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3 years ago

State the mathematical expression to define the availability of equipment over a specified time and operational availability?

Engineering

1 answer:

Gelneren [198K]3 years ago

7 0

Answer:

$Availability=\dfrac{Up\ time}{Down\ time+Up\ time}$

Explanation:

Availability:

It define as the probability of system which perform desired task before showing any failure .

The availability can be define as follows

$Availability=\dfrac{Up\ time}{Down\ time+Up\ time}$

Or we can say that

$Availability=\dfrac{Up\ time}{total\ time}$

Availability can also be express as

$Availability=\dfrac{MTBF}{MTBF+MTTR}$

Where MTBF is the mean time between two failure.

MTTR is the mean time to repair.

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Can someone explain it to me!

Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0

3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0

3 years ago

A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surfac

UkoKoshka [18]

Answer:

864 KN

Explanation:

Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.

Please kindly check attachment for the step by step solution of the given problem.

3 0

3 years ago

HW6P2 (20 points) The recorded daily temperature (°F) in New York City and in Denver, Colorado during the month of January 2014

Maurinko [17]

Answer & Explanation:

function Temprature

NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];

DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];

%AVERAGE CALCULATION AND ROUND TO NEAREST INT

avgNYC=round(mean(NYC));

avgDEN=round(mean(DEN));

fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);

fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);

%part B

count=1;

NNYC=0;

NDEN=0;

while count<=length(NYC)

if NYC(count)>avgNYC

NNYC=NNYC+1;

end

if DEN(count)>avgDEN

NDEN=NDEN+1;

end

count=count+1;

end

fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);

fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);

%part C

count=1;

highDen=0;

while count<=length(NYC)

if NYC(count)>DEN(count)

highDen=highDen+1;

end

count=count+1;

end

fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);

end

%output

check the attachment for additional Information

8 0

4 years ago

If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total ac

rjkz [21]

Answer:

8.62m/s²

Explanation:

the particle is experiencing both translational and circular motion

v=4t²

$a_{t}$ = $\frac{dv}{dt}$ =8t

at t=1s, $\frac{dv}{dt}$ =8(1)=8m/s²

$a_{c}$ = $\frac{v^{2} }{r}$

at t=1, v= 4(1)² = 4m/s

$a_{c}$ =4²/5

$a_{c}$ =3.2m/s²

∴ magnitude of total acceleration, a

a= $\sqrt{a_{t} ^{2} + a_{c} {2} }$

a= $\sqrt{8^{2} +3.2^{2} }$

a= $\sqrt{64+10.24}$

a= $\sqrt{74.24}$

a=8.62m/s²

5 0

3 years ago

Read 2 more answers