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# Unit5

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### Unit5

1. 1. NON-FLOW PROCESS J2006/5/1 UNIT 5 NON-FLOW PROCESS OBJECTIVESGeneral Objective : To understand and apply the concept of non-flow process in thermodynamicsSpecific Objectives : At the end of the unit you will be able to:  define and calculate the following non-flow processes: • polytropic • constant volume • constant pressure
2. 2. NON-FLOW PROCESS J2006/5/2 INPUT5.0 NON-FLOW PROCESS Processes, which are undergone by a system when the working fluid cannot cross the boundary, are called non-flow process. In a close system, although energy may be transferred across the boundary in the form of work energy and heat energy, the working fluid itself never crosses the boundary. Any process undergone by a close system is referred to as non-flow process. The equation for non-flow process is given as follows: U1 + Q = U2 + W or, U2 – U1 = Q –W In words, this equation states that in a non-flow process, the change in the internal energy of the fluid is equal to the nett amount of heat energy supplied to the fluid minus the nett amount of work energy flowing from the fluid. This equation is known as the non-flow energy equation, and it will now be shown how this may apply to the various non-flow processes.
3. 3. NON-FLOW PROCESS J2006/5/35.1 Polytropic process (pVn = C) This is the most general type of process, in which both heat energy and work energy cross the boundary of the system. It is represented by an equation in the form pVn = constant (5.1) If a compression or expansion is performed slowly, and if the piston cylinder assembly is cooled perfectly, then the process will be isothermal. In this case the index n = 1. If a compression or expansion is performed rapidly, and if the piston cylinder assembly is perfectly insulated, then the process will be adiabatic. In this case the index n = γ. If a compression or expansion is performed at moderate speed, and if the piston cylinder assembly is cooled to some degree, then the process is somewhere between those discussed above. Generally, this is the situation in many engineering applications. In this case the index n should take some value, which is between 1 and γ depending on the degree of cooling. Some practical examples include: compression in a stationary air compressor (n = 1.3) compression in an air compressor cooled by a fan (n = 1.2) compression in a water cooled air compressor (n = 1.1) P 1 pVn=C W P1 2 P2 W v1 v2 v Qloss Figure 5.1 Polytropic process
4. 4. NON-FLOW PROCESS J2006/5/4 Equation 5.1 is applied at states 1 and 2 as: p1V1n = p 2V2n or n p 2  V1  =  (5.2) p1 V2  Also, for a perfect gas, the general property relation between the two states is given by p1V1 p 2V2 = (5.3) T1 T2 By the manipulation of equations 5.2 and 5.3 the following relationship can be determined: n −1 n −1 T2  p 2  n V  (5.4) =  =  1 T1  p1   V2  By examining equations 5.2 and 5.4 the following conclusions for a polytropic process on a perfect gas can be drawn as: An increase in volume results in a decrease in pressure. An increase in volume results in a decrease in temperature. An increase in pressure results in an increase in temperature. Work transfer: Referring to the process represented on the p-V diagram (Fig.5.1) it is noted that the volume increases during the process. In other words the fluid is expands and the expansion work is given by 2 W = ∫ pdV 1 2 c =∫ dV (since pVn = C, a constant) 1 Vn 2 dV = c∫ n 1 V
5. 5. NON-FLOW PROCESS J2006/5/5 p1V1 − p 2V2 = [larger pV- small pV] (5.5) n −1 Note that after expansion p2 is smaller than p1. In the p – V diagram, the shaded area under the process represents the amount of work transfer. Since this is an expansion process (i.e. increase in volume), the work is done by the system. In other words, the system produces work output and this is shown by the direction of the arrow representing W as shown in Fig. 5.1. Heat transfer: Energy balance is applied to this case (Fig.5.1) as: U1 – Qloss - W = U2 Qloss = (U1 – U2) – W or W = (U1 – U2) - Qloss Thus, in a polytropic expansion the work output is reduced because of the heat loses. Referring to the process represented on the p–V diagram (Fig.5.1) it is noted that during this process the volume increases and the pressure decreases. For a perfect gas, equation 5.4 tells us that a decrease in pressure will result in a temperature drop. For adiabatic process: W= For polytropic process: W=
6. 6. NON-FLOW PROCESS J2006/5/6 Example 5.1 The combustion gases in a petrol engine cylinder are at 30 bar and 800oC V2 8.5 before expansion. The gases expand through a volume ratio ( ) of ( ) and V1 1 occupy 510 cm3 after expansion. When the engine is air cooled the polytropic expansion index n = 1.15. What is the temperature and pressure of the gas after expansion, and what is the work output? Solution to Example 5.1 V2 = 510 cm3 P1= 30 bar p2 = ? t1 = 800oC Qloss t2 = ? W State 1 State 2 Data: p1 = 30 bar; T1 = 800 + 273 = 1073 K; n = 1.15 V2 = 8.5; V2 = 510 cm3; V1 t2 = ? p2 = ? W=? Considering air as a perfect gas, for the polytropic process, the property relation is given by equation 5.4 as: n −1 V  T2 = T1  1   V2  1.15−1  1  = 1073 x    8.5  = 778.4 K = 505.4oC
7. 7. NON-FLOW PROCESS J2006/5/7 From equation 5.2 n V  p 2 = p1  1  V2  1.15  1  = 30 x    8.5  = 2.56 bar Now, V2 = 510 cm3 = 510 x 10-6 m3 and, V2 = 8.5 V1 Then, 510 x10 −6 V1 = 8.5 = 60 x 10-6 m3 Work output during polytropic expansion is given by equation 5.5 as: p1V1 − p 2V2 W = [larger pV- small pV] n −1 (30 x10 5 )(60 x10 −6 ) − ( 2.56 x10 5 ) − (510 x10 −6 ) = 1.15 − 1 = 330 J = 0.33 kJ
8. 8. NON-FLOW PROCESS J2006/5/8 Activity 5A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 5.1 0.112 m3 of gas has a pressure of 138 kN/m2. It is compressed to 690 kN/m2 according to the law pV1.4 = C. Determine the new volume of the gas. 5.2 0.014 m3 of gas at a pressure of 2070 kN/m2 expands to a pressure of 207 kN/m2 according to the law pV1.35 = C. Determine the work done by the gas during expansion. 5.3 A cylinder containing 0.07 kg of fluid has a pressure of 1 bar, a volume of 0.06 m3 and a specific internal energy of 200 kJ/kg. After polytropic compression, the pressure and volume of the fluid are 9 bar and 0.011 m 3 respectively, and the specific internal energy is 370 kJ/kg. Determine a) the amount of work energy required for the compression b) the quantity and direction of the heat energy that flows during the compression.
9. 9. NON-FLOW PROCESS J2006/5/9 Feedback To Activity 5A 5.1 Since the gas is compressed according to the law pV1.4 = C, then, p1V11.4 = p 2V21.4 1.4 1 / 1.4 p1 V2  V p  ∴ =  or 2 =  1  p 2  V1  V1  p 2    from which, 1 / 1.4 p  p1 V2 = V1  1  p  = V1 1.4  2 p2 138 = 0.012 x 1.4 690 = 0.0348 m3 5.2 The work done during a polytropic expansion is given by the expression: p1V1 − p 2V2 W = [larger pV- small pV] n −1 In this problem V2 is, as yet, unknown and must therefore be calculated. Now p1V1n = p 2V2n 1/ n p  ∴ V2 = V1  1  p   2 1 / 1.35  2070  or V2 = 0.014 x    207  V2 = 0.077 m3
10. 10. NON-FLOW PROCESS J2006/5/10 (2070 x10 3 x0.014 − 207 x10 3 x0.077) ∴ Work done = 1.35 − 1 = 37.3 x 103 Nm = 37.3 x 103 J = 37.3 kJ 5.3 a) For a polytropic process, p1V1n = p 2V2n In the given case 1 x 0.06n = 9 x 0.011n n  0.06    =9 ∴  0.011  n = 1.302 p1V1 − p 2V2 W = n −1 (1x10 5 x 0.06) − (9 x10 5 x 0.0111) = 1.302 − 1 = -13.2 kJ The negative sign indicates that work energy would flow into the system during the process. b) The non-flow energy equation gives Q – W = U2 – U1 Q – (- 13.2) = ( 370 x 0.07 ) – ( 200 x 0.07 ) ∴ Q = - 1.3 kJ The negative sign indicates that heat energy will flow out of the fluid during the process.
11. 11. NON-FLOW PROCESS J2006/5/11 INPUT5.2 Constant volume process If the change in volume during a process is very small then that process may be approximated as a constant volume process. For example, heating or cooling a fluid in a rigid walled vessel can be analysed by assuming that the volume remains constant. Q p 2 p 1 1 2 Q v v a) Heating b) Cooling Figure 5.2 Constant volume process (V2=V1) The general property relation between the initial and final states of a perfect gas is applied as: p1V1 p 2V2 = T1 T2 If the volume remain constant during the process, V2 = V1 and then the above relation becomes p1 p 2 = T1 T2 or T2 p = 2 (5.6) T1 p1 From this equation it can be seen that an increase in pressure results from an increase in temperature. In other words, in constant volume process, the temperature is proportional to the pressure.
12. 12. NON-FLOW PROCESS J2006/5/12 Work transfer: Work transfer (pdV) must be zero because the change in volume, dV, during the process is zero. However, work in the form of paddle-wheel work may be transferred. Heat transfer: Applying the non flow energy equation Q – W = U2 – U1 gives Q – 0 = U2 – U1 i.e. Q = U2 – U1 (5.7) This result, which is important and should be remembered, shows that the nett amount of heat energy supplied to or taken from a fluid during a constant volume process is equal to the change in the internal energy of the fluid.5.3 Constant pressure process If the change in pressure during a process is very small then that process may be approximated as a constant pressure process. For example, heating or cooling a liquid at atmospheric pressure may be analysed by assuming that the pressure remains constant. P W p 1 2 W v2 – v1 v v1 Q v2 Figure 5.3 Constant pressure process
13. 13. NON-FLOW PROCESS J2006/5/13 Consider the fluid in the piston cylinder as shown in Figure 5.2. If the load on the piston is kept constant the pressure will also remain constant. The general property relation between the initial and final states of a perfect gas is applied as: p1V1 p 2V2 = T1 T2 If the pressure remain constant during the process, p2 = p1 and then the above relation becomes V1 V2 = T1 T2 or T2 V2 = (5.8) T1 V1 From this equation it can be seen that an increase in volume results from an increase in temperature. In other words, in constant pressure process, the temperature is proportional to the volume. Work transfer: Referring to the process representation on the p-V diagram it is noted that the volume increases during the process. In other words, the fluid expands. This expansion work is given by 2 W = ∫ pdV 1 2 = p ∫ dV (since p is constant) 1 = p (V2 – V1) (larger volume – smaller volume) (5.9) Note that on a p-V diagram, the area under the process line represents the amount of work transfer. From Figure 5.3 W = area of the shaded rectangle = height x width = p (V2 – V1) (larger volume – smaller volume) This expression is identical to equation 5.9
14. 14. NON-FLOW PROCESS J2006/5/14 Heat transfer: Applying the non flow energy equation Q – W = U2 – U1 or Q = (U2 – U1) + W (5.10) Thus part of the heat supplied is converted into work and the remainder is utilized in increasing the internal energy of the system. Substituting for W in equation 5.10 Q = (U2 – U1) + p(V2 – V1) = U2 – U1 + p2 V2 – p1 V1 (since p2 = p1 ) = (U2 + p2 V2) – (U1 + p1 V1) Now, we know that h = u + pv or H = U + pV Hence Q = H2 – H1 (larger H – smaller H) (5.11) Referring to the process representation on the p-v diagram shown in Figure 5.3, it is noted that heating increases the volume. In other words, the fluid expands. For a perfect gas, equation 5.8 tells us that an increase in volume will result in corresponding increase in temperature. For constant volume process: W=0 For constant pressure process: W = p (V2 – V1)
15. 15. NON-FLOW PROCESS J2006/5/15 Example 5.2 The specific internal energy of a fluid is increased from 120 kJ/kg to 180 kJ/kg during a constant volume process. Determine the amount of heat energy required to bring about this increase for 2 kg of fluid. Solution to Example 5.2 The non flow energy equation is Q – W = U2 – U1 For a constant volume process W=0 and the equation becomes Q = U2 – U1 ∴ Q = 180 – 120 = 60 kJ/kg Therefore for 2 kg of fluid Q = 60 x 2 = 120 kJ i.e. 120 kJ of heat energy would be required.
16. 16. NON-FLOW PROCESS J2006/5/16 Example 5.3 2.25 kg of fluid having a volume of 0.1 m3 is in a cylinder at a constant pressure of 7 bar. Heat energy is supplied to the fluid until the volume becomes 0.2 m3. If the initial and final specific enthalpies of the fluid are 210 kJ/kg and 280 kJ/kg respectively, determine a) the quantity of heat energy supplied to the fluid b) the change in internal energy of the fluid Solution to Example 5.3 Data: p = 7.0 bar; V1 = 0.1 m3 ; V2 = 0.2 m3 a) Heat energy supplied = change in enthalpy of fluid Q = H2 – H1 = m( h2 - h1 ) = 2.25( 280 – 210 ) = 157.5 kJ b) For a constant pressure process W = P(V2 – V1) = 7 x 105 x ( 0.2 – 0.1) = 7 x 104 J = 70 kJ Applying the non-flow energy equation Q – W = U2 – U1 gives U2 – U1 = 157.5 – 70 = 87.5 kJ
17. 17. NON-FLOW PROCESS J2006/5/17 Activity 5B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 5.4 The pressure of the gas inside an aerosol can is 1.2 bar at a temperature of 25o C. Will the aerosol explode if it is thrown into a fire and heated to a temperature of 600o C? Assume that the aerosol can is unable to withstand pressure in excess of 3 bar. 5.5 0.05 kg of air, initially at 130o C is heated at a constant pressure of 2 bar until the volume occupied is 0.0658 m3. Calculate the heat supplied and the work done. 5.6 A spherical research balloon is filled with 420 m3 of atmospheric air at a temperature of 10o C. If the air inside the balloon is heated to 80oC at constant pressure, what will be the final diameter of the balloon?
18. 18. NON-FLOW PROCESS J2006/5/18 Feedback To Activity 5B 5.4 Data: p1 = 1.2 bar; T1= 25 + 273 = 298 K T2 = 600 + 273 = 873 K; p2 = ? We can idealize this process at constant volume heating of a perfect gas. Applying the general property relation between states 1 and 2 p1V1 p 2V2 = T1 T2 in this case V2 = V1 p1 p 2 Hence, = T1 T2 T or p 2 = p1 2 T2 873 = 1.2 x 298 = 3.52 bar Since the aerosol cannot withstand pressures above 3 bar, it will clearly explode in the fire.
19. 19. NON-FLOW PROCESS J2006/5/19 5.5 Data: m = 0.5 kg; p = 2 bar; V2 = 0.0658 m3; T1 = 130 + 273 =403 K Using the characteristic gas equation at state 2 p 2V2 T2 = mR 2 x 10 5 x 0.0658 = 0.05 x 0.287 x 10 3 = 917 K For a perfect gas undergoing a constant pressure process, we have Q = mcp(T2 – T1) i.e. Heat supplied = 0.05 x 1.005(917 – 403) = 25.83 kJ W = p (V2 – V1) From equation pV = RT ∴ Work done = R (T2 – T1) = 0.287(917 – 403) i.e. Work done by the mass of gas present = 0.05 x 0.287 x 514 = 7.38 kJ
20. 20. NON-FLOW PROCESS J2006/5/20 5.6 Data: T1 = 10 + 273 = 283 K; T2 = 80 + 273 = 353 K V1 = 420 m3; V2 = ? Applying the general property relation between states 1 and 2 p1V1 p 2V2 = T1 T2 Since the air is heated at constant pressure p1 = p2 Then, V1 V2 = T1 T2 T2 or V2 = V1 T1 353 = 420 x 283 = 523.9 m3 4 3 Since the balloon is a sphere, V2 = πr 3 where r = radius of the balloon Hence, 4 3 523.9 = πr 3 Solving gives r=5m Final diameter of balloon, d = 2r = 2 x 5 = 10 m
21. 21. NON-FLOW PROCESS J2006/5/21 SELF-ASSESSMENT You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self- Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck. 1. A receiver vessel in a steam plant contains 20 kg of steam at 60 bar and 500oC. When the plant is switched off, the steam in the vessel cools at constant volume until the pressure is 30 bar. Find the temperature of the steam after cooling and the heat transfer that has taken place. 2. 0.25 kg of combustion gas in a diesel engine cylinder is at temperature of 727oC. The gas expands at constant pressure until its volume is 1.8 times its original value. For the combustion gas, R = 0.302 kJ/kgK and cp = 1.09 kJ/kgK. Find the following: a) temperature of the gas after expansion b) heat transferred c) work transferred 3. A quantity of gas has an initial pressure and volume of 0.1 MN/m2 and 0.1 m3, respectively. It is compressed to a final pressure of 1.4 MN/m2 according to the law pV1.26 = constant. Determine the final volume of the gas. 4. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is compressed polytropicly at 7 bar following the law pV1.25 = C. Determine the following: a) Intial volume b) final volume c) work transfer d) heat transfer e) change in internal energy
22. 22. NON-FLOW PROCESS J2006/5/22 Feedback to Self-Assessment Have you tried the questions????? If “YES”, check your answers now. 1. 233.8oC; 14380 kJ rejected 2. 1527oC; 218 kJ added; 60.4 kJ output 3. 0.01235 m3 4. 158.9oC; 12390 cm3; 6.82 kJ input; 2.56 kJ rejected; 4.26 kJ increase CONGRATULATION S!!!!…May success be with you always…