A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loops use Freon as the working fluid. The instrumentation must be kept as a temperature greater than negative 67−67 degrees Fahrenheit [°F] to avoid damage. The temperature in the area of Mars where the rover is exploring is 189189 kelvins [K]. If the system must remove 47.447.4 British Thermal Units [BTU] of energy, what volume of Freon is needed in units of liters [L]?

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creamydhaka creamydhaka · Answer 1 · 2019-11-14T11:24:54+01:00

Answer:

[tex]V=2.0757\times 10^{-4} \times 1000 = 0.20757 \,liter[/tex]

Explanation:

Given that:

quantity of heat to be removed, [tex]Q= 47.4\,BTU[/tex]

we know,

[tex]1\,BTU= 1055.06\,J[/tex]

density of freon, [tex]\rho=1490\,kg.m^{-3}[/tex]

latent heat of vaporization of freon, [tex]L=161.7\times 10^3 J.kg^{-1}[/tex]

So,

heat to be removed in joules,

[tex]Q=47.4\times 1055.06[/tex]

[tex]Q=50009.844\,J[/tex]

Now, the quantity of freon required to remove the above mentioned heat:

[tex]m=\frac{Q}{L}[/tex]

[tex]m=\frac{50009.844}{161.7\times 10^3}[/tex]

[tex]m=0.3093\,kg[/tex]

Now, volume

[tex]V=\frac{m}{\rho}[/tex]

[tex]V=\frac{0.3093}{1490}[/tex]

[tex]V=2.0757\times 10^{-4}m^3[/tex]

[tex]V=2.0757\times 10^{-4} \times 1000[/tex]

[tex]V = 0.20757 \,liter[/tex]