NON-FLOW PROCESS J2006/4/1 UNIT 4NON - 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 describe the differences between the flow and the non-flow processes identify heat and work in reversible process define and calculate the following non-flow processes : • constant temperature (Isothermal) • adiabatic
NON-FLOW PROCESS J2006/4/2 INPUT4.0 INTRODUCTION What is a non flow process? O nce a fluid has entered a system, it may be possible for it to undergo a series of processes in which the fluid does not flow. An example of this is the cylinder of an internal combustion engine. In the suction stroke, the working fluid flows into the cylinder in which it is then temporarily sealed. Whilst the cylinder is sealed, the fluid is compressed by the piston moving into the cylinder, after which heat energy is supplied so that the fluid possesses sufficient energy to force the piston back down the cylinder, causing the engine to do external work. The exhaust valve is then opened and the fluid is made to flow out of the cylinder into the surroundings. Processes which are undergone by a system when the working fluid cannot cross the boundary are called non-flow process. This process occurs during the compression and the working stroke as mentioned in the above example (refer to Fig. 4.0).4.1 Differences Between The Flow and Non-flow processes 4.1.1 Flow Process SUCTION WORKING STROKE COMPRESSION STROKE EXHAUST STROKE STROKE Figure 4.0 The cycle of an internal combustion engine
NON-FLOW PROCESS J2006/4/3 In an open system, not only the energy transfers take place across the boundary, the fluid may also cross the boundary. Any process undergone by an open system is called a flow process. This process may be sub-divided into an unsteady flow process and steady flow process. The general equation is shown below, C12 C2 gZ1 + u1 + P1v1 + + Q = gZ 2 + u 2 + P2 v 2 + 2 + W 2 2 4.1.2 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 the non-flow process. If the fluid is undergoing a non-flow process from state (1) to state (2) then the terms from the general equation for p1V1 and p2V2 (which represent the amount of work energy required to introduce and expel the fluid from the system) will be zero, since the fluid is already in the system, and will still be in the system at the end of the process. For the same reason, the changes in kinetic and potential energies of the fluid will also be zero. Thus the equation becomes U1 + Q = U2 + W or, U2 – U1 = Q –W (4.1) 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.4.2 Constant temperature (Isothermal) process (pV = C) If the change in temperature during a process is very small then that process may be approximated as an isothermal process. For example, the slow expansion or compression of fluid in a cylinder, which is perfectly cooled by water may be analysed, assuming that the temperature remains constant.
NON-FLOW PROCESS J2006/4/4 P 1 W 2 W v v1 Q v2 Figure 4.2 Constant temperature (Isothermal) process The general relation properties between the initial and final states of a perfect gas are applied as follows: p1V1 p 2V2 = T1 T2 If the temperature remains constant during the process, T1 = T2 and the above relation becomes p1V1 = p 2V2 From the equation we can know that an increase in the volume results in a decrease in the pressure. In other words, in an isothermal process, the pressure is inversely proportional to the volume. Work transfer: Referring to the process represented on the p – V diagram in Fig.4.2 it is noted that the volume increases during the process. In other words the fluid is expanding. The expansion work is given by 2 W = ∫ pdV 1 2 c =∫ dV (since pV = C, a constant) 1 V
NON-FLOW PROCESS J2006/4/5 2 dV = c∫ 1 V 2 dV = p1V1 ∫ 1 V V2 larger volume = p1V1 ln smaller volume V1 V2 = mRT1 ln (since p1V1 = mRT1) V1 p1 V2 p = mRT1 ln (since = 1) (4.2) p2 V1 p2 Note that during expansion, the volume increases and the pressure decreases. On the p – V diagram, the shaded area under the process line represents the amount of work transfer. Since this is an expansion process (i.e. increasing 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. Heat transfer: Energy balance to this case is applied: U1 + Q = U2 + W For a perfect gas U1 = mcvT1 and U2 = mcvT2 As the temperature is constant U1 = U2 Substituting in the energy balance equation, Q=W (4.3) Thus, for a perfect gas, all the heat added during a constant temperature process is converted into work and the internal energy of the system remains constant. For a constant temperature process W==Q or W==Q
NON-FLOW PROCESS J2006/4/6 4.3 Adiabatic process (Q = 0) If a system is thermally well insulated then there will be negligible heat transfer into or out of the system. Such a system is thermally isolated and a process within that system may be idealised as an adiabatic process. For example, the outer casing of steam engine, steam turbines and gas turbines are well insulated to minimise heat loss. The fluid expansion process in such machines may be assumed to be adiabatic. P 1 W 2 W v v1 Thermal insulation v2
NON-FLOW PROCESS J2006/4/7 Figure 4.3 Adiabatic (zero heat transfer) process For a perfect gas the equation for an adiabatic process is pVγ = C Cp where γ = ratio of specific heat = Cv The above equation is applied to states 1 and 2 as: p1V1γ = p 2V2γ γ p 2 V1 = (4.4) p1 V2 Also, for a perfect gas, the general property relation between the two states is given by the equation below p1V1 p 2V2 = (4.5) T1 T2 By manipulating equations 4.4 and 4.5 the following relationship can be determined: γ −1 γ −1 T2 p 2 γ V (4.6) = = 1 T1 p1 V2 By examining equations 4.4 and 4.6 the following conclusion for an adiabatic process on a perfect gas can be drawn: 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.4.3) it is noted that the volume increases during the process.
NON-FLOW PROCESS J2006/4/8 In other words, the fluid expanding and the expansion work is given by the formula: 2 W = ∫ pdV 1 2 c =∫ γ dV (since pVγ = C, a constant) 1V 2 dV = c∫ γ 1 V p1V1 − p 2V2 = [larger pV- small pV] (4.7) γ −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. As 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 4.3). Heat transfer: In an adiabatic process, Q = 0. Applying an energy balance to this case (Fig.4.3): U1 - W = U2 W = U1 – U2 Thus, in an adiabatic expansion the work output is equal to the decrease in internal energy. In other words, because of the work output the internal energy of the system decreases by a corresponding amount. For a perfect gas, U1 = mcvT1 and U1 = mcvT1 On substitution W = mcv(T1-T2) [larger T- smaller T] (4.8) We know cp- cv = R or R cv = γ −1
NON-FLOW PROCESS J2006/4/9 Substituting in equation 4.8 mR (T1 − T2) W = (4.9) γ −1 But, mRT2 = p2V2 and mRT1 = p1V1 Then the expression for the expansion becomes p1V1 − p 2V2 W = (4.10) γ −1 Referring to the process represented on the p-V diagram it is noted that during this process the volume increases and the pressure decreases. For a perfect gas, equation 4.6 tells that a decrease in pressure will result in a temperature drop. Example 4.1 In an industrial process, 0.4 kg of oxygen is compressed isothermally from 1.01 bar and 22o C to 5.5 bar. Determine the work done and the heat transfer during the process. Assume that oxygen is a perfect gas and take the molecular weight of oxygen to be M = 32 kg/kmole. Solution to Example 4.1 Data: m = 0.4 kg; p1= 1.01 bar; t1= 22oC p2 = 5.5 bar; W=? Q=? From the equation R R= 0 M 8314 = 32 = 260 J/kgK = 0.260 kJ/kgK
NON-FLOW PROCESS J2006/4/10 For an isothermal process Work input, p2 W = mRTln p1 5.5 = 0.4 x 0.260 x ( 22 + 273) ln 1.01 = 52 kJ In an isothermal process all the work input is rejected as heat. Therefore, heat rejected, Q = W = 52 kJ Example 4.2 In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20oC is compressed into one sixth of its original volume. Determine the pressure and temperature of the air after compression. If the compressor cylinder contains 0.05 kg of air, calculate the required work input. For air, take γ = 1.4 and cv = 0.718 kJ/kgK. Solution to Example 4.2 Data : p1 = 0.98 bar; T1= 20 + 273 = 293 K V2 1 = ; m = 0.05 kg; W = ? V1 6 As the cylinder is well insulated the heat transfer is negligible and the process may be treated as adiabatic. Considering air as a perfect gas γ p 2 V1 From equation 4.4, = p1 V2 p2 = 0.98 x 61.4
NON-FLOW PROCESS J2006/4/11 = 12 bar γ −1 T2 V 1 From equation 4.6, = T1 V2 T2 = 293 x 60.4 = 600 K = 327oC Re-writing equation 4.8 for an adiabatic compression process W = mcv(T2-T1) [larger T- smaller T] = 0.05 x 0.718 (600-293) = 11 kJ
NON-FLOW PROCESS J2006/4/12 Activity 4 TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 4.1 In the cylinder of a large engine, 1.0 kg of carbon dioxide at 527o C and 20 bar expands isothermally to a pressure of 1.4 bar. What is the final volume of the gas? Take R = 189 Nm/kgK for carbon dioxide. 4.2 1 kg of nitrogen (molecular weight 28) is compressed reversibly and isothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and the heat flow during the process. Assume nitrogen to be a perfect gas. 4.3 Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015 m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate the final temperature, the final volume, and the work done on the mass of air in the cylinder.
NON-FLOW PROCESS J2006/4/13 Feedback to Activity 4 4.1 Data: m = 1.0 kg; T1= 527 + 273 = 800 K p1 = 20 bar; p2= 1.4 bar; V2 = ? Carbon dioxide is a perfect gas and we can apply the following characteristic gas equation at state 1. p1V1 = mRT1 mRT1 V1 = p1 1x189 x800 = 20 x10 5 = 0.0756 m3 Applying the general property relation between state 1 and 2 p1V1 p 2V2 = T1 T2 For an isothermal process T1 = T2 Hence, p1V1 = p 2V2 20 V2 = x 0.0756 1 .4 V2 = 1.08 m3
NON-FLOW PROCESS J2006/4/14 4.2 Data: m=1kg; M= 28 kg/kmole p1 = 1.01 bar; T1 = 20 + 272 = 293 K; p2 = 4.2 bar From equation R R= 0 M 8.314 = 28 = 0.297 kJ/kgK The process is shown on a p-v diagram below. When a process takes place from right to left on a p-v diagram the work done by the fluid is negative. That is, work is done on the fluid. p 4.2 pV=C 1.01 v From equation 4.2 p1 W = mRT1 ln p2 1.01 = 1 x 0.297x293x ln 4.2 = -124 kJ/kg For an isothermal process for a perfect gas, Q = W = -124 kJ/kg
NON-FLOW PROCESS J2006/4/15 4.3 Data: p1=1.02 bar; T1=22 + 273 = 295 K; v1= 0.015 m3; p2= 6.8 bar From equation 4.6 γ −1 T2 p 2 γ = T1 p1 (1.4 −1) / 1.4 6.8 T2 = 295 x 1.02 = 507.5 K (where γ for air = 1.4) i.e. Final temperature = 507.5 – 273 = 234.5oC From equation 4.4 γ 1/ γ p 2 V1 v1 p 2 = or = p1 V2 v 2 p1 1 / 1.4 0.015 6.8 ∴ = v2 1.02 i.e. Final volume v2 = 0.0038 m3 For an adiabatic process, W = u1 – u2 and for a perfect gas, W = cv(T1- T2) = 0.718(295-507.5) = - 152.8 kJ/kg i.e. work input per kg = 152.8 kJ The mass of air can be found using equation pV = mRT p1v1 1.02 x10 5 x0.015 ∴ m= = = 0.018kg RT1 0.287 x10 3195 i.e. Total work done = 0.0181 x 152.8 = 2.76 kJ
NON-FLOW PROCESS J2006/4/16 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. 0.05 m3 of a perfect gas at 6.3 bar undergoes a reversible isothermal process to a pressure of 1.05 bar. Calculate the heat flow to or from the gas. 2. Nitrogen (molecular weight 28) expands reversibly in a perfectly thermally insulated cylinder from 2.5 bar, 200oC to a volume of 0.09 m3. If the initial volume occupied was 0.03 m3, calculate the work done during the expansion. Assume nitrogen to be a perfect gas and take cv = 0.741 kJ/kg K. 3. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is compressed adiabatically to 5 bar. Determine the following: a) final temperature b) final volume c) work transfer d) heat transfer e) change in internal energy 4. A quantity of gas occupies a volume of 0.3 m3 at a pressure of 100 kN/m2 and a temperature of 20oC. The gas is compressed isothermally to a pressure of 500 kN/m2 and then expanded adiabatically to its initial volume. For this quantity of gas determine the following: a) the heat received or rejected (state which) during the compression, b) the change of internal energy during the expansion, c) the mass of gas.
NON-FLOW PROCESS J2006/4/17 Feedback to Self-Assessment Have you tried the questions????? If “YES”, check your answers now. 1. 56.4 kJ 2. 9.31 kJ 3. 222.7oC, 14230 cm3, 6.56 kJ input, 0 kJ, 6.56 kJ increase. 4. – 48.3 kJ (heat rejected), -35.5 kJ, 0.358 kg CONGRATULATIONS!!!! You can now move on to Unit 5…