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bixtya [17]
3 years ago
14

A cog system on the beginning segment of a roller coaster needs to get 29 occupied cars up a 120-m vertical rise over a time int

erval of 60 s. Each car experiences a gravitational force of 5900 N. The cars start at rest and end up moving at 0.50 m/s.
Physics
1 answer:
Rudik [331]3 years ago
8 0

Answer:

Hello your question has some missing part attached below is the missing part

How much work is done by the cog system on the cars?

<em>answer</em> : 2.053 * 10^7 J

Explanation:

Total weight of car ( wT )  = 5900 * 29 = 171100 N

mass of car ( mT ) = wT / g  = 171100 / 9.81 = 17459.18 kg

Hence <u>work done by the cog system on the cars</u>

W = ( KE_{2} + PE_{2} ) - ( KE_{1} + PE_{1} )

    = ( 1/2 mv^2 + mgh ) - ( 0 )

    = ( 1/2 * 17459.18 * 0.50^2) + ( 17459.18 * 9.81 * 120 )

    = 2.053 * 10^7 J

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Saturated ethylene glycol at 1 atm is heated by a chromium-plated surface which is circular in shape and has a diameter of 200-mm and is maintained at 480 K.  

At 470 K the properties of the saturated liquid are mu = 0.38 * 10-3 N. s/m^2, Cp = 3280 J/kg. K and Pr = 8.7. The saturated vapour density is p= 1.66 kg/m^3. Take the liquid to surface constants to be Cnb = 0.010 and m=4.1.  

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The heating power requirement = 559.2 W

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The fraction of the power requirement  of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

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From the thermodynamics tables; we deduced the value for enthalpy at the pressure 1 atm and T_{sat} = 470 K   for the saturated ethylene glycol.

Value for enthalpy of formation h_{fg} = 812 kJ/kg

Density of saturated ethylene glycol \rho___l = 1111 kg/m³

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q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2}  [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3

= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )]

= 308.56 × 576.6 × 0.1

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q_{boil} = q__s }*A S

= 1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2

Thus, the heating power requirement = 559.2 W

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m= \frac{q_{boil}}{h_[fg}}

= \frac{559.2}{812*10^3}

= 6.89*10^{-4}kg/s

Thus, the rate of evaporation = 6.89*10^{-4}kg/s

To determine to what fraction in the power requirement of the maximum power is associated with the critical total flux ; we needed to first calculate the critical heat flux.

So, the  calculation for the critical heat is given as:q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}

= q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}

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= \frac{q''s}{q''max}

= \frac{1.78*10^4}{6.77*10^5}

= 0.026

Thus, the fraction of the power requirement  of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

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