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2 years ago

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

Engineering

1 answer:

Gelneren [198K]2 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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2 years ago

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2 years ago

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Gre4nikov [31]

Answer:

Option A, World War II

Explanation:

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Hence, option A is the right answer

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2 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}$\\=5.0502$

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3 years ago

Steam enters a turbine operating at steady state at 2 MPa, 323 °C with a velocity of 65 m/s. Saturated vapor exits at 0.1 MPa an

Lera25 [3.4K]

Answer:

$\dot Q_{out} = 13369.104\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0$

The rate of heat transfer between the turbine and its surroundings is:

$\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}$

The specific enthalpies at inlet and outlet are, respectively:

$h_{in} = 3076.41\,\frac{kJ}{kg}$

$h_{out} = 2675.0\,\frac{kJ}{kg}$

The required output is:

$\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW$ $\dot Q_{out} = 13369.104\,kW$

4 0

3 years ago