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Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

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Answer: A maximum of 1 hour

Explanation:

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if both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

egoroff_w [7]

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.

<h3>What is a ram air input?</h3>

A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.

This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.

In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.

Read more on pilots here: brainly.com/question/10381526

#SPJ1

Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

a) increase during a climb

b) decrease during a climb

c) remain constant regardless of altitude change

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2 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using

kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

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3 years ago