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HACTEHA [7]
3 years ago
5

If a ball is dropped from a height​ (H) its velocity will increase until it hits the ground​ (assuming that aerodynamic drag due

to the air is​ negligible). During its​ fall, its initial potential energy is converted into kinetic energy. If the ball is dropped from a height of 720 centimeters​ [cm], and the impact velocity is 39 feet per second​ [ft/s], determine the value of gravity in units of meters per second squared ​[m/s2​].
Engineering
1 answer:
enot [183]3 years ago
4 0

Answer

Explanation:

so the velocity is 39 feet per sec so the impact is 720 cm from the ground

take 720 * 39 sq

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Answer:

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Explanation:

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The constant pressure specific heat of air, c_{p} = 1.005 kJ/kg -K

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Using the steady flow energy equation:

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a) For the isentropic process:

Power developed by the turbine is given by the relation \dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})

Isentropic efficiency, \eta_{t} = 80%

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, T_{3s} = ? where T_{3s} is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }

T_{3s} = 230.9 K

To get the temperature at 80% efficiency, we will use the relation:

\eta_{t} = \frac{T_{2} - T_{3}  }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3}  }{300 - 230.9 }

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Power developed by the turbine is given by the relation:

\dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW

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