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Unformatted text preview: 1 lesson 6 II. Energy and the First Law of Thermodynamics A. Generic Statement of the First Law for a Closed System = system leaves energy rate system enters energy rate system within energy of change of rate Time System No mass enters or leaves. Heat Work out in system E E E = system in out dE E E dt = Rate form of energy balance: Integrated form of energy balance: E system + E surroundings = 0 Note: (412) (411) lesson 6 II. Energy and the First Law of Thermodynamics B. Specific Statement of First Law for a Closed System 1. Energy is conserved. 2. Energy can cross the boundary of a closed system by only two mechanisms: heat transfer and work transfer. 3. The change in energy of a closed system is equal to the net heat transferred to the system minus the net work performed by the system (417). 2 1 V ( ) 2 E mu m mgz kJ = + + ) kJ ( W Q E net , out net , in = 2 1 V 2 kJ e u gz kg = + + , , (kJ/kg) in net out net e q w = , , (kJ/kg) in net out net de q w = Energy per unit mass: in,net out,net dE Q W (kJ ) = The differential form of (417): / and / w W m q Q m = = Total energy: 2 lesson 6 II. Energy and the First Law of Thermodynamics 4. (Rate form) The rate of change in energy of a closed system is equal to the rate of heat transfer to the system, minus the rate of work performed by the system plus. ( ) kW W Q dt dE net , out net , in = = 2 1 t t dt dt dE E = 2 1 t t dt Q Q = 2 1 t t dt W W Integration gives the previous form, (417): ) kJ ( W Q E net , out net , in = lesson 6 II. Energy and the First Law of Thermodynamics 5. Example 1 (ideal gas). We have two rigid, insulated chambers connected by a valve. Chamber A is filled with air at 10 bar (gage) and 300 K and B is empty. At these conditions, the air behaves as an ideal gas. Now open the valve and allow the entire system (chambers A and B) to reach equilibrium. Find the change in U and T....
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 Summer '08
 thermo

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