Chemical Engineering Department University of Washington

Transcription

1 Chemcal Engneerng Department Unversty of Washngton ChemE 60 - Exam I July 4, Mass Flow Rate of Steam Through a Turbne (5 onts) Steam enters a turbne at 70 o C and.8 Ma and leaves at 00 ka wth a qualty of 90%. The turbne produces.5 MW of power and heat losses to the surroundngs occur at a rate of 5 kw. Neglectng changes n knetc and potental energy, determne the mass flow rate of the steam n kg/s. Data: Superheated Steam Tables Saturated Steam Tables Densty (kg/m 3 ).80 Ma (T sat 07. o C) Internal Volume Energy Enthalpy (m 3 /kg) (kj/kg) (kj/kg) Sat ressure Volume (m 3 /kg) Internal Energy (kj/kg) Enthalpy (kj/kg) (Ma) Temp. (C) Sat. Lq Sat. Vap Sat. Lq Sat. Vap Sat. Lq Sat. Vap age of

2 - Slow Leak From an Ammona Tank (35 onts) A rgd, cylndrcal storage tank contans.00 m 3 of a two-phase mxture of ammona. Intally, saturated lqud ammona at 5 o C occupes 90% of the volume of the tank. The remanng 0% of the volume s occuped by ammona vapor, also at 5 o C. The valve at the top of the tank develops a leak. Ammona vapor very slowly escapes from the tank. Heat transfer from the surroundngs nto the ammona occurs rapdly enough that the temperature of the ammona remans constant. Eventually the lqud volume drops to 45% of the volume of the tank. (3 pts) a.) Determne the mass of ammona that leaked out of the tank n kg. (4 pts) b.) Wrte the general form of the st Law and elmnate all terms that are neglgble n ths problem. Explan why each term can be neglected. (8 pts) b.) Determne the amount of heat transfer from the surroundngs n kj. 3 - olytropc Compresson of Carbon Doxde (40 onts) A pston-and-cylnder devce contans 3.5 kg of carbon doxde gas (MW 44 g/gmole) at 00 ka and 80 o C. The carbon doxde s compressed polytropcally (V constant) wth.6. The fnal temperature s 5 o C. (5 pts) a.) Determne the boundary work for ths process, n kj. (4 pts) b.) Determne the heat transfer for ths process, n kj. ( pts) c.) Determne the fnal pressure n the cylnder, n ka. Data: Shomate Equaton constants for carbon doxde: A B C D E o Remember, the Shomate Equaton yelds C n J/gmole-K. age of

3 Thermo-CD by B-Cubed - Mass Flow Rate of Steam Through a Turbne (5 onts) roblem Steam enters a turbne at 70 o C and.8 Ma and leaves at 00 ka wth a qualty of 90%. The turbne produces.5 MW of power and heat losses to the surroundngs occur at a rate of 5 kw. Neglectng changes n knetc and potental energy, determne the mass flow rate of the steam n kg/s. Data: Superheated Steam Tables ressure Volume (m 3 /kg) Internal Energy (kj/kg) Enthalpy (kj/kg) (Ma) Temp. (C) Sat. Lq Sat. Vap Sat. Lq Sat. Vap Sat. Lq Sat. Vap Saturated Steam Tables.80 Ma (T sat 07. o C) Densty (kg/m 3 ) Volume (m 3 /kg) Internal Energy (kj/kg) Enthalpy (kj/kg) Sat Read carefully : - Ths problem s an applcaton of the st Law for open systems to a SISO, SS turbne. - We are gven both Q and W, so the unknowns n the st Law are the mass flow rate of the steam and the propertes of the feed and effluent. - We know both the T and of the turbne feed, so we can obtan all of ts other propertes from the steam table data provded n the problem statement. - We know the effluent pressure and we know the state les wthn the two-phase envelope because the qualty s gven. Ths s enough nformaton for us to determne all of the - All we need to do s collect all the data and solve the st Law for the mass flow rate of the steam through the turbne. Draw a dagram : Lst gven nformaton : T 70 o C 800 ka 00 ka Turbne x 0.90 W shaft 500 kw Q -5 kw Fnd : m??? kg/s Lst all assumptons : - The process operates at steady state. - Inlet and outlet streams are n equlbrum states. - Changes n knetc and potental energy are neglgble. - The system has a sngle nlet stream and a sngle outlet stream. - The propertes of the water n the nlet and outlet streams are unform. Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

4 A good place to begn s by wrtng the st Law for open systems. n out d v g vj g Esys Q Wother + m H + + z mj Hj + + zj dt gc gc j gc gc We can smplfy ths equaton substantally because the turbne s a steady state, SISO process n whch changes n knetc and potental energy are neglgble. ( ) Q W m Hˆ m Hˆ Hˆ other We can solve ths form of the st Law for the mass flow rate of steam. Q W m ( Hˆ Hˆ ) Now, all we need to do s determne H and H. Because T > T sat at.8 Ma, we can determne H by nterpolaton n the superheated steam tables. Temperature ( o C) other Enthalpy (kj/kg) H kj/kg Because the turbne effluent s a two-phase mxture, we must use the followng equaton to determne ts enthalpy: We can look up the propertes of the saturated lqud and saturated vapor n the saturated steam tables at 00 ka. ( ) H ˆ x H ˆ + x H ˆ sat lq sat lq H sat lq kj/kg H sat vap kj/kg H 449. kj/kg Solve equatons : Fnally: m.74 kg/s Verfy assumptons : Answer questons : None of the assumptons made can be verfed based on the nformaton gven n the problem statement. The mass flow rate of steam through the turbne s.7 kg/s. Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

5 Thermo-CD by B-Cubed roblem - Slow Leak From an Ammona Tank (35 onts) A rgd, cylndrcal storage tank contans.00 m 3 of a two-phase mxture of ammona. Intally, saturated lqud ammona at 5 o C occupes 90% of the volume of the tank. The remanng 0% of the volume s occuped by ammona vapor, also at 5 o C. The valve at the top of the tank develops a leak. Ammona vapor very slowly escapes from the tank. Heat transfer from the surroundngs nto the ammona occurs rapdly enough that the temperature of the ammona remans constant. Eventually the lqud volume drops to 45% of the volume of the tank. (3 pts) a.) Determne the mass of ammona that leaked out of the tank n kg. (4 pts) b.) Wrte the general form of the st Law and elmnate all terms that are neglgble n ths problem. Explan why each term can be neglected. (8 pts) c.) Determne the amount of heat transfer from the surroundngs n kj. Read carefully : The problem s transent because the mass and energy of the contents of the tank change wth tme. The process s sothermal because heat transfer nto the ammona s fast. The vapor and lqud wthn the tank are always saturated at 5 o C and therefore have constant propertes. As a result, the propertes of the vapor leakng from the tank do not change wth tme ether. Draw a dagram : sat (5 o C) T 5 o C V Lq, 0.90 m 3 V Vap, 0.0 m 3 T 5 o C V Lq, 0.45 m 3 V Vap, 0.55 m 3 Lst gven nformaton : T 5 o C V.00 m 3 V Lq, 0.90 m 3 V Lq, 0.45 m 4 V Vap, 0.0 m 3 V Vap, 0.55 m 3 Fnd : a.) m out kg b.) Wrte st Law and cancel terms. c.) Q kj Lst all assumptons : - Both states are equlbrum states - Open system - Quas-equlbrum process - Changes n potental and knetc energy are neglgble - Unform State: The propertes of the vapor n the tank are unform and the propertes of the vapor leakng from the tank are the same as the propertes of the vapor wthn the tank. - Unform Flow: the propertes and flowrate of the vapor leakng from the tank are unform and constant wth respect to tme. Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

6 art (a) Begn by wrtng the transent form of the energy balance equaton, usng the contents of the tank as the system. m m m m sys out Next, we can determne m and m from the gven volume of lqud and vapor n the tank and the specfc volume of the vapor and the lqud. The key s that, n both the ntal and fnal states, the vapor and lqud n the tank are both saturated! VLq VVap m mlq + m m Vap Lq mvap ˆV ˆV So, all we need to do now s look up the specfc volume of saturated vapor and saturated lqud at 5 o C n the ammona tables. V sat lq m 3 /kg V sat vap m 3 /kg Solve equatons : m Lq, kg m Lq, 7.5 kg m Vap, kg m Vap, kg m kg m kg m out kg Sat'd Lq Sat'd Vap Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

7 art (b) In order to determne Q, we should begn by wrtng the ntegral form of the st Law for open systems and then smplfy t based on the assumptons lsted above. vn g vout g Usys Q Wother + mn Hn + + zn mout Hout + + zout gc gc gc gc art (c) Rearrange the smplfed form of the st Law as follows: ˆ ˆ sys out out U U U m U m U Q m H Now, solve the st Law for Q: Q m Uˆ m Uˆ + m H (Eqn ) Next, we need to determne the specfc nternal energy of the contents of the tank n both the ntal and fnal states. The key to ths s to recall that the vapor and lqud n the tank are saturated n both the ntal and fnal states. ˆ ˆ sat vap ( ) ˆ mvap U xu + x U x sat lq m + m We calculated the mass of vapor and lqud for both the ntal and fnal states n part (a), so all we need to do now s look up the specfc nternal energy of saturated lqud and saturated vapor at 5 o C n the amona tables. U sat lq kj/kg U sat vap kj/kg The only unknown left n (Eqn ) s the specfc enthalpy of the vapor leavng the tank. We know that ths vapor s saturated vapor at 5 o C, so we can look up ts enthalpy n the ammona tables. lq out out vap H out kj/kg Solve equatons : State State x x U 98.3 kj/kg U 33.0 kj/kg Q 36. MJ Verfy assumptons : None of the assumptons made can be verfed based on the nformaton gven n the problem statement. Answer questons : A whoppng 68 kg of ammona leaked out of the tank! No less than 30 MJ of heat was transferred from the surroundngs nto the ammona durng ths process to keep the ammona n the tank at 5 o C. Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

8 Thermo-CD by B-Cubed 3 - olytropc Compresson of Carbon Doxde (40 onts) roblem 3 A pston-and-cylnder devce contans 3.5 kg of carbon doxde gas (MW 44 g/gmole) at 00 ka and 80 o C. The carbon doxde s compressed polytropcally (V constant) wth.6. The fnal temperature s 5 o C. (5 pts) a.) Determne the boundary work for ths process, n kj. (4 pts) b.) Determne the heat transfer for ths process, n kj. ( pts) c.) Determne the fnal pressure n the cylnder, n ka. Data: Shomate Equaton constants for carbon doxde: A , B , C , D , E Read carefully : The key here s to assume (and verfy) that the CO s an deal gas throughout the compresson process. Ths allows us to use the Shomate Eqn to evaluate H. If we assume the process s a quas-equlbrum process, we can ntegrate along the polytropc process path to determne W b. Then, knowng H and W b, we can use the st Law to determne Q. We can determne by smultaneously solvng the polytropc process path equaton and the deal gas EOS. Draw a dagram : olytropc Compresson V.6 C Q??? kj W b??? kj Lst gven nformaton : 00 ka A T 80 o C B T 5 o C C m 3.5 kg D E Fnd : W b??? kj Q??? kj MW 44 g/gmole??? ka R 8.34 J/mole-K Lst all assumptons : - The ntal and fnal states are equlbrum states. - The gas n the cylnder behaves as an deal gas throughout the process - The process s a quas-equlbrum process. - Changes n knetc and potental energy are neglgble. - Boundary work s the only form of work that crosses the system boundary. Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

9 art (a) Assumng that the process s a quas-equlbrum process allows us to evaluate the boundary work from: V W dv We also know the path that the process follows because t s polytropc: V C Therefore: C V b V and: W b V C dv V V V V For deal gases: V n R T, therefore: All we need to do now s determne the number of moles n the cylnder. Wb nr T T m n MW ( ) Solve equatons : n gmoles W b kj art (b) The st Law s almost always mportant n Thermo problems. In ths case, we wll use the ntegral form of the st Law for closed systems, wth only boundary work crossng the system boundary. U + Ekn + Epot Q Wb U Q Wb Q U + Wb We know W b, so we need to evaluate U and then we can use the st Law to evaluate Q. The Shomate Eqn wll allow us to evaluate H from: H Then, we can relate H to U usng the defnton of enthalpy: H U + V or: H U + (V) But, the Ideal Gas EOS tells us that: V n R T Therefore: (V) n R T T T C dt o and: or: U H nr T U H R T H J/gmole U J/gmole U 4.9 kj Q 65.3 kj Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc

10 art (c) State les on the polytropc process path. We assumed that the CO n ths state s an deal gas. Therefore, the followng equatons apply. V nrt V C lug nto the polytropc path equaton: or: nrt V nrt C Therefore: C T T ( nr) C C ( nr) T T T T Solve equatons : ka Verfy assumptons : If the molar volume at state and at state are both greater than 0 L/gmole, then t s vald to assume that the CO behaves as an deal gas throughout ths process. Use the Ideal Gas EOS to estmate the molar volume at state and. RT V V 9.36 L/gmole > 0 L/gmole V 4.04 L/gmole > 0 L/gmole The deal gas assumpton s vald for ths problem. Answer questons : a.) The boundary work for ths process s: W b kj (The negatve sgn ndcates that work s done ON the system.) b.) The heat transfer for ths process s: Q 65.3 kj (The postve sgn ndcates that heat s transferred INTO the system.) c.) The pressure n the fnal state s: 38 ka Thermo-CD Soluton Copyrght 003 by B-Cubed Dr. Baratuc