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论文作者:bluesky论文属性:课程作业 Coursework登出时间:2013-03-08编辑:bluesky点击率:4106
论文字数:1722论文编号:org201303080906554522语种:英语 English地区:英国价格:$ 33
摘要:本文是关于喷气实验课程作业,可以详细的看出整个实验的过程及相关的精准数据。可以做为研究实验的一个标本参考,喷嘴压力比的设计如何形也有详细的介绍。
Variation of mass flow rate with back pressure
back pressure Pb manometer water levels in inches p1-p2 mass flow rate
0 1.9 435.61 0.000351
50 1.9 435.61 0.000351
100 1.9 435.61 0.000351
150 1.85 423.164 0.000346
200 1.9 435.61 0.000351
250 1.9 435.61 0.000351
300 1.9 435.61 0.000351
350 1.85 423.164 0.000346
400 1.6 360.934 0.00032
450 0.85 174.244 0.000222
500 -0.1 N/A N/A
And additional variables are available given the changes in the density analysis. Generally, conservation of mass conservation equation for incompressible flow can be resolved by considering, as opposed to. In general, includes the principle of conservation of energy. However, this is another variable (temperature), and the fourth equation, therefore, to describe the thermodynamic temperature of the flow that is required (e.g., ideal gas equation of state, etc.) has been introduced.
The identity politics of the density of the fluid density isentropic you, ρ0, the reference value, when you define a way to compress the stream can are stopped if that helps you compare the density. In principle, the static density is 5% or more relative to the change in density, must be analyzed and then compress the fluid flow. Ideal gas specific heat at a rate of 1.4 cases around the Mach number is greater than 0.3 occur. Below this value, however, in certain cases, may depend heavily compressed or the level of accuracy required, should be treated likely.
T = 473(1/5)^(1-1/1.2) = 361.7K
delta(s)/R = ln(p1/p2)(t2/t1)^(gamma/(gamma-1))
delta(s)/R = ln5(361.7/472)^3.5 = 0.671
delta(s) = 192.5J/Kg*K
p2 = p1+ delta(p) = 125kPa
p2/p1 = (Ma^2(gamma-1)/2+1)^(gamma/(gamma-1))=125/105
Ma = 0.634
a=rRT^0.5 = (1.4*287*294.7)^0.5 = 343
u = 217.7 = a*Ma
a1 = sqrt(rRT) = sqrt(1.3*188.95*300) = 271.46
T2/T1 = (1+(gamma-1)*M1^2)/2)/(1+(gamma-1)*M2^2)/2)
T2 = 399
u2 = a2*M2 = 201
Theoretical computation of the maximum mass flow rate
A(throat) = 3.14159*0.00479^2/4 = 1.80202E-05
M(max) = A(throat)*(1.21*287*294.4*1.21*1.4*(5/6)^6)^0.5 = 0.004339863
Conclusions
One of the most common examples of shock waves of compressible flow phenomena. Impact on the thermodynamic properties is characterized by discontinuous change. Pressure wave is suddenly removed from the series together to form a membrane separating two different regions of compression or shock-dimensional flow. This is a technique used to generate shock waves in shock tube is often. Compressible fluid nozzle has an important role in determining behavior. Subsonic and supersonic flows react differently to changes in cross section. Convergence of the duct (small diameter, large diameter in the direction of flow reduction) is increased through experience, the speed of subsonic flow, supersonic flow by redu本论文由英语论文网提供整理,提供论文代写,英语论文代写,代写论文,代写英语论文,代写留学生论文,代写英文论文,留学生论文代写相关核心关键词搜索。