ENSC 46 Tutorial, Week#6 Exergy: Control Mass Analysis An insulated piston-cylinder device contains L of saturated liquid water at a pressure of 50 kpa which is constant throughout the process. An electric resistance heater inside the cylinder is turned on, and electrical work is done on the water in the amount of 00 kj. Assuming the surroundings to be at 5C and 00 kpa, determine: a) The minimum work with which this process could be accomplished b) The entropy generated during the process Step : Draw a diagram to represent the system (show control mass of interest) The control mass boundary encloses the water in the cylinder Surroundings T 0 =5 C P =00 kpa 0 State Intermediate State W Q=0 b State Surroundings T 0 =5 C P =00 kpa 0 Sat. Liquid H O P =50 kpa V =L W e = 00 kj H 0 P =50 kpa Step : Write out what is required to solve for Find: a) The minimum work with which this process could be accomplished b) The entropy generated during the process Step 3: Property table Property State T[K] P[kPa] h[kj/kg] s[kj/kg*k] v[m 3 /kg] V[m 3 ] (sat liq) 50 0.00 50 Step 4: Assumptions Assumptions: ) Insulated piston-cylinder (Q=0) ) KE, PE 0 3) Compression/expansion processes are in quasi-equilibrium Step 5: Solve Part a) The solution for part a) involves finding the MINIMUM work with which this process could be accomplished. As stated in Cengel and Boles (Ch7), Reversible Work, W rev, is defined as the minimum work that needs to be supplied as a system undergoes a process between the specified initial and final states. Therefore, the minimum work with which this process M. Bahrami ENSC 46 (S ) Tutorial 6
can be accomplished is equal to the reversible work. Before the reversible work can be determined, the properties of the H O at the initial and final states must be determined. We can determine the properties at state using the information given in the problem (saturated liquid H O at P = 50 kpa) with Table A-5. From Table A-5 s = s f@50kpa =.4336 [kj/kg*k] v = v f@50kpa = 0.00053 [m 3 /kg] h = h f@50kpa = 467.[kJ/kg] From the specific volume at state, v, and the volume of state, V, we can determine the mass of H O in the cylinder. m V v 0.00 3 m 0.00053 kg 3 m. kg H O 9 The properties at state must now be determined (the pressure at state (50 kpa) is known because this is a constant pressure process). An energy balance on the system can be written to link state to state, as shown in Eq. E W W (Eq) e b E Using the assumption KE, PE 0, and noting that W b = P(V -V ), Eq can be re-expressed as Eq. Eq can be rearranged to Eq3. U U We P( V ) (Eq) V U PV U PV We (Eq3) Recognizing that U+PV = H, and that H = m H0 h, Eq3 can be rewritten as Eq4. m H O h h W (Eq4) e Isolating h in Eq4, and substituting in the values for W e, m H0, and h, h can be determined. M. Bahrami ENSC 46 (S ) Tutorial 6
h kj kj kj 467. kg kg kg We 00 h 65 mh O.9 Using h = 65 kj/kg and P = 50kPa with Table A-5, we find that state is within the vapor dome, in between the saturated liquid and vapor states on the 50 kpa isobar line. The properties at state can determine by first finding the quality, and then using this to interpolate in Table A-5. x h h g h f h f 65 467. 0.5 693.6 467. s = s f@50kpa + xs fg@50kpa =.4336 + 0.5(5.7897) = 4.44 [kj/kg*k] v = v f@50kpa + x(v g@50kpa -v f@50kpa) = 0.00053 + 0.5(.58) = 0.6033 [m 3 /kg] As mentioned previously, the minimum work that needs to be provided to accomplish this process represents the reversible work input, which can be computed from the difference in the system exergy at state and state, as expressed in Eq5. W rev (Eq5) The system of exergy at states and can be determined from the expressions in Eq6 and Eq7 respectively. ( 0 0 0 0 V0 U U ) T ( S S ) P ( V ) (Eq6) ( 0 0 0 0 V0 U U ) T ( S S ) P ( V ) (Eq7) Substituting Eq6 and Eq7 into Eq5, an expression for the reversible work can be obtained as shown in Eq8. Wmin Wrev ( U U) T0 ( S S) P0 ( V V ) (Eq8) Note how the dead state properties U 0, S 0, and V 0 cancel each other out in Eq8. Substituting the previously obtained expression for internal energy represented in Eq into Eq8 and bringing the m HO outside the bracket to convert the properties to specific properties, Eq8 can be written as Eq9. W rev P P v v W m T ( s ) (Eq9) e H O 0 0 s M. Bahrami ENSC 46 (S ) Tutorial 6 3
Substituting the known property values into Eq9, we can calculate the minimum work input. kj kj Wmin 00[ kj ].9[ kg] (50 00)(0.6033 0.00053) (98)(4.44.4336) kg kg = 440.6 kj Answer a) Part b) The exergy can be calculated from the expression shown in Eq0. T0S gen (Eq0) To find the entropy generated during the process we can perform an entropy balance on the system as shown in Eq. S in S S S (Eq) out gen system For a closed system, entropy can only be carried across the system boundary by heat transfer. Since the system is insulated, S in = 0 and S out = 0. The change in the system entropy is the difference in entropy between state and state i.e. S system = m HO (s -s ). Using this information with Eq, the (Irreversibility) can be determined from Eq. T m ( s ) (Eq) 0 H O s Substituting in the know values into Eq, the (Irreversibility) can be calculated. kj T0 S gen T0mH O ( s s) (98[ K])(.9[ kg])(4.44.4336) kg K = 70. kj Answer b) Step 6: Concluding Statement and Remarks The minimum work with which this process could be accomplished is 440.6kJ. The amount of exergy during the actual process is 70. kj. One may question why an input of 00kJ is required in the actual process if the process requires a minimum of 440.6kJ to accomplish it with 70.kJ of exergy in the actual process (which sums to 4.8 kj). In the actual process we still have an excess amount of work out of the system above that which is required to expand M. Bahrami ENSC 46 (S ) Tutorial 6 4
against the surroundings pressure. This excess work can be expressed as (P-P 0 )m HO (v -v ), which is equal to 57.kJ. Summing these three components together accounts for the 00kJ input. Minimum Work Input = 440.6 kj + Exergy Destroyed = 70. kj + Excess Work Output = 57. kj = Actual Work Input = 00 kj M. Bahrami ENSC 46 (S ) Tutorial 6 5