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schepotkina [342]
3 years ago
6

3. A particle is projected to the right from the position S = 0, when an initial velocity of 8 m/s. If the acceleration of the p

article is defined by the relation a = -0.5 v3/2, where a in m/s2 and v in m/s. Determine a) the distance the particle will have traveled when its velocity is 5 m/s b) the time when v = 1m/s c) the time require for the particle to travel 8m
Engineering
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

a) 3.5 m

b) 14 secs

c) 1.4 secs

Explanation:

<u>a)  Determine the distance the particle will travel</u>

given velocity ( final velocity ) = 5 m/s

v^2 = u^2 + 2as

s = ( v^2 - u^2 ) / 2a

  = ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )

  = 3.5 m

<u>b) Determine the time when v = 1m/s</u>

V = u + at

1 = 8 + (  -0.5 * 1^3/2 ) * t

∴ t = 14 secs

c) Determine the time required for particle to travel 8 m

<em>we will employ both equations above </em>

V^2 = u^2 + 2as

s = 8 m , V = unknown , u = 8 m/s   back to equation

V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8

∴ V^2 + 8V^3/2 - 64 = 0

resolving the above equation

V = 3.478 m/s

now using the second equation

V = u + at

3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t

hence : t = 1.4 secs

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4.In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is gener
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Answer:

The rate of irreversible loss will be "55.22 MW".

Explanation:

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Flow of water,

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Now,

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The power generated by turbine will be:

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On substituting the values, we get

⇒     =1000\times 9.8\times 100\times 120

⇒     =117.72 \ MW

Power generated in actual will be:

= \frac{50}{0.8}

= 62.5 \ MW

Hence,

Throughout the piping system,

The rate of irreversible loss is:

= Power \ generated \ by \ turbine-Power \ generated  \ in \ actual

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A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an
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Answer:

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Explanation:

The given values are:

P_4=50\ kPa

h_4=0.7(2304.7)+340.5

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and,

P_3=15 \ mPa

h_3=hg

    =2610.8 \ KJ/Kg

s_3=sg

    =5.3108 \ KJ/Kgh

At 45,

⇒ x_{45} = \frac{5.3108-1.0912}{6.5019}

          =0.66

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or,

V_f=0.001030 \ m^3/Kg

then,

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        =355.94 \ kJ/kg

hence,

The isentropic efficiency of turbine will be:

⇒ n_T=\frac{h_3-h_4}{h_3-h_{45}}

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         =84.818 (%)

The thermal efficiency of cycle will be:

⇒ n_C=\frac{W_T-W_P}{2_{in}}

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         =28.45 (%)  

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