Two kilograms of air is contained in a rigid wellinsulated tank with a volume of 0.6 m3 . The tank is fitted with a paddle wheel (stirrer) that transfers energy to the air at a constant rate of 10 W for 1h. If no changes in kinetic or potential energy occur, determine a) The specific volume at the final state, in m3 /kg. b) The energy transfer by work, in kJ. c) The change in specific internal energy of the air, in kJ/kg.

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boffeemadrid boffeemadrid · Answer 1 · 2021-02-20T06:33:40+01:00

Answer:

[tex]0.3\ \text{m}^3/\text{kg}[/tex]

[tex]36\ \text{kJ}[/tex]

[tex]18\ \text{kJ/kg}[/tex]

Explanation:

V = Volume of air = [tex]0.6\ \text{m}^3[/tex]

P = Power = 10 W

t = Time = 1 hour

m = Mass of air = 2 kg

Specific volume is given by

[tex]v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}[/tex]

The specific volume at the final state is [tex]0.3\ \text{m}^3/\text{kg}[/tex]

Work done is given by

[tex]W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}[/tex]

The energy transfer by work, is [tex]36\ \text{kJ}[/tex]

Change in specific internal energy is given by

[tex]\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}[/tex]

The change in specific internal energy of the air is [tex]18\ \text{kJ/kg}[/tex]