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Unformatted text preview: 2.2. EXERGY 25 h + (1/2) v v + g z h + gz o . q = - q cv H,Carnot q L,Carnot w Carnot Carnot Engine T o o Control Volume Figure 2.2: Sketch of control volume balance for exergy discussion. Now take the “in” state to be simply a generic state with no subscript, and the “out” state to be the reference state, so 0 = ˙ Q cv − ˙ W cv + ˙ m parenleftbigg h − h o + 1 2 ( v · v ) + g ( z − z o ) parenrightbigg . (2.81) Let us next scale by ˙ m so as to get 0 = q cv − w cv + h − h o + 1 2 v · v + g ( z − z o ) . (2.82) Here we have taken q cv ≡ ˙ Q cv / ˙ m and w cv = ˙ W cv / ˙ m . Now, let us insist that w cv = 0, and solve for q cv to get q cv = − parenleftbigg h − h o + 1 2 v · v + g ( z − z o ) parenrightbigg . (2.83) Now, we imagine the working fluid to be at an elevated enthalpy, velocity, and height relative to its rest state. Thus in the process of bringing it to its rest state, we will induce q cv < 0. By our standard sign convention, this means that thermal energy is leaving the0....
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.
- Fall '11