2015-苏大纳米学院化工原理-期中试卷-A卷-答案

期 中 试卷 （ A ） 共 12 页

考试形式（开/闭卷） 开卷 2015 年 5 月

学院（部） 纳米学院 年级 2012级（化学方向）专业 学号 姓名 成绩

题号 一 二 三 四 五 总分 评卷人 签字 得分

Problem 1. Important Concepts (5 point each, 5x8=40 points) Explain the following concepts. If you use symbols, it is required that you list the meaning and the physical unit of each symbol that you use. 1.1 Drag Coefficient

1.2 Net Positive Suction Head (NPSH) 1.3 Hydraulic radius 1.4 Darcy’s Law

1.5 Logarithmic mean temperature difference (LMTD) 1.6 Thermal diffusivity 1.7 Prandtl number 1.8 Nusselt number

1 / 12

( 装订 线 内 不 答 题 和 打 分 ) 答 题 和 打 分 ) ………………………………装………………………………………………订………………… ……………………线……………………………………线……………………… …………………………………线……………………………………线………………………Solution.

1.1 Drag Coefficient is defined as the ratio of the average drag per unit projected area to the product of the density of the fluid and the velocity head. 𝐶𝐷≡

𝐹𝐷⁄𝐴𝑝ρ𝑢2⁄2

, where 𝐶𝐷 is the drag coefficient (dimensionless), 𝐹𝐷

is the total drag (unit: N), 𝐴𝑝 is the projected area (unit: m2), 𝜌 is the fluid density (unit: kg/m3), and 𝑢 is the velocity of the approaching stream (unit: m/s).

1.2 Net Positive Suction Head (NPSH) is defined as the difference between the absolute stagnation pressure in the flow at the pump suction and liquid vapor pressure and given by

2

𝑝2𝑢2𝑝𝑣

NPSH=+−

𝜌𝑔2𝑔𝜌𝑔where 𝑝𝑣 is the vapor pressure of fluid at the given temperature (unit: N/m2), 𝑝2 is the pressure at the pump suction (unit: N/m2), 𝑣2 is the flow velocity at the pump suction (unit: m/s), 𝜌 is the fluid density (unit: kg/m3), and 𝑔 is the gravitational constant (unit: m/s2).

1.3 Hydraulic radius (denoted by 𝑟𝐻, unit m) is defined as the ratio of the cross-sectional area of the channel (denoted by 𝑆, unit m2) to the wetted perimeter of the channel (denoted by 𝐿𝑝, unit m), i.e., 𝑟𝐻=𝑆⁄𝐿𝑝. 1.4 Darcy’s Law states that the flow is proportional to the pressure drop and inversely proportional to the fluid viscosity. It is often used to describe flow of liquids through porous media.

1.5 Logarithmic mean temperature difference (LMTD) is the logarithmic

2 / 12

mean of the terminal point temperature differences, ∆𝑇1 and ∆𝑇2 (also known as approaches, both in unit of K) in a heat exchanger:

∆𝑇1−∆𝑇2

∆𝑇𝑚=

ln(∆𝑇1⁄∆𝑇2)where ∆𝑇𝑚 (unit: K) is the LMTD.

1.6 Thermal diffusivity is defined as the thermal conductivity divided by density and specific heat capacity at constant pressure,

𝑘α= 𝜌𝑐𝑝

where α (unit: m2/s) is the thermal diffusivity, 𝑘, in unit of W/(mK), is the thermal conductivity, 𝜌 is the density (unit: kg/m3), and 𝑐𝑝 is the specific heat capacity, in unit of J/(kgK).

1.7 Prandtl number (denoted by 𝑃𝑟, dimensionless) is defined as the ratio of the diffusivity of momentum (kinematic viscosity, 𝜈=𝜇⁄𝜌, in unit of m2/s, where 𝜇 is the viscosity in unit of Pas, and 𝜌 is the density with unit kg/m3) to the thermal diffusivity, α (unit: m2/s).

1.8 Nusselt number (denoted by 𝑁𝑢, dimensionless) is defined as the heat transfer coefficient (ℎ, in unit of W/(m2K)) divided by the ratio between thermal conductivity (𝑘, in unit of W/(mK)) and the characteristic length such as the pipe diameter (𝐷, in unit of m):

ℎ

𝑁𝑢=

𝑘⁄𝐷The Nusselt number can be understood as the ratio of the tube diameter of the equivalent thickness of the laminar layer.

3 / 12

( 装订 线 内 不 答 题 和 打 分 ) …………………………………………装………………………………………………订………………………………………线……………………………………线……………………… Problem 2. Fluid Flow in Microfluidic Systems (10 points) Microfluidics is the science that deals with the flow of liquid inside micrometer-size channels. At least one dimension of the channel is in the order of a micrometer or tens of micrometers in order to consider it microfluidics. Microfluidics can be considered both as a science (study of the behaviour of fluids in micro-channels) and a technology (manufacturing of microfluidics devices for applications such as lab-on-a-chip). In the past two decades, microfluidics research has seen phenomenal growth, with many new and emerging applications in fields ranging from chemistry, physics, and biology to engineering. With the emergence of nanotechnology, microfluidics is currently undergoing dramatic changes, embracing the rising field of nanofluidics. Answer the following questions. 2.1 What is the Reynolds Number? (3 points)

2.2 What is the Reynolds Number in a typical microfluidic channel (assume water, velocity ~ 1 mm/s, channels of size 10 μm and room temperature)? (3 points) What type of flow do we have in that case? (2 points)

2.3 If the microfluidic flow is driven by pressure differences, how does the pressure drop vary with flow velocity? (2 points)

4 / 12

Solution.

2.1 The Reynolds number is defined as

𝑑𝑉𝜌𝑑𝑉𝜌𝑉2inertial force𝑅𝑒====

𝑉𝜇𝜈Viscous force𝜇𝑑 where 𝑅𝑒= Reynolds number (dimensionless)

𝑑= Characteristic length of the system (unit: m)

( 装订 线 内 不 答 题 和 打 分 ) …装………………………………………………订………………………………………线……………………………………线……………………… 𝑉= Characteristic flow velocity (unit: m/s) 𝜌= Density of the fluid (unit: kg/m3)

𝜇= Viscosity of the fluid (unit: Pas)

𝜈=𝜇⁄𝜌=Kinematic viscosity of the fluid (unit: m2/s)

2.2 we have 𝑉= 1 mm/s = 10-3 m/s, 𝑑= 10μm = 10-5 m, 𝜌= 1000 kg/m3, and 𝜇=10-3 Pas. Therefore,

𝑑𝑉𝜌10−5×10−3×103

𝑅𝑒===0.01≪1≪2100 −3𝜇10We expect a laminar flow in a typical microfluidic channel. 2.3 As the flow is laminar, we expect a linear relationship between pressure drop and flow velocity, ∆𝑝~𝑉.

5 / 12

( 装订 线 内 不 答 题 和 打 分 ) …………………………………………装………………………………………………订………………………………………线……………………………………线……………………… Problem 3. Incompressible Flow in a Pipe (15 points)

Water passes through a long straight pipe of inner diameter d = 4 mm with the average velocity 0.4 m/s.

3.1 What is the pressure drop in Pa when water flows over a pipe length of L = 2 m? (5 points)

3.2 Find the maximum velocity and radial distance r at which it occurs. (2 points)

3.3 Find the radial distance r at which the average flow velocity equals the local flow velocity. (3 points)

3.4 If kerosene flows through this pipe instead of water, how do the answers to the above questions (3.1, 3.2, 3.3) change? (5 points) Use 0.001 Pas and 1000 kg/m3 for the viscosity and density of water, and use 0.003 Pas and 800 kg/m3 for the viscosity and density of kerosene.

Solution.

3.1 We have tube inner diameter 𝑑=4 mm=4×10−3 m, average flow velocity 𝑉=0.4 m⁄s, 𝜌= 1000 kg/m3, and 𝜇=10-3 Pas. The Reynolds number is thus given by

𝑑𝑉𝜌(4×10−3 m)×(0.4 m⁄s)×(103 kg⁄m3)𝑅𝑒===1600−3𝜇10 Pa∙s<2100

Thus the flow is laminar, and we can use the Hagen-Poiseuille equation to obtain the pressure drop over a pipe length of L = 2 m:

6 / 12

32𝐿𝑉𝜇32×(2 m)×(0.4 m⁄s)×(10−3 Pa∙s)∆𝑝===1600 Pa

(4×10−3 m)2𝑑23.2 For laminar flow in a pipe, the velocity profile is parabolic:

𝛥𝑝𝑅2𝑟2

𝑢𝑧(𝑟)=[1−()]

4𝜇𝐿𝑅We find the maximum flow velocity 𝑢𝑚𝑎𝑥 at the center where 𝑟=0. The maximum flow velocity is twice of the average velocity, thus

𝑢𝑚𝑎𝑥=0.8 m/s

3.3 For laminar flow in a pipe, the velocity profile can be written as

𝑟2𝑟2

𝑢𝑧(𝑟)=𝑢𝑚𝑎𝑥[1−()]=2𝑉[1−()]

𝑅𝑅To have the average flow velocity equals the local flow velocity, we obtain from 𝑉=2𝑉[1−()] that 𝑟⁄𝑅=1⁄√2≈0.7071. Thus,

𝑅𝑟=0.7071 𝑅=0.7071×0.5×4 mm=1.4142 mm.

3.4 If we consider kerosene flows through this pipe instead of water, the Reynolds number becomes

𝑑𝑉𝜌(4×10−3 m)×(0.4 m⁄s)×(800 kg⁄m3)𝑅𝑒===426.7

𝜇0.003 Pa∙s<2100

The flow is still laminar. Now the pressure drop becomes

…………………线……………………… 𝑟2

32𝐿𝑉𝜇32×(2 m)×(0.4 m⁄s)×(0.003 Pa∙s)

∆𝑝===4800 Pa 2−32()𝑑4×10 mThe answers to questions (3.2) and (3.3) stay the same, because they do not dependent on the fluid properties as long as the flow remains laminar.

7 / 12

( 装订 线 内 不 答 题 和 打 分 ) 题 和 打 分 ) ………………装………………………线…………………………订 ………………………………………线……………………………………线……………………… ……………………………线Problem 4. Motion of Particles through Fluids (10 points)

A spherical quartzose particle with a density of 2650 kg/m3 settles freely in 20℃ air (density 1.205 kg/m3, viscosity 1.80×10-5 Pas).

4.1 Determine the maximum particle diameter for which the settling motion obeys Stokes’ Law. (5 points)

4.2 Determine the minimum diameter for which the settling motion obeying Newton’s Law. (5 points) Solution.

We have particle density 𝜌𝑝=2650 kg⁄m3, fluid density 𝜌=1.205 kg⁄m3, fluid viscosity 𝜇=1.80×10−5 Pa∙s. Consider the criterion for settling regimes, the K-parameter

𝜌(𝜌𝑝−𝜌)𝑔

𝐾=𝑑𝑝[]2𝜇1⁄3

4.1 To have the settling motion obeys Stokes’ Law, 𝐾<2.6. Substituting the known variable values, we obtain

𝜌(𝜌𝑝−𝜌)𝑔[]𝜇21⁄3

=45885 m−1

and 𝑑𝑝<5.67×10−5 m=56.7 μm. Thus, the maximum particle diameter for which the settling motion obeys Stokes’ Law is 56.7 μm. 4.2 To have the settling motion obeying Newton’s Law, the range of 𝐾 should be 68.9<𝐾<2360. Thus,

0.0015 m<𝑑𝑝<0.051 m

Thus the minimum particle diameter should be 0.0015 m = 1.5 mm.

8 / 12

( 装订 线 内 不 答 题 和 打 分 ) …………………………………………装………………………………………………订………………………………………线……………………………………线……………………… Problem 5. Heat Transfer (25 points) Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4-mm-thick layers of glass (thermal conductivity, k=0.78 W•m-1•℃-1) separated by a 10-mm-thick stagnant air space (thermal conductivity, k=0.026 W•m-1•℃-1), as is shown in the Figure below.

5.1 Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface (T1) for a day during which the room is maintained at 20 ℃ while the temperature of the outdoor is -10 ℃. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be hi=10 W•m-2•℃-1 and

9 / 12

……………………………………线………………………………… ) 分… …打订 和… 题… 答… 不…… 内… 线… 订…装… (……… …………………………………装………线…………………………………………………………ho=40 W•m-2•℃-1, respectively, which includes also the effects of radiation. (15 points)

5.2 Redo your calculation for the case where the two 4-mm-thick glasses that enclose a stagnant air space is replaced by a single 8-mm-thick window glass while other parameters stay the same. (10 points)

Solution.

Surface area of the window, 𝐴=0.8 m ×1.5 m=1.2 m2. The room temperature, 𝑇∞1=20 ℃ The temperature outdoor, 𝑇∞2=−10 ℃

Thermal conductivity of glass, 𝑘1=𝑘3=0.78 W•m-1•℃-1 Thermal conductivity of air, 𝑘2=0.026 W•m-1•℃-1 Thickness of glass layer, 𝐵1=𝐵3= 4 mm = 0.004 m Thickness of the air layer, 𝐵2= 10 mm = 0.01 m

Heat transfer coefficients on the inner surfaces of the window, ℎ𝑖=10 W∙m−2∙℃−1

Heat transfer coefficients on the outer surfaces of the window, ℎ𝑜=40 W∙m−2∙℃−1

5.1 Thermal resistors in series

𝑅=1𝑖ℎ=(10 W∙m−2∙℃−1 ×1.2 m2)−1=0.08333 ℃⁄W

𝑖𝐴 10 / 12

𝐵10.004 m𝑅1===0.00427 ℃⁄W −1−12𝑘1𝐴0.78 W∙m∙℃ ×1.2 m𝐵20.01 m𝑅2===0.32051 ℃⁄W −1−12𝑘2𝐴0.026 W∙m∙℃ ×1.2 m𝑅3=𝑅1=0.00427 ℃⁄W

1𝑅𝑜==(40 W∙m−2∙℃−1 ×1.2 m2)−1=0.02083 ℃⁄W

ℎ𝑜𝐴The total thermal resistance

𝑅𝑡𝑜𝑡=𝑅𝑖+𝑅1+𝑅2+𝑅3+𝑅𝑜=0.43321 ℃⁄W

The rate of heat transfer (from the room to outdoor) at steady state is then:

𝑇∞1−𝑇∞230 ℃𝑞===69.25 W

𝑅𝑡𝑜𝑡0.43321 ℃⁄WThe temperature of the inner surface of the windows is then

𝑇1=𝑇∞1−𝑞𝑅𝑖=20 ℃−69.25 W×0.08333 ℃⁄W=14.2 ℃

5.2 If the two 4-mm-thick glasses that enclose a stagnant air space is replaced by a single 8-mm-thick window glass while other parameters stay the same, we have

𝑅𝑡𝑜𝑡=𝑅𝑖+𝑅1+0+𝑅3+𝑅𝑜=0.1127 ℃⁄W

The rate of heat transfer (from the room to outdoor) at steady state is then:

𝑇∞1−𝑇∞230 ℃𝑞===266.2 W

𝑅𝑡𝑜𝑡0.1127 ℃⁄WThe temperature of the inner surface of the windows is then 𝑇1=𝑇∞1−𝑞𝑅𝑖=20 ℃−266.2 W×0.08333 ℃⁄W=−2.2 ℃

11 / 12

………………………线……………………………………线………… …) 分…… 打… 和…… 题… 答…… 不… 内…订 线… 订…装…… (……………………………………装…………………………………

12 / 12