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schepotkina [342]
3 years ago
15

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

s.
Engineering
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by  \frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}

So availability of system =\frac{900}{900+100}=\frac{900}{1000}=0.9

So the availability of system will be 0.9

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The component of a regenerative vapor power cycle that permits only liquid to pass through to a region of lower pressure is a Valve/trap.

<h3>What is vapor?</h3>
  • In physics, a substance in the gas phase at a temperature lower than its critical temperature is referred to as a vapor or vapor.
  • This means that the vapor can be condensed to a liquid by increasing the pressure on it without decreasing the temperature.
  • An aerosol is distinct from a vapor.
  • A suspension of minute liquid, solid, or both types of particles in a gas is known as an aerosol.
<h3>Why does vapor form?</h3>
  • Evaporation or sublimation are two processes that can be used to create it.
  • Unlike clouds, fog, or mist, which are only suspended drops of liquid water in the atmosphere, watevaporur is a gas and cannot be seen.
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Learn more about vapor here:

brainly.com/question/14578189

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3 0
2 years ago
The ________ is the part of the drill press that holds and rotates the cutting tool.
lana66690 [7]

Answer:

Spindle

Explanation:

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8 0
4 years ago
Read 2 more answers
I will mark as brainliest !
Sliva [168]

Answer:

7.8 Mph

Explanation:

Rate of cycling = 1.1 rev/s

Rear wheel diameter = 26 inches

Diameter of sprocket on pedal = 6 inches

Diameter of sprocket on rear wheel = 4 inches

Circumference of rear wheel =  \pi d=26\piπd=26π

Speed would be

\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s

Converting to mph

1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph

134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph

The Speed of the bicycle is 7.8 mph

3 0
3 years ago
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its
kow [346]
<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil (N_{p}) to the number of turns in its secondary coil (N_{s}) is equal to the ratio of the input voltage (V_{p}) to the output voltage (V_{s}) of the transformer. i.e

\frac{N_p}{N_s} = \frac{V_p}{V_s}            ----------------(i)

<em>From the question;</em>

N_{p} = number of turns in the primary coil = 8 turns

N_{s} = number of turns in the secondary coil = 515 turns

V_{p} = input voltage = 117V

<em>Substitute these values into equation (i) as follows;</em>

\frac{8}{515} = \frac{117}{V_s}

<em>Solve for </em>V_{s}<em>;</em>

V_{s} = 117 x 515 / 8

V_{s} = 7532V

Therefore, the output voltage (in V) of the transformer is 7532

6 0
3 years ago
You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your
olchik [2.2K]

Solution:

The given formula,

x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}

y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}

\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}

From the table,

1) \(\quad x=2 \cdot 87, \quad y=0.96\)\\\(\frac{x}{y}=\frac{2187}{0.96}\)22895\\\\2) \(x=3 \cdot 466 \cdot y=6 \cdot 998\)\\\(\frac{x}{y}=\frac{3 \cdot 466}{0.898}\)\(=3 \cdot 4729\)

3\(x=3 \cdot 229, y=1 \cdot 08\)\\\(\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}\)\\=2.9898\)\\\\4) \(x=6.525, y=1.094\)\\\(\frac{x}{y}=\frac{5.625}{1.094}\)\\=5.0502

8 0
4 years ago
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