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quiz sol 1

# quiz sol 1 - = ∆ m 4 points 3 We start with the first law...

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Washington State University ME 402: Thermal System Design Quiz 1 Solution 1. We start with the first law of thermodynamics for an open system + + + + = + in out out in out in gz v h m gz v h m W W Q Q 2 2 2 2 The system has the following configuration: As we know in an open system the same mass that comes in must leave the system, therefore we can use the following simplification: m m m out in = = Since there is also no change in height the potential energy is zero. 0 = PE Finally there is no heat coming into the system or leaving the system and there is also no work done by the system. Therefore the first law of thermodynamics simplifies to: + = + + + = in in in out out out v h v h v h m v h m 2 2 2 2 0 2 2 2 2 Since in out in out h h v v < > Final answer (a) in out h h < 3 points in in v A , out out v A ,

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2. L 2 1 = L 1 2 = P = constant P = 1MPa Q For the first part of the problem we have to use the boundary work: a) ()( ) = = × × = = = = 2 1 2 3 3 3 6 1 2 1 10 10 2 1 10 kJ m m N P d P Pd W B b) Since the system is a closed system there is no change in mass of the system
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Unformatted text preview: = ∆ m 4 points 3. We start with the first law of thermodynamics: ∑ ∑ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = − + − in out out in out in gz v h m gz v h m W W Q Q 2 2 2 2 & & & & & & Since there is no gain or loss of heat in the system and there is no change in potential energy and also the system produces no work. The change in kinetic energy is so small tat it could be neglected, therefore the first law of thermodynamics will be simplified to: ∑ ∑ − = in out in h m h m W & & & There mass coming into the system and leaving the system are the same, therefore: m m m out in & & & = = ( ) in out in h h m W − = & & The work in per unit mass could be written as: in out in h h m W − = & & 3 points in m & out m &...
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quiz sol 1 - = ∆ m 4 points 3 We start with the first law...

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