hot and cold streams.
B. A higher z value indicates that the temperature change of the hot stream is larger than that of the cold stream from inlet to outlet, implying that more heat is transferred to the hot stream than the cold stream receives.
C. At a fixed Y value, the correction factor decreases with increasing z because transferring more heat to the hot stream enhances shell-side induced vortices, making the heat-transfer process less ideal.
D. When Y increases, the correction factor decreases more dramatically at higher z values than at lower z values because a smaller heat-capacity coefficient (mCp) of the hot stream causes a smaller change in Y for a given increase in heat-transfer rate.
E. For this 1-2 shell-and-tube heat exchanger, the correction factor is always less than unity because this configuration inherently contains regions of both counterflow and parallel flow within the exchanger.
F. For the same correction factor value, a smaller Y implies that the hot-stream temperature can be changed more easily.
G. The decrease in the correction factor is a direct consequence of a larger fraction of the maximum possible heat being transferred within the exchanger.
H. Increasing the inlet temperature of the hot stream reduces the value of Y, thereby increasing the correction factor.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 5 頁

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這一題的完整詳解

核心觀念

1–2 shell-and-tube heat exchanger 的實際對數平均溫差為

ΔTlm,actual=FΔTlm,counterflow,\Delta T_{\mathrm{lm,actual}} =F\Delta T_{\mathrm{lm,counterflow}},

因此熱傳量為

Q=UAFΔTlm,counterflow.Q=UA F\Delta T_{\mathrm{lm,counterflow}}.

其中 FF 是相對於理想逆流熱交換器的修正因子,1–2 管殼式熱交換器因同時含有逆流與並流區域,故通常有 F<1F<1。

由能量平衡:

Q=CH(TH1−TH2)=CC(TC2−TC1),Q=C_H(T_{H1}-T_{H2}) =C_C(T_{C2}-T_{C1}),

其中 C=m˙CpC=\dot m C_p。因此

z=TH1−TH2TC2−TC1=CCCH.z=\frac{T_{H1}-T_{H2}}{T_{C2}-T_{C1}} =\frac{C_C}{C_H}.

而

Y=Tt,out−Tt,inTs,in−Tt,inY=\frac{T_{t,\mathrm{out}}-T_{t,\mathrm{in}}} {T_{s,\mathrm{in}}-T_{t,\mathrm{in}}}

代表 tube-side 流體的無因次溫升;在固定 zz 下,YY 越大表示熱傳量及熱回收程度越高。圖中顯示:

  • 固定 zz 時,YY 增加,FF 下降。
  • 固定 YY 時,zz 增加,FF 下降。
  • 固定 FF 時,較小的 YY 對應較大的 zz。

選項分析

A:錯誤。
雖然固定 zz 時,YY 增加會使 FF 降低,但這不代表實際熱傳量降低。由

Q=CCY(TH1−TC1),Q=C_CY(T_{H1}-T_{C1}),

在其他條件固定時,YY 增加反而表示 QQ 增加。較低的 FF 只表示相同端溫條件下,相較於理想逆流的有效平均溫差較小。

B:錯誤。
較大的 zz 表示熱流體的溫降相對於冷流體的溫升較大,也就是熱流體的熱容量率較小:

z=CCCH.z=\frac{C_C}{C_H}.

但熱流體放出的熱量必然等於冷流體吸收的熱量:

QH=QC.Q_H=Q_C.

不能因此判定傳給熱流體的熱量多於冷流體接收的熱量。

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第 5 題

Consider a counterflow cooling tower. Air at temperature TybT_{yb} and enthalpy HybH_{yb} enters the bottom of the tower and leaves at the top with temperature TyaT_{ya} and enthalpy HyaH_{ya}. Water enters the top at temperature TxaT_{xa} and leaves at the bottom at temperature TxbT_{xb}. The mass velocity of the air is GyG_y (the mass of vapor-free air per hour per unit cross-section of the tower). The mass velocities of water at the inlet and outlet are, respectively, GxaG_{xa} and GxbG_{xb}. At a distance Z from the bottom of the contact zone, the air and water temperatures are TyT_y and TxT_x, and the air enthalpy is HyH_y.
(a) Derive the operating line for the tower, on a plot of air enthalpy versus water temperature. Please indicate the slope of the operating line. You may assume GxG_x to be constant (i.e., Gxb=GxaG_{xb} = G_{xa}) with a constant specific heat of water (CLC_L) (5%).
(b) The tower operates with Txa=95∘FT_{xa} = 95^\circ\text{F} and Txb=75∘FT_{xb} = 75^\circ\text{F} when the air has dry-bulb and wet-bulb temperatures, respectively, of 70∘F70^\circ\text{F} and 60∘F60^\circ\text{F}. The enthalpy of air at the bottom (HyH_y) is estimated at 18.7 Btu/lb18.7 \text{ Btu/lb}. The tower has 4 ft4 \text{ ft} of stacked plastic fill, and the flow rates are Gy=2,000 lb/h⋅ft2G_y = 2,000 \text{ lb/h} \cdot \text{ft}^2 and Gx=2,200 lb/h⋅ft2G_x = 2,200 \text{ lb/h} \cdot \text{ft}^2. Determine the number of transfer units (NoyN_{oy}), the height of a transfer unit based on the overall gas-phase driving force (HoyH_{oy}), and the temperature approach. For 75∘F≤Tx≤95∘F75^\circ\text{F} \le T_x \le 95^\circ\text{F}, CLC_L is taken to be 1 Btu/lb⋅∘F1 \text{ Btu/lb} \cdot ^\circ\text{F}, and the equilibrium line is approximated by: H=1.2T−59H = 1.2T-59 (10%).

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這一題的完整詳解

核心觀念

冷卻塔的操作線由空氣與水之間的微分能量衡算決定。以無水空氣質量為基準,空氣焓的增加等於水的顯熱損失:

Gy dHy=GxCL dTxG_y\,dH_y=G_x C_L\,dT_x

此處 GxG_x、GyG_y 分別為水與無水空氣的質量速度。將操作線畫在「空氣焓 HyH_y 對水溫 TxT_x」圖上,其斜率為 GxCL/GyG_xC_L/G_y。

計算整體氣相傳質單元數時,使用空氣的整體氣相推動力 Hy∗−HyH_y^*-H_y:

Noy=∫HybHyadHyHy∗−HyN_{oy}=\int_{H_{yb}}^{H_{ya}}\frac{dH_y}{H_y^*-H_y}

其中 Hy∗H_y^* 是與該位置水溫達氣液平衡時的空氣焓。

解題方法

(a) 操作線與斜率

由題設假定 GxG_x 與 CLC_L 為常數,將能量衡算從塔底積分至任意位置:

∫HybHydHy=GxCLGy∫TxbTxdTx\int_{H_{yb}}^{H_y}dH_y = \frac{G_xC_L}{G_y}\int_{T_{xb}}^{T_x}dT_x

因此操作線為

Hy=Hyb+GxCLGy(Tx−Txb)H_y=H_{yb}+\frac{G_xC_L}{G_y}(T_x-T_{xb})

在 HyH_y 對 TxT_x 圖上的斜率為

dHydTx=GxCLGy\boxed{\frac{dH_y}{dT_x}=\frac{G_xC_L}{G_y}}

(b) 傳質單元數、傳質單元高度與溫度接近值

先由塔的整體能量衡算求塔頂空氣焓。水由 95∘F95^\circ\mathrm{F} 冷卻至 75∘F75^\circ\mathrm{F},所以空氣焓增加量為

Hya−Hyb=GxCLGy(Txa−Txb)=2200(1)2000(95−75)=22 Btu/lbH_{ya}-H_{yb} = \frac{G_xC_L}{G_y}(T_{xa}-T_{xb}) = \frac{2200(1)}{2000}(95-75) = 22\ \mathrm{Btu/lb}

已知 Hyb=18.7 Btu/lbH_{yb}=18.7\ \mathrm{Btu/lb},故

Hya=18.7+22=40.7 Btu/lbH_{ya}=18.7+22=40.7\ \mathrm{Btu/lb}

操作線斜率為

GxCLGy=2200(1)2000=1.10 Btu/lb∘F\frac{G_xC_L}{G_y} = \frac{2200(1)}{2000} = 1.10\ \frac{\mathrm{Btu/lb}}{^\circ\mathrm{F}}

因此操作線可寫為

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第 6 題

For mass transfer in fixed-bed adsorption, a solute material balance for a section dLdL of the bed is written as:
ϵ∂c∂t+(1−ϵ)ρp∂W∂t=−u0∂c∂L\epsilon \frac{\partial c}{\partial t} + (1-\epsilon) \rho_p \frac{\partial W}{\partial t} = -u_0 \frac{\partial c}{\partial L}
where ϵ\epsilon is the external void fraction of the bed; cc is the solute concentration; ρp\rho_p is the density of the particle; WW is the adsorbate loading; u0u_0 is the superficial velocity of the fluid; LL is the distance through the bed. The rate of accumulation in the fluid and in the solid is the difference between input and output flows. For adsorption from a gas or a dilute solution, the accumulation in the fluid is usually negligible compared to the accumulation on the solid. The transfer process is approximated using an overall volumetric coefficient (KaK_a) and an overall driving force:
ρp(1−ϵ)∂W∂t=Ka(c−c∗)\rho_p (1-\epsilon) \frac{\partial W}{\partial t} = K_a (c - c^*)
where c∗c^* is the concentration in equilibrium with the average concentration WW in the solid.
(a) Consider irreversible adsorption with a constant mass-transfer coefficient, in which case the rate of mass transfer is just proportional to the fluid concentration. Derive the concentration at the end of the bed, assuming that the first portion of the bed has not been saturated (3%).
(b) After the first portion of the bed is saturated, the concentration profile moves steadily down the bed, keeping the same shape. The concentration profile can be expressed using dimensionless terms:
CC0=N(τ−1)N(τ)−1\frac{C}{C_0} = \frac{N(\tau-1)}{N(\tau)-1}
where NN represents the overall number of transfer units; tt is the ratio of the time (t)(t) to the ideal time t∗t^* (t=t/t∗t = t/t^*). Determine NN for a 20-cm bed with 60% utilization at the break point (when c/c0=0.05c/c_0 = 0.05) (4%).
(c) Predict NN when the bed length is increased to 40 cm (3%).

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這一題的完整詳解

核心觀念

本題考查**固定床吸附(Fixed-Bed Adsorption)**之質傳動力學與破出曲線(Breakthrough Curve)分析,出自經典化工單元操作教科書(McCabe, Smith & Harriott, Unit Operations of Chemical Engineering):

  1. 固定床物料平衡與質傳速率方程式:
    • 固-液/固-氣兩相流動吸附的連續方程式:
      ϵ∂c∂t+(1−ϵ)ρp∂W∂t=−u0∂c∂L\epsilon \frac{\partial c}{\partial t} + (1-\epsilon) \rho_p \frac{\partial W}{\partial t} = -u_0 \frac{\partial c}{\partial L}
    • 當處理稀薄流體時,流體孔隙內的溶質累積量遠小於固相吸附劑的累積量,流體累積項可忽略(ϵ∂c∂t≈0\epsilon \frac{\partial c}{\partial t} \approx 0)。
    • 總質傳推動力與質傳速率:
      ρp(1−ϵ)∂W∂t=Ka(c−c∗)\rho_p (1-\epsilon) \frac{\partial W}{\partial t} = K_a (c - c^*)
  2. 不可逆吸附(Irreversible Adsorption):
    • 吸附等溫線極為陡峭,未飽和之固相吸附劑表面對應的平衡濃度 c∗=0c^* = 0,質傳速率僅正比於流體本體濃度 cc。
  3. 等型態前鋒(Constant-Pattern Frontier)與破出曲線:
    • 吸附區形成穩態移行波(Traveling wave / Constant pattern),濃度剖面外形維持不變沿床層向下推移。
    • 描述不可逆吸附之無因次濃度曲線關係式為:
      N(τ−1)=1+ln⁡(cc0)⟺cc0=exp⁡[1+N(τ−1)]N(\tau - 1) = 1 + \ln\left(\frac{c}{c_0}\right) \quad \Longleftrightarrow \quad \frac{c}{c_0} = \exp\left[ 1 + N(\tau - 1) \right]
    • 質傳單元數(Number of Transfer Units, NTU):N=KaLu0=LHTUN = \dfrac{K_a L}{u_0} = \dfrac{L}{HTU},且 N∝LN \propto L。
    • 床層利用率(Utilization at break point)在破出點定義為無因次時間 τb=tbt∗\tau_b = \dfrac{t_b}{t^*}。

解題方法與詳細推導

(a) 推導吸附床出口之流體濃度(當前端床層尚未達到飽和)

  1. 基本假設與簡化:

    • 流體相之累積項可忽略:ϵ∂c∂t≈0\epsilon \dfrac{\partial c}{\partial t} \approx 0。
    • 吸附為不可逆(Irreversible adsorption),在固相未飽和的床層區段中,固相平衡濃度 c∗=0c^* = 0。
    • 質傳係數 KaK_a 為常數。
  2. 聯立物料平衡與質傳速率式:
    將流體累積項忽略後的物料平衡式為:
    (1−ϵ)ρp∂W∂t=−u0∂c∂L(1-\epsilon) \rho_p \frac{\partial W}{\partial t} = -u_0 \frac{\partial c}{\partial L}
    將質傳速率式 ρp(1−ϵ)∂W∂t=Ka(c−c∗)=Kac\rho_p (1-\epsilon) \dfrac{\partial W}{\partial t} = K_a (c - c^*) = K_a c 代入上式,消去固相累積項:
    −u0∂c∂L=Kac-u_0 \frac{\partial c}{\partial L} = K_a c

  3. 分離變數並積分:
    在準穩態(pseudo-steady state)下,流體濃度隨床長變化之常微分方程式為:
    dcc=−Kau0dL\frac{dc}{c} = -\frac{K_a}{u_0} dL
    邊界條件(Boundary Conditions):

    • 床層入口處(L=0L = 0):c=c0c = c_0
    • 床層出口處(L=LL = L):c=c(L)c = c(L)

    兩邊進行定積分:
    ∫c0c(L)dcc=−∫0LKau0dL\int_{c_0}^{c(L)} \frac{dc}{c} = -\int_{0}^{L} \frac{K_a}{u_0} dL

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