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Why must fire extinguishers be routinely maintained

sukhopar [10]

Answer:

Fire extinguishers controls the amount of pollution caused by smoke and burning. Although they have no expiring date, they won't last forever. Extinguishers should work for 5 to 15 years, but fire extinguishers be routinely maintained to make sure they remain effective.

Explanation:

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7 0

3 years ago

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What is clearance? What is backlash? What is interference? Explain briefly.

Anton [14]

Explanation:

Clearance:

For easy matching and dis matching of hole and shaft we use size of hole little bit more than the size of shaft and this difference in size is called clearance.

Backlash:

It is the clearance between the two mating gear to avoids failure of gears.Actually when temperature of gears increases then at the same time the size of gear also increases ,due this there is a possibility foe jamming of gears so to avoids this backlash is provides.

Interference:

When two gears are matting then addendum of one gear inters into the deddendum of another gear and due to this gears get jam .This phenomenon is called interference.

5 0

3 years ago

Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre

sergiy2304 [10]

Answer: $\dot m_{in} = 23.942 \frac{kg}{s}$ , $\dot H_{out} = 39632.62 kW$

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to $\dot V = 1 \frac{m^{3}}{s}$ . Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

$\dot m_{in} - \dot m_{out} = 0$

First Law of Thermodynamics

$- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0$

This 2 x 2 System can be reduced into one equation as follows:

$-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0$

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

$\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}$

The mass flow rate can be calculated by using this expression:

$\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}$

$\dot m_{in} = 23.942 \frac{kg}{s}$

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

$h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}$

$h_{out} = 1655.36 \frac{kJ}{kg}$

The enthalpy rate at outflow is:

$\dot H_{out} = \dot m \cdot h_{out}$

$\dot H_{out} = 39632.62 kW$

3 0

3 years ago

1. A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with S = 1.5. The manometer readi

julia-pushkina [17]

Answer:

1) The pressure difference is 4.207 kilopascals.

2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

Explanation:

1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:

$\Delta P = \frac{S\cdot \rho_{w}\cdot g \cdot \Delta h}{1000}$ (Eq. 1)

Where:

$S$ - Relative density, dimensionless.

$\rho_{w}$ - Density of water, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta h$ - Height difference in the U-tube manometer, measured in meters.

$\Delta P$ - Gas pressure difference, measured in kilopascals.

If we know that $S = 1.5$ , $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.286\,m$ , then the pressure difference is:

$\Delta P = \frac{1.5\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.286\,m)}{1000}$

$\Delta P = 4.207\,kPa$

The pressure difference is 4.207 kilopascals.

2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: ( $p = 2.5\,psi$ )