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zavuch27 [327]
3 years ago
10

A large plate is at rest in water at 15?C. The plate is suddenly translated parallel to itself, at 1.5 m/s. The resulting fluid

movement is not exactly like that in a b.l. because the velocity profile builds up uniformly, all over, instead of from an edge. The governing transient momentum equation, Du/Dt = ?(?2u/?y2), takes the form 1 ? ?u ?t = ?2u ?y2
`Determine u at 0.015 m from the plate for t = 1, 10, and 1000 s. Do this by first posing the problem fully and then comparing it with the solution in Section 5.6. [u 0.003 m/s after 10 s.]
Engineering
1 answer:
ratelena [41]3 years ago
5 0

Answer:

Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s

   

Explanation:

du/u² = dt = ∫du/2.3183 = ∫dt

0.4313 u = t + c

(a) t = 0, u= 15m/s, c = 0.647

u = t+c/0.4313 = t + 0.647/0.4313

(a) when t= 1   u = 1+ 0.647/0.4313 = 3.8187m/s

(b) when t= 10   u = 10 + 0.647/0.4313 = 24.0858m/s

(c)when t= 1000  u = 1000 + 0.647/0.4313 = 3220.071m/s

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Here: \gamma is the specific heat ratio. Substitute the values in above equation.

\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}}  (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm

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Calculate the temperature at throat.

\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K

Calculate the velocity at exit.  

V*=M*\sqrt{ \gamma R T*}

Here: R is the gas constant.  

V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}

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Calculate the density of air at throat using isentropic flow equation.  

\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}

Calculate the mass flow rate.  

m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}

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