# HW 4 - Prob 4.3 PROBLEM STATEMENT: One kilogram of water is...

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Prob 4.3 PROBLEM STATEMENT: One kilogram of water is contained in a sealed kettle with a volume of 3 m3 at T=20C. An electrical immersion heater inside the kettle is turned on and remains on until the kettle is filled with saturated vapor. Electricity costs \$0.15 per kilowatt-hour. (a) Draw a sketch defining the system you choose for analysis. Also sketch a P-v diagram of the process undergone by that system. (Show the saturation dome on your sketch.) Note any necessary assumptions. (b) Find the final pressure in the kettle. (c) Find the cost of the electricity necessary to carry out the process. DIAGRAM DEFINING SYSTEM AND PROCESS: State 1: Liquid-Vapor Phase State 2: Saturated Vapor T1=20C 3 V2=V 1=3 m3 V 1=3 m m=1 kg QIN,1-2 Water, V=3 m3 State 1: Liquid-Vapor, T1=20C State 2: Saturated Vapor (i.e. x2 = 1) P2 (kPa) and the electricity cost (\$) GIVEN: FIND: ASSUMPTIONS: Quasi- static and isometric (constant volume), no heat loss from kettle surface. GOVERNING RELATIONS: First Law, QIN,1-2 - WOUT,1-2 = U2 - U1 QUANTITATIVE SOLUTION: a). P-v Diagram P P2 P1 1 2 T2 T1=20 v 3 m3/kg b). In the initial state, State 1, we know V1 = 3 m3, m = 1 kg, so v 1=3 m3/kg. For T1 = 20C, Table 10s shows that the initial state is in the saturation region and P1 is 2.3388 kPa. Since v is constant throughout the process (sealed kettle), we can check Table 11s to find the state in which v=3 m3/kg for a saturated vapor. For a PR04- 03af.doc 04/08/04 Chk:MA Rev. PSS saturated vapor, v=3 m3/kg falls in the 50-60 kPa region. Using P1=2.3388 kPa and interpolating Table 11s between 50 and 60 kPa to find P2: 3(m3 / kg) - 3.2408(m3 / kg) P2 = 50kpa + (60kpa - 50kpa) 3 3 2.7324(m / kg)-3.2408(m / kg) P2 = 54.74 kPa c). We write the First Law as: QIN,1-2 - WOUT,1-2 = U2 - U1 For a constant volume process, WOUT,1-2 = 0. The energy added by the electrical heater is in the form of heat transfer to the water in the kettle, and the first law lets us determine QIN,1-2 if we can determine U in the initial and final states. At state 1, U1=mu1, where u1 is calculated as follows, using Table 10s: v - vL u1 = uL +x 1 (uV - uL ) = uL + (u V - uL ) v V - vL m3 m3 3 -0.001002 kJ kg kg ( 2402.0-83.833 ) kJ = 83.833 + kg 3 3 kg 57.7781 m -0.001002 m kg kg u1 = 204.2 kJ / kg Since the fluid is a saturated vapor at state 2, the value of u2 is simply uV at P2 = 54.74 kPa, or u2

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## This note was uploaded on 04/05/2012 for the course ME 326 taught by Professor Schmidt during the Spring '07 term at University of Texas at Austin.

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HW 4 - Prob 4.3 PROBLEM STATEMENT: One kilogram of water is...

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