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2 years ago

What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

Engineering

1 answer:

yanalaym [24]2 years ago

7 0

Answer:

$Re=\dfrac{\rho\ v\ l}{\mu }$

Explanation:

Reynolds number:

Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

$Re=\dfrac{F_i}{F_v}$

For plate can be given as

$Re=\dfrac{\rho\ v\ l}{\mu }$

Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

$Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.$

$Reynolds\ number\is \$

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2 years ago

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2 years ago

A driver traveling at 65 mi/h rounds a curve on a level grade to see a truck overturned across the roadway at a distance of 350

horrorfan [7]

Answer:

When the driver first sees the overturned truck, he will continue to move at 65mi/h during this reaction time. During this time, the vehicle will travel 1.47*65*t feet, or 95.5f ft.

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d=(si²-sf²)/30(F +/- 0.01G)

In this case, the initial speed (Si) is 65mi/h. The friction factor, F, is related to the deceleration rate, and is computed by dividing the deceleration rate by the deceleration rate due to gravity, or 10/32.2 =0.31. The grade is level, i.e., G = 0. The braking distance is 350 – 95.5t.

Therefore,

350-95t=(65²-sf²)/(30*0.31)

This equation is solved for various values of t from 0.50 to 5.00 s. Note that at the point where the reaction distance becomes more than 350ft, the final speed is a constant 65mi/h, and the braking distance is essentially “0.”

Check the attachment below for the table and graph.

*COMMENT*

The vehicle is going to hit the overturned truck in any event. For reaction times over approximately 3.75s, the vehicle will hit the truck at full speed, 65mi/h.

6 0

2 years ago

Consider a plane composite wall that is composed of two materials of thermal conductivities kA 0.1 W/mK and kB 0.04 W/mK and thi

Triss [41]

Answer:

(a) 761.9 W

(b) 184.762 °C

(d) see figure

Explanation:

Data

$k_A = 0.1 W/mK$

$k_B = 0.04 W/mK$

$L_A = 0.010 m$

$L_B = 0.020 m$

resistance, $R = 0.30 (m^2 K)/W$

$T_1 = 200 C$

$h_1 = 10 W/m^2 K$

$T_2 = 40 C$

$h_2 = 20 W/m^2 K$

area, $A = 2.5 m \times 2 m = 5 m^2$

(a)

The rate of heat transfer is calculated as

$Q = A \, \frac{1}{R_t} \, (T_1 - T_2) (1)$

Total flux resistance is

$R_t = 1/h_1 + 1/h_2 + L_A/k_A + L_B/k_B + R$

$R_t = (m^2 K)/10 W + (m^2 K)/20 W + 0.010 m (mK)/0.1 W+ 0.020 m (mK)/0.04 W + 0.30 (m^2 K)/W$

$R_t = 1.05 (m^2 K)/W$

From equation 1

$Q = 5 m^2 \, \frac{1}{1.05 (m^2 K)/W} \, (200 - 40) K$

$Q = 761.9 W$

(b)

Between ambient next to material A and material A heat flux is

$Q = A \, h_1 \, (T_1 - T_A)$

$T_A = T_1 - \frac{Q}{A \, h_1}$

$T_A = 200 C - \frac{761.9 W}{5 m^2 \, 10 W/m^2 C}$

$T_A = 184.762 C$

(c)

Between material B and ambient next to material B heat flux is

$Q = A \, h_2 \, (T_B - T_2)$

$T_B = \frac{Q}{A \, h_2}+ T_2$

$T_B = \frac{761.9 W}{5 m^2 \, 20 W/m^2 C} + 40 C$

$T_B = 55.238 C$

(d)

See figured attached

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2 years ago

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Slav-nsk [51]

Answer:

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Explanation:

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6 0

2 years ago