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lesantik [10]
2 years ago
6

In order to lift a lighter object on the other side, a boy placed 155 N of

Engineering
1 answer:
const2013 [10]2 years ago
4 0

Answer:

0.58 m

Explanation:

In order to find the vertical distance the load lifts, we first need to find the magnitude of the resistance force. So, Rd = Fd' where R = resistance force = ?, d = resistance distance = 3.5 m, F = effort force = 155 N and d' = effort distance = 1.5 m.

So R = Fd'/d = 155 N × 1.5 m/3.5 m = 232.5 Nm/3.5 m = 66.43 N  

Now, the vertical work done by the effort = vertical work done by load

Fy = Ry' where y = vertical distance moved by effort = 0.25 m and y' = vertical distance moved by resistance force

So, Fy = Ry'

y' = Fy/R

= 155 N 0.25 m/66.43 N

= 38.75 Nm/66.43 N

= 0.58 m

So, the vertical distance the lighter object moves is y' = 0.58 m

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Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

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Energy Balance

-q_{loss} - w_{out} + h_{in} - h_{out} = 0

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\nu_{in} = 0.055665\,\frac{m^{3}}{kg}

h_{in} = 3650.6\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mix)

\nu_{out} = 5.89328\,\frac{m^{3}}{kg}

h_{out} = 2500.2\,\frac{kJ}{kg}

a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

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\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

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Answer:

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