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seraphim [82]
2 years ago
3

Water at 25°C and 1 atm is flowing over a long flat plate with a velocity of 27 m/s. Determine the distance from the leading edg

e of the plate where the flow becomes turbulent, and the thickness of the boundary layer at that location. The density and dynamic viscosity of water at 1 atm and 25°C are rho = 997 kg/m3 and μ = 0.891 × 10–3 kg/m·s.

Engineering
1 answer:
Mama L [17]2 years ago
7 0

Answer / Explanation

To properly answer this question, we will be making few assumptions in respect to the question to make the learning process easier.

These assumptions are:

1) We assume that the flow is steady and in-compressible.

2) We assume that the critical Reynolds number is  Recr = 5 × 10⁵

3) The surface of the plate is smooth.

We also assume according to the question that the properties, density and dynamic viscosity of water at 1 atm and 25°C are ρ = 997 kg/m3   and

μ = 0.891 × 10⁻³ kg/m⋅s

We move forward by representing the question in the diagram attached below:

On analyzing the diagram attached below, we see that the distance from the leading edge of the  plate where the flow becomes turbulent is the distance  xcr where the Reynolds number becomes equal to the  critical Reynolds number,

Therefore:

Recalling the Reynolds equation, we have

Recr = ρ VXcr / μ

If we go ahead to make Xcr the subject of the formula, we have:

Xcr = μRecr /  ρV

Now if we insert the values representing the parameter in the equation, we have:

Xcr = ( 0.891 x 10 ⁻³kg/m·s) (5 × 10⁵) / (997 kg/m3 ) (8m/s)

calculating the above further, we derive:0.056 meters

If we convert to centimeter, we then have 5.6 cm.

Thus,

The thickness of the boundary layer at that location is obtained by substituting this value  of x into the laminar boundary layer thickness relation. Therefore, we have:

δcr = 5x / Reₓ⁰.⁵ = δcr = 5xcr / Reₓ⁰.⁵

= 5 (0.56m / 5 × 10⁵)⁰.⁵

= 0.00040 m = 0.4 mm

In respect to the answer gotten for the thickness of the boundary above, we can now say that the flow becomes turbulent after about 5 cm from the leading edge of the  plate and the thickness of the boundary layer at that location is 0.4 mm.  Also, from the calculation thus far, we can clearly state that when the flow becomes turbulent, the boundary layer thickness starts to  increase, and the value of its thickness can be determined from the boundary layer  thickness relation for turbulent flow.

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(b) Change in specific entropy=0.0044\frac{KJ}{Kg-K}

Explanation:

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        h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}

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Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

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