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3 years ago

List the three main methods employed in dimensional analysis

Engineering

1 answer:

valina [46]3 years ago

8 0

Answer:

Rayleigh method, The Buckingham π Theorem, common pi group

Explanation:

Rayleigh method - The Rayleigh method is a basic technique for discovering a functional relationship in reference to a parameter of interest.

The dimensional analysis method describes a functional relationship between certain variables in context of an exponential equation

The Buckingham π Theorem The theorem notes that if a factor A1 relies on independent variable i.e. A2, A3,. An, functional bond can then be set to zero in the form f(A1, A2, A3,.–, An)= 0. If all these n parameters can be defined as m dimensional units, the pi (π) theorem states that they are grouped in respect of n-m dimensional.

There are some common pi group method which are based on Buckingham π theorem.

They are named as froude number, Reynold's number, weber number etc.

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it is used to meusure the amount of electric current. A.clamp meter. B .micrometer C steel rule D. electric meter

Marianna [84]

Answer:

Option D 

Electric meter is used to measure the amount of electric current

6 0

3 years ago

End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf

sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = ( $\frac{1}{12}$ ) × m × L² ................2

I = ( $\frac{1}{12}$ ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

total force at A will be

Fa = $\sqrt{a(x)^2+a(y)^2}$ ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$

Fa = 57.32 N

3 0

4 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

Take Re=100,000:

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

Take Re=500,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

Take Re=1,000,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

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3 years ago

Can be used to eliminate rubbing friction of wheel touching frame. 1.Traction 2.Thrust washer

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Answer:

thrust washer

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3 years ago

How does the engine thermostat regulate the engine temperature?

Inga [223]

Answer:

It's job is to block the flow of coolant to the radiator until the engine has warmed up. Once the engine reaches its operating temperature (generally about 200 degrees F, 95 degrees C), the thermostat opens. By letting the engine warm up as quickly as possible, the thermostat reduces engine wear, deposits and emissions.

6 0

4 years ago