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vazorg [7]
3 years ago
11

Consider two different flows over geometrically similar airfoil shapes, one airfoil being twice the size of the other. The flow

over the smaller airfoil has freestream properties given by T[infinity] = 200 K, rho[infinity] = 1.23 kg/m3 , and V[infinity] = 100 m/s. The flow over the larger airfoil is described by T[infinity] = 800 K, rho[infinity] = 1.739 kg/m3, and V[infinity] = 200 m/s. Assume that both μ and a are proportional to T1/2. Are the two flows dynamically similar?
Engineering
1 answer:
yulyashka [42]3 years ago
5 0

Answer

No;

The two flows are not dynamically similar

 Explanation:  Given

T∞,1 = 800k

V∞,1 = 200m/s

p∞,1 = 1.739kg/m³

T∞,2 = 200k

V∞,2 = 100m/s

p∞,2 = 1.23kg/m³

Size1 = 2 * Size2 (L1 = 2L2)  Assumptions Made

α ∝√T

μ∝√T  Two (2) conditions must be met if the two flows are to be considered similar.

Condition 1: Similar Parameters must be the same for both flows

Condition 2: The bodies and boundaries must be genetically true.  Condition 2 is true

Checking for the first condition...

Well need to calculate Reynold's Number for both flows

And Check if they have the same Reynold's Number  Using the following formula

Re = pVl/μ

Re1 = p1V1l1/μ1  Re2 = p2V2l2/μ2  Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2

Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2

Re1/Re2 = p1V1l1μ2/p2V2l2μ1

Re1/Re2 = p1V1l1√T2 / p1V1l1√T1

Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)

Re1/Re2 = 9837.2/3479

Re1/Re2 = 2,828/1

Re1:Re2 = 2.828:1

Re1 ≠ Re2,

So condition 1 is not satisfied  Since one of tbe conditions is not true, the two flows are not dynamically similar

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Answer:

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The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

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From the question we are given that the diameter of A is = 1 cm = 0.01 m

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  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

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The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

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The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

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         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

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          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

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Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

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