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