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il63 [147K]
3 years ago
7

Can you determine the critical distance along a flat surface?

Engineering
1 answer:
Keith_Richards [23]3 years ago
6 0

Explanation:

Consider a fluid of density, ρ moving with a velocity, U over a flat plate of length, L.

Let the Kinematic viscosity of the fluid be ν.

Let the flow over the fluid be laminar for a distance x from the leading edge.

Now this distance is called the critical distance.

Therefore, for a laminar flow, the critical distance can be defined as the distance from the leading edge of the plate where the Reynolds number is equal to 5 x 10^{5}

And Reynolds number is a dimensionless number which determines whether a flow is laminar or turbulent.  

Mathematically, we can write,

    Re = \frac{\rho .U.x}{\mu }

or 5 x 10^{5} =  \frac{\rho .U.x}{\mu } ( for a laminar flow )

Therefore, critical distance

x=\frac{5\times 10^{5}\times \mu }{\rho \times U}

So x is defined as the critical distance upto which the flow is laminar.

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An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

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