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homework_27.1_sol

# homework_27.1_sol - Solution to Homework 27.1 22.5...

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Solution to Homework 27.1 22.5 Efficiency of this unit is 98%, so the electrical power required is: 98% electrical heat p Q Q VC T r = = D & . The volume flux is: 3 3 3 4 2 gal 3.7854 10 m 1min m 1.2618 10 min gal 60 s s V - - - = = & and the temperature difference is: 5 (120 60) 60 33.3 9 o o T F C = - = = = D constant-pressure heat capacity is: o o cal 4.184 J 1000 g 4184 J 1 g C cal 1 kg kg C p C = = so the electrical power required is: 3 3 4 o 3 o 98% kg m 4184 J 1 10 1.2618 10 33.3 C m s kg C 0.98 17580 W 17939 W 17.9 kW 0.98 p electrical VC T Q r - D = ° - = - & 22.6 Efficiency of this unit is 70%, so the natural gas burning energy required is: 70% NG heat Q Q = . From the result in (22.5), 17580 W heat Q = , and the natural gas burning energy is: 55.509 MJ ˆ kg NG methane Q m H m - =- =- - - D & & So the mass consumption rate is: ( ) m 6 m ˆ 70% 2.205 lb 17580 W 3600 s 0.7 55.509 10 J/kg kg hr 3.59 lb /hr heat methane Q m H = - � D = - - &

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22.9 traveling miles total combustion energy required energy fuel economy = tank volume tank volume mile specific heat of combustion total mass required energy tank volume mile specific heat of combustion densit = - = = required energy y mile o
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homework_27.1_sol - Solution to Homework 27.1 22.5...

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