Kemiteknik - Värme- och strömningsteknik Processteknikens grunder (PTG) 05 Räkneöningar a 5,.9.05 kl. -5 kurs-assistent C. Haikarainen 0 frågor; -6 utan sar; 7-0 med sar. För äldigt länge sedan spenderade stora dinosaurier största delen a sitt li i attnet, såsom illustrerats i figuren intill. Anta att dessa dinosaurier sträckte på halsen för att andas från ett djup på m, och beräkna hur mycket arbete (i kj) ett andetag (för att flytta bort attnet omkring den) på 0,05 m luft i så fall skulle ha krät. Vattnets densitet: 000 kg/m ; atmosfäriskt tryck: bar. A ery long time ago, dinosaurs spent most of their time in the water, as illustrated by the figure aboe. Assume that these animals stretched their necks to breathe from a depth of m, and calculate how much work (in kj) one breath (in order to push away the water surrounding it) of 0,05 m air would require in that case. Density of water: 000 kg/m ; atmospheric pressure: bar. PTG exam question March 007 (4 points of 0). KJ05 Q-8 A chamber is diided equally in two parts by a membrane. One side contains H at a pressure of 0 kpa and the other side is eacuated. The total chamber olume is 0.004 m. At time t = 0, the membrane ruptures and the hydrogen expands freely into the eacuated side. If the chamber is considered adiabatic, find the final pressure. Assumptions:. Hydrogen is an ideal gas at these conditions.. The chamber is perfectly insulated.. Figuren ger trycket - olym, p V - diagrammet för en ideal gas, med specifik ärme id konstant olym c = ½ R (J /mol K) med R = 8,4 (J/mol K), som kan ara på tillstånd A, B, C och D i någon process. I tillstånd A, temperatur T = K. a. Beräkna mängden gas, n, i mol b. Beräkna arbetet W processen genererar för de tre olika ägar ACB (ds från A till B ia C), för ADB och för isoterm AB, respektie. c. Beräkna förändringen a den inre energi ΔU a gasen för dessa tre ägar ACB, ADB och isoterm AB, respektie. d. Beräkna ärmen Q som upptas eller ages a processen för de tre olika ägar ACB, ADB och isoterm AB, respektie. e. Beräkna de temperaturer för punkterna C och D, respektie.
The figure aboe gies the pressure - olume, p V diagram for an ideal gas, with specific heat at constant olume c = ½ R (J/mol K) with R = 8.4 (J/mol K), that can be at states A, B, C and D in some process. In state A, temperature T = K. a. Calculate the amounts of moles of gas, n. b. Calculate the work W the process generates for the three different paths ACB (from A to B ia C), for ADB and for isotherm AB, respectiely. c. Calculate the change of the internal energy ΔU of the gas for these three paths ACB, ADB and isotherm AB, respectiely. d. Calcultate the heat Q taken up or gien off by the process for the three different paths ACB, ADB and isotherm AB, respectiely. e. Calculate the temperatures for points C and D, respectiely. PTG exam question December 0 (8 points of 0) 4. Ett ätskeflöde O a olja id temperaturen 50 C kyls ner med 5 C atten W. En ärmeäxlare, med den totala ärmeöerföringskoefficienten U = 00 W/(m² K), anänds. Det nedkylda oljeflödets temperatur regleras genom att låta en del a flödet passera förbi ärmeäxlaren (by-pass). En termostat TC reglerar flödet a olja förbi ärmeäxlaren, resten strömmar genom den. Processen beskris a flödesschemat och datatabellen nedan. a. Beräkna temperaturen TO, ut, tot (i C, decimal) för det totala flödet olja som lämnar processenheten. b. Beräkna temperaturen TO, ut, (i C, decimal) för den delen a oljeflödet som lämnar ärmeäxlaren. c. Beräkna ärmeäxlarens ärmeöerföringsyta A (i m ). d. Beräkna ärmeäxlarens erkningsgrad ( effectieness ) ε (i %) för dessa förhållanden. T = 50 C T = 78, C Q TC T = 5 C O W Mass flow ṁ (kg/s) Massaström Specific heat c p (kj/(kg K) Specifik ärmekapacitet Temperature in T i ( C) Temperatur in Temperature out T u ( C) Temperatur ut Water / Vatten 45,4 4,9 5,0 78, Oil / Olja: Total 00,0,00 50,0 Oil / Olja: By-pass 5,0,00 A liquid oil stream O with temperature 50 C is cooled using 5 C cooling water W. A heat exchanger is used with oerall heat transfer coefficient U = 00 continues on next page.
W/(m² K). The temperature of the cooled oil stream is controlled by by-passing part of the oil stream; a temperature control TC controls the oil flow through the by-pass; the rest of the oil flows through the heat exchanger. A process schematic and a process data table are gien aboe. a. Calculate the temperature TO, out, total (in C, decimal) of the total oil stream that leaes this process unit. b. Calculate the temperature TO, out, HX (in C, decimal) of the oil stream that leaes the heat exchanger. c. Calculate the heat exchanging surface area A (in m ) of the heat exchanger. d. Calculate the effectieness, ε (in %), of the heat exchanger for these conditions. PTG exam question February 007 (8 points of 0) 5. En iss process behöer ett stabilt flöde a tryckluft id 0, kg/s, 600 kpa och 50 C. För detta syfte komprimeras luft från omginingens tillstånd (00 kpa och 0 C) först till 600 kpa i en adiabatisk kompressor, sedan matas luften till en ärmeäxlare där den kyls ner till 50 C id ett konstant tryck. Som kylmedel anänds atten som kommer in i ärmeäxlaren id 5 C och lämnar den id 40 C. Fastställ massflödet atten, ṁw (kg/s), då hastigheten a arbetet gjort a kompressorn är W s = 44 kj/ s. Anta att luft är en ideal gas. Den specifika ärmen för luft och atten kant antas ara,0 kj/(kg K) och 4, kj/(kg K), respektie. A certain process requires a steady supply of compressed air at 600 kpa and 50 C at the rate of 0.kg/s. For this purpose, air at ambient conditions (00 kpa and 0 C) is first compressed to 600 kpa in an adiabatic compressor, then it is fed to a heat exchanger where it is cooled to 50 C at constant pressure. As cooling medium, water is used: it enters the heat exchanger at 5 C and leaes at 40 C. Determine the mass flow rate of water, ṁw (kg/s), if the rate of work done on the compressor is W s =44 kj/ s. Assume air to be an ideal gas. The specific heat of air and water can be taken to be,0 kj/(kg K) and 4, kj/(kg K), respectiely. PTG exam question December 0 (4 points of 0)
6. KJ05 Q-6 A small oil refinery uses rier water to cool some of the fluid streams in the refinery. Consider a two shell pass, four tube pass heat exchanger that uses 5 kg/s of rier water at 0 C on the shell side to cool 0 kg/s process fluid (cp = 00 J/kg K) from 80 C to 5 C. If the oerall heat transfer coefficient is 600 W/m K, determine, using the LMTD method: a. the outlet temperature of the coolant (in C) b. the heat transfer area required (in m ). The system is steady. Potential and kinetic energy effects are negligible. No work is done on or by the system. Both fluids are ideal with constant specific heats. Sources: Old PTG exams, and: KJ05 = Kaminski, Jensen Introduction to Thermal and Fluids Engineering, Wiley, 005 SEHB06 = Schmidt, Ezekoye, Howell & Baker Thermodynamics Wiley, 006
Prob. 8/7/04 PROBLEM. PROBLEM STATEMENT: A rigid insulated tank whose olume is 0. m contains pure oxygen initially at 00 kpa and 7 C. A fan inside the tank, drien by a shaft which passes through the tank wall, is used to stir the oxygen. After a while, the pressure in the tank rises to 50 kpa. How much internal energy (kj) did the fan add to the oxygen? The specific heats of oxygen can be assumed constant and can be ealuated at the mean temperature for this process. DIAGRAM DEFINING SYSTEM AND PROCESS: W s GIVEN: O, V=0. m P =00 kpa, T =7 C=00 K P =50 kpa FIND: U (kj) ASSUMPTIONS: Ideal gas (O ), no heat transfer (insulated tank), constant olume (rigid), OK to assume constant specific heats. u GOVERNING RELATIONS: Ideal gas law, PV=RT, definition of c : c = T QUANTITATIVE SOLUTION: Ideal gas law: P =RT where R=R M RT 8.44 (kpa m /kmol K) 00 (K) = = = 0.779 m /kg MP (kg/kmol) 00 (kpa) V 0. (m ) m = = = 0.85 kg 0.779 (m /kg) The temperature at state is computed as follows: ( υ = υ Qconstant olume) P 50 (kn m ) 0.779 (m /kg) T = = =450 K R 8.44 kn m kg K u du Since the whole process takes place at contant olume c = =, T dt therefore du=c dt, and since we can assume c constant, u -u = c (T -T ). Take c,mean at T = 450+00 = 75 K =0 C ( ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.
Prob. 8/7/04 From Table 7s, c 0.675 kj/kg K. Thus, the gain in internal energy is U -U =m(u -u )= mc (T -T) = 0.85 (kg) 0.675 (kj/kg K) (450-00)(K) =9 kj DISCUSSION OF RESULTS: Een though there is no PdV work in this case (since there is no dv), shaft work can be put into the system to change its internal energy. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.
-8 In a new process, a thin metal film is produced when ery high elocity particles strike a surface, melt, and adhere to the surface. Imagine an aluminum particle with a diameter of 40 µm ( µm = 0-6 m) at a temperature of 0 o C. The particle strikes a cold aluminum surface, also at 0 o C. The particle energy is just high enough so that the particle and a portion of the surface with the same mass as the particle completely melt. What is the elocity of the particle? Assume pure aluminum with a constant specific heat of 46 J/kg K. The latent heat of fusion (the amount of heat needed to melt kg of aluminum) is 404 kj/kg. Approach: Write the first law for a constant pressure process of a closed system and eliminate all terms except the kinetic energy change and the enthalpy change. Include both sensible and latent heat of fusion in ealuating the enthalpy term. Assumptions:. There is no potential energy change.. There is no heat transfer from the surface during the process.. Specific heat is constant. Solution: Choose the particle and the part of the surface which melts as the control olume. The process begins just before the particle hits the surface and ends as the particle is brought to a complete halt and the aluminum has melted. The process occurs at constant pressure, therefore the first law may be written in the form KE+ PE+ H = Q We assume no change in potential energy and no heat transfer to the air from the surface, therefore: KE + Η = 0 The enthalpy change has two components: the enthalpy rise as the particle temperature changes from its initial alue to the melting point of aluminum and the enthalpy rise as the solid aluminum and the particle melt. Thus mv mv + mc T T + mh 0 ( ) p where h fs is the latent heat of fusion of aluminum. The final elocity, is zero. Soling for ( ) ( ) = fs V = 4cp T T + 4hfs = cp T T + hfs Substituting the melting temperature of aluminum from Table A- J kj 000 J V = 46 ( 9 7 0) K + 404 kg K kg kj = m s V V V Answer - 8
-46 Carbon monoxide is expanded slowly in a well-insulated, frictionless piston-cylinder assembly from 00 cm, 5ºC to 400 cm. Find the final temperature Approach: To find the final temperature use T T = k Assumptions:. Carbon monoxide behaes like an ideal gas under these conditions.. The process is a quasi-equilibrium process.. Specific heat is constant. 4. The process is adiabatic. Solution: Since the process is slow and frictionless, it may be assumed quasi-static. Since it is also well-insulated, it maybe be assumed adiabatic. For an ideal adiabatic process of an ideal gas k T = T Values for the specific heat ratio, k, are aailable in Table A-8. The final temperature is not known, but it is expected to be less than the initial temperature. As shown in the table, k is ery insensitie to temperature, being almost the same at 50, 00, and 50 K. Therefore, k =.4 will do, and 00 T = ( 5 + 7) 400 (.4 ) o = 66 K = 7 C Answer - 46
Prob.4 8/7/04 PROBLEM.4 PROBLEM STATEMENT: Air in a piston-cylinder apparatus undergoes a sequence of two processes (note this is not a complete cycle): Process -: Compression from P =00 kpa and olume of = 0.04 m /kg =0.0 m /kg, along a process path to a specific. P = constant. Process -: Constant pressure expansion back to the original olume, = Sketch the process on a P - diagram and determine the oerall work per unit mass (w OUT,- ) in kj/kg. Specify whether the work is done by the air or on the air. DIAGRAM DEFINING SYSTEM AND PROCESS: W OUT Air GIVEN: Air in a piston-cylinder system State : P =00 kpa, = 0.04 m /kg State :. = 0.0 m /kg, P = constant State : P =P, = FIND: w OUT,- (kj/kg) ASSUMPTIONS: Frictionless piston GOVERNING RELATION: For P n = constant (polytropic process) P -P w OUT,- = Pd=. -n QUANTITATIVE SOLUTION: Since P. = constant,.... P 00 ( 0.04) P = P P = = = 46. kpa.. ( 0.0) P - diagram P (kpa) 46. 00 0.0 0.04 (m /kg) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.
Prob.4 8/7/04 The work per unit mass for Process -, P -P w OUT,- = Pd= -n [(46. 0.0)-(00 0.04) ]( kn/m )(m /kg) = -. w = -.08 kj/kg work is into the system OUT,- ( ) The work per unit mass for Process -, w = Pd = P ( - ) ( Qconstant pressure) OUT,- OUT,- = 46. (kn/m ) (0.04-0.0) ( m /kg) ( the system) w = +4.9 kj/kg work is done by Thus, the oerall work per unit mass is w OUT,- = w - OUT,- + w OUT,- = (.08)+ 4.9 = +.84 kj/kg work is done by the system DISCUSSION OF RESULTS: While the representation of a process by the "polytropic" relationship P n =constant may seem rather arbitrary at this point, it will take on special meaning when we get to Chapter 6 and beyond. This type of relationship between pressure and olume is, in fact, characteristic of many gas compression and expansion processes, such as in the cylinder of an internal combustion engine. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.