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2 years ago

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By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer

equations. For steady, incompressible, and two-dimensional flow, neglecting gravity, the result is delta u/ delta x + delta v/ delta y= 0; u delta u/ delta x +v delta u/ delta y= -1/p(delta u/ delta x)+ v delta^2 u/ delta y^2 Use L and V0 as characteristic length and velocity, respectively. Non-dimensionalize these equations and identify the similarity parameters that result.

1 answer:

Mademuasel [1]2 years ago

7 0

Answer: Attached below is the well written question and solution

answer:

i) Attached below

ii) similar parameter = $\frac{V}{VoL } = 1 / Re$

Explanation:

Using ; L as characteristic length and Vo as reference velocity

i) Nondimensionalize the equations

ii) Identifying similarity parameters

the similar parameters are = $\frac{V}{VoL } = 1 / Re$

Attached below is the detailed solution

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The following displacement (cm), output (V) data have been recorded during a calibration of an LVDT. Displacement (cm), output (

nadya68 [22]

Answer:

2.08V/cm

Explanation:

Plot the points on a graph. Draw the line of best fit. Calculate the gradietn of line of best fit.

Attached is the graph plotted on excel.

The equation of the line is

Votlage= 2.08× distance + 0.276

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3 years ago

A cantilever beam AB of length L has fixed support at A and spring support at B.

atroni [7]

The force in the spring will be $F =\dfrac{KPl^3}{3EI}$ .

The deflection of the beam will be $\rho$ = $0.15(\dfrac{KPL^3}{3EI})$

<h3>What is a cantilever beam?</h3>

A rigid, horizontally extending structural member known as a cantilever is supported at only one end. Typically, it extends from a solidly affixed flat vertical surface, such as a wall.

Given that:-

A cantilever beam AB of length L has fixed support at A and spring support at B.
The spring behaves in a linearly elastic manner with stiffness k. If a concentrated load P is applied at B.

The spring force will be calculated as:-

$F = kx$

Deflection will be given by:-

$x = \dfrac{PL^3}{3EI}$

The spring force will be calculated by:-

$F = \dfrac{KPL^3}{3EI}$

The deflection of the beam will be given as:-

$\rho = \dfrac{0.15KPL^3}{3EI}$

Therefore the force in the spring will be $F =\dfrac{KPl^3}{3EI}$ ..The deflection of the beam will be $\rho$ = $0.15(\dfrac{KPL^3}{3EI})$

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1 year ago

3Px=y−y2p2<br><br>first order higher dgree

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Answer:

the order higher is 3p79g5t88=yv5379

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2 years ago

A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa

Fed [463]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

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3 years ago

Read 2 more answers

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

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3 years ago