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3 years ago

Describe three differences between liquids and gases in fluid power systems. Help !!!

1 answer:

3 0

Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

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A teenage brain is already fully developed to enable us to manage risks effectively.

Kisachek [45]

Answer:

false

Explanation:

the brain is only really fully devolved by age 26

3 0

2 years ago

Read 2 more answers

A mixing basin in a sewage filtration plant is stirred by a mechanical agitator with a power input/WF L T=. Other parameters de

MakcuM [25]

Answer: π= G[√(u.V/W)]

STEP 1

Given parameters:

Power Input W= FL/T,

Absolute Viscosity u= FT/L²

Basin volume V= V/L³

Velocity gradient G= V/L³

STEP 2

We start by expressing the velocity gradient G as a function of W, u, V

G= G(W,u,V)

To get the pii terms, we use the dimension number formula n=k - r

where n and k are natural numbers representing number of fundamental dimensions and variable present respectively.

n= 4-3=1

STEP 3:

We expressed the pii terms as

π= G.W^a.u^b.V^c

The three fundamental F L T

We can write as

Fⁿ.Lⁿ.Tⁿ= 1/T. (FL/T)^a.(FT/L²)^b.(L³)

Using the exponential rule and by comparing coefficient on both sides;

Fⁿ.Lⁿ.Tⁿ= F^a+b. L^a-2b+3c. T^-a+b-1

Fⁿ= F^a+b = a+b= 0..............I

Lⁿ= L^a-2b+3c=0 = a-2b+3c=0...........ii

Tⁿ=L^-a+b-1=0. -a+b-1=0............iii

From the above equations we have,

a+b =0: b=-a...........iv

putting eq. iv into iii , we have

-a-a-1=0: -2a-1=0: a= -1/2

substituting the above value of a into eq iv, we have

b= 1/2

substituting the value of b above into eq 2, we have,

-1/2-2(1/2)+3c=0

c=1/2.

Lastly, from the pii terms given above we can obtain dimensionless relationship,

π=G(W^-1/2.u^1/2.V^1/2)

We can write this as

π= G[ √1/W.√u. √1/2] = G[(√u.V/√W)] or G[√(u.V/W)].... final answer.

5 0

3 years ago

Ayo, how do I change my username on here?

nydimaria [60]

Answer:

I'm not sure

Explanation:

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4 0

3 years ago

An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th

sp2606 [1]

Answer:

- 14.943 W/m^2K ( negative sign indicates cooling )

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

Calculate the overall heat loss coefficient

Note : heat lost by water = heat loss through convection

m*Cp*dT = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp ( integrate the relation )

In ( $\frac{49-8}{60-8}$ ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

= - 14.943 W/m^2K ( heat loss coefficient )

7 0

2 years ago

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

Nata [24]

Explanation:

For series

$\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)$

$$I=I_{1}=I_{2}=I_{n}$ (current is the same)$

$V=I R(\text {voltage is directly proportional to } R)$

$R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }$

For parallel

$\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })$

$I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})$

$I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)$

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}$

3 0

3 years ago