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2 years ago

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Consider two different flows over geometrically similar airfoil shapes, one airfoil being twice the size of the other. The flow

over the smaller airfoil has freestream properties given by T[infinity] = 200 K, rho[infinity] = 1.23 kg/m3 , and V[infinity] = 100 m/s. The flow over the larger airfoil is described by T[infinity] = 800 K, rho[infinity] = 1.739 kg/m3, and V[infinity] = 200 m/s. Assume that both μ and a are proportional to T1/2. Are the two flows dynamically similar?

1 answer:

yulyashka [42]2 years ago

5 0

Answer

No;

The two flows are not dynamically similar

Explanation: Given

T∞,1 = 800k

V∞,1 = 200m/s

p∞,1 = 1.739kg/m³

T∞,2 = 200k

V∞,2 = 100m/s

p∞,2 = 1.23kg/m³

Size1 = 2 * Size2 (L1 = 2L2) Assumptions Made

α ∝√T

μ∝√T Two (2) conditions must be met if the two flows are to be considered similar.

Condition 1: Similar Parameters must be the same for both flows

Condition 2: The bodies and boundaries must be genetically true. Condition 2 is true

Checking for the first condition...

Well need to calculate Reynold's Number for both flows

And Check if they have the same Reynold's Number Using the following formula

Re = pVl/μ

Re1 = p1V1l1/μ1 Re2 = p2V2l2/μ2 Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2

Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2

Re1/Re2 = p1V1l1μ2/p2V2l2μ1

Re1/Re2 = p1V1l1√T2 / p1V1l1√T1

Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)

Re1/Re2 = 9837.2/3479

Re1/Re2 = 2,828/1

Re1:Re2 = 2.828:1

Re1 ≠ Re2,

So condition 1 is not satisfied Since one of tbe conditions is not true, the two flows are not dynamically similar

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An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs

creativ13 [48]

Answer:

ΔKE=-347.278 kJ

ΔPE= 441.45 kJ

Explanation:

given:

mass m=900 kg

the gravitational acceleration g=9.81 m/s^2

the initial velocity $V_{1}$ =100 km/h-->100*10^3/3600=27.78 m/s

height above the highway h=50 m

h1=0m

the final velocity $V_{F}$ =0 m/s

<u>To find:</u>

the change in kinetic energy ΔKE

the change in potential energy ΔPE

<u>assumption:</u>

We take the highway as a datum

<u>solution:</u>

ΔKE=5*m*( $V_{F}$ ^2- $V_{1}$ ^2)

=-347.278 kJ

ΔPE=m*g*(h-h1)

= 441.45 kJ

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2 years ago

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt

= $\frac{190}{0.0003317+0.0000458+0.0110+0.0004976}$ watt

q = 15999.86 watt

The total amount of daily heat transfer = 15999.86 × 86400

= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = $\frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}$

15999.86 $=\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}$

$T_{3}$ - 10 = 7.96

$T_{3}$ = 17.96 ° C.

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2 years ago

A water initially contains 40 mg · L−1 of Mg2+. The pH of the water is increased until the concentration of hydroxide ion (OH−)

Darya [45]

Answer:

concentration of Mg ion = 0.0122 g/L

Explanation:

Given data;

initial concentration of Magnesium in water is 40 mg/l

concentration of $(OH^-) = 0.001000$

we have dissociation reaction Magnesium dioxide

$Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^{-}$

from above reaction we can conclude

concentration of $Mg(OH)_2 = \frac{OH}{2} = \frac{.0010}{2} = 0.0005 M$

Mass of magnesium ion is calculated as = Mg mole * molar mass of magnesium

concentration of Mg ion = 0.0005*24.305 g/mol = 0.0122 g/L

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2 years ago

. A belt drive is desired to couple the motor with a mixer for processing corn syrup. The 25-hp electric motor is rated at 950 r

forsale [732]

Answer:

Hello the table which is part of the question is missing and below are the table values

For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1

Answers:

belt size = 140 in with diameter of 20.1n

actual speed of belt = 288.49 in/s

actual center distance = 49.345 in

Explanation:

Given data :

Electric motor (driver sheave) speed (w1) = 950 rpm

Driven sheave speed (w2) = 250 rpm

pick D1 ( diameter of driver sheave) = 5.8 in ( from table )

To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first

VR = $\frac{w1}{w2}$ = 950 / 250

also since the speed of belt would be constant then ;

Vb = w1r1 = w2r2 ------- equation 1

r = d/2

substituting the value of r into equation 1

equation 2 becomes : $\frac{w1}{w2} = \frac{d2}{d1}$ = VR

Appropriate belt size ( d2) can be calculated as

d2 = $\frac{w1d1}{w2}$ = $\frac{950 * 5.8}{250}$ = 22.04

From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value

next we have to determine the belt length /size

$L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}$

inputting all the values into the above equation including the value of C as calculated below

L ≈ 140 in

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we use this equation to get the ideal center distance

$d2< C_{ideal} < 3( d1 +d2)$

22.04 < c < 3 ( 5.8 + 20.1 )

22.04 < c < 77.7

the center distance is between 22.04 and 77.7 but taking an average value

ideal center distance would be ≈ 48 in

To calculate the actual center distance we use

$C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}$ -------- equation 3

B = $4L -2\pi (d2 + d1 )$

inputting all the values into (B)

B = 140(4) - 2 $\pi$ ( 20.01 + 5.8 )

B ≈ 399.15 in

inputting all the values gotten Back to equation 3 to get the actual center distance

C = 49.345 in ( actual center distance )

Calculating the actual belt speed

w1 = 950 rpm = 99.48 rad/s

belt speed ( Vb) = w1r1 = w1 * $\frac{d1}{2}$

= 99.48 * 5.8 / 2 = 288.49 in/s

3 0

2 years ago

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Bas_tet [7]

Answer:

The correct code is given below:-

print("Predictions are hard.")

print("Especially about the future.")

user_num = 5

print("user_num is:", user_num)

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