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vazorg [7]
2 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]2 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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You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
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Answer:

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Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

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inductance of solenoid

L = N²μA / l

A = πd²/4  

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L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

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L = 6 mH

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b)

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Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

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