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vaieri [72.5K]
3 years ago
14

Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s

Engineering
1 answer:
zlopas [31]3 years ago
5 0

Answer:

4m/s

Explanation:

We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted

Mathematically

Power_{motor} } =\frac{Energy }{time}\

We also know that Power = force x velocity      ..................(i)

The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load

=> power = mg x Velocity........(ii)

While hoisting the load at at constant speed only the potential energy of the mass increases

Thus Potential energy = Mass x g x H...................(iii)

where

g = accleration due to gravity (9.81m/s2)

H = Height to which the load is hoisted  

Equating equations (ii) and (iii) we get

m x g x v = \frac{mgh}{t}

thus we get v = H/t

Applying values we get

v = 6/1.5 = 4m/s

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Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ  F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ  F$

Change in temperature, $\Delta T = T_2-T_1$

                                         $\Delta T = -2.43-87.745$

                                           $\Delta T=-90.175^\circ  F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

   $=11.401+0.32198(82.898)$

   $=38.09297 \ Btu/lbm$

Change in internal energy  

$\Delta u= u_2-u_1$

   = 38.09297-40.5485

  = -2.4556        

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Answer:

increases by a factor of 6.

Explanation:

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8 0
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Answer:

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