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1、传输原理试卷（Transmission principle test paper）08 grade material processing metallurgy transmission principle examination volume (B volume) answer and scoring standardMarking of test papers1. Refer to the scoring standard of problem-solving steps given in this scoring standard.2, the problem-solving metho
2、d is correct, the calculation results are wrong, as appropriate, buckle 12 points.3, the problem solving ideas are basically correct, but the analysis is not comprehensive, not entirely correct, according to the problem solving steps score points as appropriate.4, the error of the calculation result
3、s in front affect the calculation results, as long as the subsequent calculation formula and method is correct, no duplication of points.5, do not ask students to solve the problem steps and the answer is exactly the same, as long as you can show that they know the correct solution and the results a
4、re correct, you can score.6, problem-solving methods, calculation results are correct, but the intermediate steps in solving problems are incomplete, as appropriate, deduct 01 points.7, no physical unit of quantity or unit error deducted 1 points.First, multiple-choice questions (2 points per subjec
5、t, totaling 10 points)1, the fluid continuum model implies B.The gap between molecules of A. fluid is very small. The physical parameter describing fluid motion is continuous function by B.C. fluid is incompressible, and the physical properties of D. fluid are constant2. The flow of A is easy to occ
6、ur when the viscous fluid flows at low velocity in a tube with very small inner diameter.C. steady flow D. unsteady flow in A. laminar B. turbulent flow3. Siphon water between two open water tanks. The maximum pressure at the top of the siphon is B, and the atmospheric pressure is strong.A. greater
7、than B., less than C. equal to D. is greater than or equal to4, according to the heat conduction mechanism, in the three states of water gas, liquid and solid, the thermal conductivity of A is the smallest.A. gas B. liquid C. solid D. gas and liquid5. The emissivity of an object depends on the surfa
8、ce characteristics of the object (material type, surface condition and temperature), and the external conditions A.A. independent B. relatedC. has little to do with D., sometimes related, sometimes irrelevant, depending on the specific circumstances;Two, fill in the blanks (1 points per minute, a to
9、tal of 15 points)1. When the fluid flows in the pipe, the expression of Reynolds number Re is as follows: its physical meaning is: the ratio of inertia force to viscous force.2. Ideal fluid is fluid that does not exist viscous force or its action can be neglected.3, when the fluid flows in the pipel
10、ine, according to the thickness of the laminar boundary layer and the relative roughness of the pipeline surface, the pipeline can be divided into two types: Hydraulic smooth pipe and hydraulic rough pipe.4, the actual fluid in the flow process, due to resistance caused by energy loss. Among them, b
11、ecause of the viscous fluid itself or (and) the resistance loss of the micro fluid mass transfer, pulsation caused by called friction; and because of the channel change or (and) to form some of the obstacles flow resistance loss is called local resistance.5, the time of transient heat conduction in
12、a semi infinite plate is called the inertia time when the unsteady heat conduction is not affected by the surface temperature change. The inertia time is independent of the surface temperature, and is proportional to the square of the distance from the surface to the surface, inversely proportional
13、to the thermal diffusivity of the material.6. When convective heat transfer, the thin layer of fluid velocity changes rapidly near the solid wall is called velocity boundary layer, and the layer with rapid change of temperature is called temperature boundary layer.Three, the title (5 points per day,
14、 a total of 20 points)1. Briefly describe the definition of velocity boundary layer and how to divide the thickness of velocity boundary layerAnswer: when a fluid flows through a solid wall, a thin layer of fluid with a large velocity gradient near the solid wall is called the velocity boundary laye
15、r. (2 points)The thickness can be determined as follows: the distance from the wall to the surface is 0.99 times as fast as the velocity at zero, the velocity at the incoming stream (or the velocity of the main stream). (3 points)2. Briefly describe the expression of Prandtl criterion number (Pr) an
16、d its physical significance.Answer: the expression of Prandtl number is; (2 points)Pr represents the ratio of momentum transfer capability to heat transfer capability. From the concept of boundary layer, it can be considered as the index of the relative thickness between the velocity boundary layer
17、and the thermal boundary layer. (3 points)3, try to explain the heat dissipation process of indoor radiator, including heat transfer mode A case study of hot water cooling in heating pipes.Answer: there are the following heat exchange links and heat transfer mode:(1) from the hot water to the inner
18、wall of the radiator pipe, the heat transfer mode is the convection heat transfer (forced convection); (2 points)(2) from the inner wall of the radiator pipe to the outer wall; the heat transfer mode is heat conduction; (1 points)(3) from the outer wall of the radiator to the indoor environment and
19、air, the heat transfer mode has the radiation heat transfer and convection heat transfer. (2 points)4. The definition of angular coefficient and the property of angle coefficient are briefly described.Answer: the percentage of radiant energy falling on the surface 2 on the surface 1 is called the an
20、gular coefficient of the surface 1 pairs and the non surface 2. The angular coefficient is purely geometric, and depends only on geometrical properties (shape, size, and relative position of the object). (2 points)The properties of angular coefficients include interchangeability (1 points)Integrity
21、(1 points)Separable (additive) (1 points)Four, calculation questions (total 55 points)1, as shown on the right. The air in the container A is partially withdrawn and the lower end is placed in the tank. Because the container A is negative, the water rises along the glass tube height of h=0.5 meters,
22、 the density of water is 1000 kg / m 3. If the local atmospheric pressure is 1.013 x 105Pa, the absolute pressure PA = (5 points) in the vessel A is testedSolution: the absolute pressure PA in the container A should satisfy the following relations:Among them, the atmospheric pressure (2 points)(3 po
23、ints)Answer: the absolute pressure in the container A is 96400Pa.2, as shown in the figure of variable diameter pipe section AB, known dA=0.2m, dB=0.4m, height difference h=1.0m, pressure gauge measured pA=7 x 104Pa, pB=4 x 104Pa, flow Q=10m3/min. Try to determine the flow direction of the water in
24、the pipe and calculate the head loss of the AB section. (10 points)Solution: first, the average velocity of water in the pipe is calculated(2 points)The continuity equation and the diameter of A and B can be seen:(1 points)Based on the 1-1 section, the total head of A and B two is calculated by the
25、surface pressure.(2 points)(2 points)Comparing the total head of A and B two, we can see that the direction of water flow is from A to B. (1 points)The head loss of AB section is hAB:(2 points)3, the level of pipeline inside diameter of 100 mm to transport oil density is 915kg/m3, the transmission d
26、istance is 16 kilometers, the oil viscosity is 1.86 * 10-4m2/s, calculate the flow of 50 tons / hour head loss along the way and the power consumed. (10 points)Solution: first, calculate the volume flow and mean velocity(2 points)(1 points)The oil flows into the laminar flow in the oil pipeline. (2
27、points)Head loss along the way: (3)Power consumption: (2 points)A: the head loss is about 188.52 meters, and the power consumption is about 25.7kW.4. A flat wall with a thickness of 20mm, whose thermal conductivity is 1.3w/ (M? C). For every square meters of wall loss less than 1500W and the outside
28、 surface of the wall is covered with a layer of lambda 0.1w/ (M? C) of the composite insulation materials, known both sides of the wall surface temperature were 750 degrees and 55 degrees, determine the thickness of thermal insulation layer. (10 points)Solution: This belongs to multi-layer flat wall
29、 heat conduction problem, according to the known, q = 1500W/m2; 1=1.3W/ lambda (M? C),Delta 1=20mm; lambda 2=0.1W/ (M? C), requiring lambda 2=?The calculation formula of heat flux is given according to the heat conduction of multilayer flat wall (2 points)So (2 points)(5 points)Answer: the insulatio
30、n layer thickness must be greater than or equal to 45mm. (1 points)5, as shown in the figure,There is a hemispherical container with a radius of r=1m. The surface area of the bottom of the blackbody is 1 of the blackbody surface and 200 of the blackbody surface with a temperature of 40. They are 2 o
31、f the area of the bottom circle, and 3 of the vessel wall is an adiabatic surface. The radiation heat exchange between the surface 1 and the surface 2 is calculated. (10 points)Solution: the area of the surface 1 is F1, and the area of the surface 2 is F2. So F1=F2=3.14/2 m2. (1 points)Because the F
32、1 and F2 are on the same plane, the angular coefficient X12=X21=0 (1)Because the surface 1 and the surface 2 are wrapped by the surface 3, the angular coefficient X13=X23=1 (2 points)Because the surface 3 is an adiabatic heavy radiation surface, so (3 points)(2 points)Answer: the radiation heat exch
33、ange between the surface 1 and the surface 2 1800W. (1 points)In Experiment 6, using hot wire anemometer flow velocity, the electric wire 0.1mm and flow direction perpendicular to the diameter of the steady state, the measured flow temperature is 25 DEG C, wire temperature is 55 DEG C, the electric
34、power consumption of the wire is 20 W / m. It is assumed that except for convective heat transfer, other heat losses can be neglected, and the velocity v = = (10) is determined at this pointNote: the criterion relation of the average surface heat transfer coefficient is known as:.The physical proper
35、ties of the air and the relation between C, N and Re in the relation formula can be seen in the following table.The relationship between C, N and ReThermophysical properties of dry airSolution to the problem of forced convection heat transfer in a circular cylinder over the air. By the l=20W/m, acco
36、rding to Newton cooling formulaL=hA (tw-tf) =h PI D (tw-tf) (1 points)The solution is h= L / PI D (tw-tf) =20/PI x 0.1 x 10-3 x (55-25)=2122 W/ (M2? K) (1 points)The qualitative temperature was tm= (tw-tf) /2= (55+25) /2=40 (1)Physical air value: lambda = 0.0276W/ (M? K); V =16.96 * 10-6m2/s; Pr=0.699 (1 points)So (2 points)Suppose the Re number is 404000, and C=0.683, n=0.466, can be found in the table above, so.(2 points)That is, Re=233.12 is in line with the above assumptions. (1 points)* (1 points)Answer: at this point the velocity of airflow is u =39.54m/s.