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hjlf
4 years ago
7

The correct statement about the relation between the shear stress (force) at the surface for laminar and turbulent flow over a f

lat plate is. a.The laminar shear stress is greater because the fluid viscosity is not important in turbulent flow. b.The turbulent shear stress is greater because the mixing dissipates more energy, producing a greater stress at the wall. c.The laminar and turbulent shear stresses are equal at the same position (x) along the plate, but since turbulence occurs farther down the plate it produces a higher shear stress. d.The laminar shear stress is less because the boundary is thinner in the laminar regime.
Engineering
1 answer:
-Dominant- [34]4 years ago
7 0

Answer:

b.The turbulent shear stress is greater because the mixing dissipates more energy, producing a greater stress at the wall.

Explanation:

For a fluid, shear stress is related to the velocity gradient (derivative), not the absolute velocity.

Following the Poiseuille model, flow profile will appear in a channel. In that case, the flow velocity is maximum in the center, but the gradient will be zero in the center. At the wall, the flow will be zero, but the gradient will have a maximum value, therefore maximum shear stress.

Since stress depends highly on the velocity gradient near to the wall of the plate, and turbulence occurs in very high velocity flows, with high Reynold's number, then the turbulent shear stress is greater because the mixing dissipates more energy, producing a greater stress at the wall.

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(a) The maximum volume flow rate for which the flow will be laminar is 0.0190 cubic meter per second

(b) The pressure drop required to deliver the maximum flow rate is 148962.96 Pascal

(c) The corresponding wall shear stress is 7600 Pascal

Explanation:

Reynolds number = 2299, density of water = 1000kg/m^3, diameter of needle = 0.27mm = 0.00027m, Length of needle = 50mm = 0.05m, viscosity of water = 0.00089kg/ms, area = 0.05m × 0.05m = 0.0025m^2, coefficient of friction = 64 ÷ Reynolds number = 64 ÷ 2299 = 0.028

Velocity = (Reynolds number × viscosity) ÷ (density × diameter) = (2299 × 0.00089) ÷ (1000 × 0.00027) = 2.046 ÷ 0.27 = 7.58m/s

(a) Maximum volume flow rate = velocity × area of needle = 7.58 × 0.0025 = 0.0190 cubic meter per second

(b) Pressure drop = ( coefficient of friction × length × density × velocity^2) ÷ (2 × diameter) = (0.028 × 0.05 × 1000 × 7.58^2) ÷ (2 × 0.00027) = 80.44 ÷ 0.00054 = 148962.96 Pascal

(c) Wall shear stress = (density × volume flow rate) ÷ area = (1000 × 0.0190) ÷ 0.0025 = 7600 Pascal

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3 years ago
In a tempering process, glass plate, which is initially at a uniform temperature Ti, is cooled by suddenly reducing the temperat
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Answer:

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Explanation:

Given that:

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(m_{1}, m_{2}, m_{3}) = (3, -6, 3)

<u>Explanation:</u>

Given data,

p=m_{1} x^{2}+m_{2} x+m_{3}

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0(P(0)), 1(P(1)), 2(P(2)), 3(P(3)), 4(P(4))

Bob receives are

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P(0) = 3

P(1) = 0

P(2) = 3

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The given equation is

p=m_{1} x^{2}+m_{2} x+m_{3}

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p(0)=m_{1}(0)+m_{2}(0)+m_{3} = 3

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4 m_{1}+2 m_{2} = 0

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P(3)=9 m_{1}+3 m_{2}+m_{3}  = 0

3 m_{1}+m_{2}+3=0

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m_{2} = -6

(m_{1}, m_{2}, m_{3}) = (3, -6, 3)

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