苏大纳米学院化工原理期中试卷A卷答案.pdf

《苏大纳米学院化工原理期中试卷A卷答案.pdf》由会员分享，可在线阅读，更多相关《苏大纳米学院化工原理期中试卷A卷答案.pdf（12页珍藏版）》请在装配图网上搜索。

1、 1 / 12 苏州大学 化工原理 (Principles of Chemical Engineering) 课程 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 requi

2、red 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 Darcys Law 1.5 Logarithmic mean temperature difference (LMTD) 1.6 Thermal diffusivity 1.7 Prandtl number 1.8 Nusselt number ( 装订 线 内 不 答

3、 题 和 打 分 ) 装订 线线 ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 2 / 12 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. 2 2 , where is the drag coefficient (dimensionless), is the total drag (unit: N), i

4、s the projected area (unit: m2), is the fluid density (unit: kg/m3), and 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 giv

5、en by NPSH = 2 + 2 2 2 where is the vapor pressure of fluid at the given temperature (unit: N/m2), 2 is the pressure at the pump suction (unit: N/m2), 2 is the flow velocity at the pump suction (unit: m/s), is the fluid density (unit: kg/m3), and is the gravitational constant (unit: m/s2). 1.3 Hydra

6、ulic radius (denoted by , unit m) is defined as the ratio of the cross-sectional area of the channel (denoted by , unit m2) to the wetted perimeter of the channel (denoted by , unit m), i.e., = . 1.4 Darcys Law states that the flow is proportional to the pressure drop and inversely proportional to t

7、he fluid viscosity. It is often used to describe flow of liquids through porous media. 1.5 Logarithmic mean temperature difference (LMTD) is the logarithmic ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 3 / 12 mean of the terminal point temperature differences, 1 and 2 (also known as appro

8、aches, both in unit of K) in a heat exchanger: = 1 2ln( 1 2 ) where (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, = where (unit: m2/s) is the thermal diffusivity, , in unit of W/(mK), is the t

9、hermal conductivity, is the density (unit: kg/m3), and is the specific heat capacity, in unit of J/(kgK). 1.7 Prandtl number (denoted by , dimensionless) is defined as the ratio of the diffusivity of momentum (kinematic viscosity, = , in unit of m2/s, where is the viscosity in unit of Pas, and is th

10、e density with unit kg/m3) to the thermal diffusivity, (unit: m2/s). 1.8 Nusselt number (denoted by , dimensionless) is defined as the heat transfer coefficient (, in unit of W/(m2K) divided by the ratio between thermal conductivity (, in unit of W/(mK) and the characteristic length such as the pipe

11、 diameter (, in unit of m): = The Nusselt number can be understood as the ratio of the tube diameter of the equivalent thickness of the laminar layer. 4 / 12 Problem 2. Fluid Flow in Microfluidic Systems (10 points) Microfluidics is the science that deals with the flow of liquid inside micrometer-si

12、ze 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 dev

13、ices 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 u

14、ndergoing 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 m and room temperature)? (3 points

15、) 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) ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 5 / 12 Solution. 2.1 The Reynolds number is defined as = = =

16、2 = inertial forceViscous force where = Reynolds number (dimensionless) = Characteristic length of the system (unit: m) = Characteristic flow velocity (unit: m/s) = Density of the fluid (unit: kg/m3) = Viscosity of the fluid (unit: Pas) = =Kinematic viscosity of the fluid (unit: m2/s) 2.2 we have =

17、1 mm/s = 10-3 m/s, = 10 m = 10-5 m, = 1000 kg/m3, and =10-3 Pas. Therefore, = = 10 5 103 103 103 = 0.01 1 2100 We 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, . ( 装订 线 内 不 答 题 和 打 分 ) 装订线

18、线 6 / 12 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 radia

19、l 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

20、s and 1000 kg/m3 for the viscosity and density of water, and use 0.003 Pas and 800 kg/m3 for the viscosity and density of kerosene. Solution. 3.1 We have tube inner diameter = 4 mm = 4103 m, average flow velocity = 0.4 m s , = 1000 kg/m3, and = 10-3 Pas. The Reynolds number is thus given by = = (410

21、 3 m)(0.4 m s )(103 kg m3 ) 103 Pa s = 1600 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: ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 7 / 12 = 322 = 32(2 m)(0.4 m s )(10 3 Pa s) (4103 m)2 = 1600 Pa 3.2 For laminar flow in a pip

22、e, the velocity profile is parabolic: () = 2 4 1( ) 2 We find the maximum flow velocity at the center where = 0. The maximum flow velocity is twice of the average velocity, thus = 0.8 m/s 3.3 For laminar flow in a pipe, the velocity profile can be written as () = 1() 2 = 21() 2 To have the average f

23、low velocity equals the local flow velocity, we obtain from = 21()2 that = 1 2 0.7071. Thus, = 0.7071 = 0.70710.54 mm = 1.4142 mm. 3.4 If we consider kerosene flows through this pipe instead of water, the Reynolds number becomes = = (410 3 m)(0.4 m s )(800 kg m3 ) 0.003 Pa s = 426.7 2100 The flow is

24、 still laminar. Now the pressure drop becomes = 322 = 32(2 m)(0.4 m s )(0.003 Pas)(4103 m)2 = 4800 Pa The 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. ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 8 / 12 Problem 4. Motio

25、n 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 Pas). 4.1 Determine the maximum particle diameter for which the settling motion obeys Stokes Law. (5 points) 4.2 Determine the mini

26、mum diameter for which the settling motion obeying Newtons Law. (5 points) Solution. We have particle density = 2650 kg m3 , fluid density = 1.205 kg m3 , fluid viscosity = 1.80105 Pa s . Consider the criterion for settling regimes, the K-parameter = ( )2 1 3 4.1 To have the settling motion obeys St

27、okes Law, 2.6. Substituting the known variable values, we obtain ( )2 1 3 = 45885 m1 and 5.67105 m = 56.7 m . Thus, the maximum particle diameter for which the settling motion obeys Stokes Law is 56.7 m. 4.2 To have the settling motion obeying Newtons Law, the range of should be 68.9 2360. Thus, 0.0

28、015 m 0.051 m Thus the minimum particle diameter should be 0.0015 m = 1.5 mm. ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 9 / 12 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

29、, k=0.78 Wm-1 -1) separated by a 10-mm-thick stagnant air space (thermal conductivity, k=0.026 Wm-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 ma

30、intained 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 Wm-2 -1 and ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 ( 装订 线 内 不 答 题 和 打 分 ) 装订线线 10 / 12 ho=40 Wm-2 -1, respectively, which includes also the effec

31、ts 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, = 0.8 m 1.5 m = 1.2 m2. The r

32、oom temperature, 1 = 20 The temperature outdoor, 2 = 10 Thermal conductivity of glass, 1=3=0.78 Wm-1 -1 Thermal conductivity of air, 2=0.026 Wm-1 -1 Thickness of glass layer, 1=3= 4 mm = 0.004 m Thickness of the air layer, 2= 10 mm = 0.01 m Heat transfer coefficients on the inner surfaces of the win

33、dow, = 10 Wm2 1 Heat transfer coefficients on the outer surfaces of the window, = 40 Wm2 1 5.1 Thermal resistors in series = 1 = (10 Wm2 1 1.2 m2)1 = 0.08333 W ( 装订 线 内 不 答 题 和 打 分 ) 装 订 线 线 线 11 / 12 1 = 1 1 = 0.004 m0.78 Wm1 1 1.2 m2 = 0.00427 W 2 = 2 2 = 0.01 m0.026 Wm1 1 1.2 m2 = 0.32051 W 3 = 1

34、 = 0.00427 W = 1 = (40 Wm2 1 1.2 m2)1 = 0.02083 W The total thermal resistance = +1 +2 +3 + = 0.43321 W The rate of heat transfer (from the room to outdoor) at steady state is then: = 1 2 = 30 0.43321 W = 69.25 W The temperature of the inner surface of the windows is then 1 = 1 = 20 69.25 W0.08333 W

35、 = 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 = +1 +0+3 + = 0.1127 W The rate of heat transfer (from the room to outdoor) at steady state is then: = 1 2 = 30 0.1127 W = 266.2 W The temperature of the inner surface of the windows is then 1 = 1 = 20 266.2 W0.08333 W = 2.2 12 / 12 ( 装订 线 内 不 答 题 和 打 分 ) 装订线线