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Earlier Lecture Basics o Rerigeration/Liqueaction, coeicient o perormance and importance o Carnot COP. Throttling, heat exchanger, compression/expansion systems. Deinition o a rerigerator, liqueier and a combination o these two systems. J T expansion is an isenthalpic process and initial temperature o the gas should be less than T INV to have a cooling eect. T For an ideal gas = 0 p. Hence, µ JT = 0. h 2
Outline o the Lecture Topic : Gas Liqueaction and Rerigeration Systems (contd) J T expansion o a real gas Adiabatic expansion Comparison o J T and Adiabatic expansions Ideal Thermodynamic Cycle 3
Introduction We know that work input is needed to generate and maintain low temperatures. As T L decreases, Carnot COP decreases stating that more work input is required to maintain very low temperatures. Hence, a knowledge o perormance o various rerigeration/liqueaction cycles is necessary to design the system with maximum COP. 4
Joule Thompson Coeicient Enthalpy (h) is the sum o the internal energy (u) and pv work. h = u + pv µ µ JT JT T T h = = p h p h p T Substitution o Enthalpy (h) in above expression o J T coeicient, we get ( pν ) 1 u = + cp p p T T In order that the J T expansion results in cooling, the bracket should be negative. 5
Joule Thompson Coeicient µ JT ( pν ) 1 u = + cp p p T T The irst term represents the departure rom Joule s law. At low pressures, the molecules are pulled apart. This results in increase in the potential energy. As a result, the kinetic energy decreases to keep the total energy constant and hence the temperature. Thereore, the irst term is always negative or a real gas. 6
Joule Thompson Coeicient µ JT ( pν ) 1 u = + cp p p T T The second term can either be positive, negative or zero. It represents the departure rom the Boyle's law. At high pressures, the molecules are squeezed together and hence, are less compressible than the Boyle's law prediction. 7
Joule Thompson Coeicient µ JT ( pν ) 1 u = + cp p p T T The second term is negative at low pressures and low temperatures, where the gases are more compressible than the Boyle s law. For a real gas, J T coeicient depends upon the relative magnitude o both the terms. 8
Equation o State van der Waals equation o state or a real gas is as given below. a p + 2 ( v b) = RT v where, a and b are constants, which gives the measure o intermolecular orces and size o the particles respectively. For an ideal gas both a and b are 0. Rearranging the terms, we get RT a p = v b v ( ) 2 9
Equation o State Upon dierentiating the ollowing equation at constant pressure (only T and v are variables), we get RT a p = v b v ( ) 2 R T RT 2a 0 = + ( v b) v ( v b) 2 v 3 Rearranging the terms, we have R v ( v b = ) T RT 2a p v b v p ( ) 2 3 10
Equation o State v Substituting T p in the J T Coeicient equation, we have ( v b) v = T RT p 2a ( v b) 2 v 3 For real gas, we get µ JT = c p R 2a b 1 b RT ν 2a b 1 1 νrt ν 2 2 µ JT 1 v = T v cp T p For large speciic volumes ν >>> ab, 1 2a µ JT = b c RT p 11
Joule Thompson Coeicient For a real gas with large speciic volumes µ JT 1 2a = b c RT p µ JT >0 <0 =0 Cond. 2a b > RT 2a b < RT 2a b = RT a= b= 0 0 0 0 T T < T > T = 2a Rb 2a Rb 2a Rb Ideal 12
Isenthalpic lines on T s Chart p h Typical isenthalpic lines are as shown in T s diagram. The drop in temperature obtained ater isenthalpic expansion at lower temperatures is very high. s This is because the gases are imperect at very low temperatures. 13
Maximum Inversion Temp. 2 h=const 3 1 The igure shows the J T expansion on a T s diagram. When the luid expands rom state 2 to state 3, the temperature rises. s This occurs because the initial temperature at state 2 is above the inversion temperature. 14
Maximum Inversion Temp. Gas T inv (K) Helium 45 Hydrogen 205 Neon 250 Nitrogen 621 Air 603 Argon 794 Oxygen 761 Methane 939 For the gases like He, Hydrogen and Neon, in order to experience J T eect, they have to be precooled below T INV. While the other gases show J T cooling when expanded at room temperature. 15
Isentropic Expansion Enthalpy (h) and Entropy (s) are the two thermodynamic state properties o matter which are unctions o pressure and temperature. When the gases are expanded at constant enthalpy, as in a J T expansion, it is called as an Isenthalpic expansion. On the similar lines, when the high pressure gases are expanded at constant entropy, it is called as an Isentropic expansion or a Reversible Adiabatic expansion. 16
Isentropic Expansion The commonly used expansion devices are turbo expanders and reciprocating expanders. This is a work producing process as shown in the schematics. W e T The ratio is called as an p Isentropic Expansion Coeicient. s W e 17
The enthalpy (s) is a unction o both pressure (p) and temperature (T). s= ( pt, ) Using the calculus, the ollowing can be derived. s p T = 1 p T s s p T Isentropic Expansion Rearranging the terms, we have T T s µ s = = p s p s p T 18
Isentropic Expansion For the same variables, entropy (s), temperature (T) and pressure (p), using the calculus, we can arrive at the ollowing. s s ds = dt + dp T p s s Tds = T dt + T dp T p p T p T c p ν T p Maxwell s Equation µ s T T s = = p s p s p T µ s T v = + c T p p 19
Isentropic Expansion For an ideal gas, the equation o state is ν = RT p Dierentiating w.r.t T at constant p, we get On substitution, we get T ν T ν ν µ s = = = c T c T c p p p p ν R ν = = T p T p For an ideal gas µ s 0, unlike the case in the J T expansion µ JT = 0. It means that the ideal gas does exhibit a cooling eect, when it undergoes an isentropic expansion. 20
Isentropic Expansion T v µ s = + c T The derivative term represents the variation o volume with temperature at constant pressure. p p This term is called as the volumetric coeicient and is always positive and hence the isentropic expansion coeicient. It is clear that the isentropic expansion results in cooling irrespective o its initial state, unlike the J T expansion. 21
Equation o State As derived in the earlier slide, dierentiating the van der Waals equation, we get RT v ( v b) = T RT 2a p v b v µ s Substituting in, we get µ s = c p ( ) 2 3 b ν 1 ν 2a b 1 1 νrt ν 2 22
For real gas µ s = c p Equation o State b ν 1 ν 2a b 1 1 νrt ν For large speciic volumes ν >>> µ = JT ab, v c p 2 23
Comparative Study J T Expansion Adiabatic Expansion It has a condition o T INV. No such condition exists. It produces no work. This is an Internal Work process. The device is simple in construction. Normally used or a phase change o luids. The clogging o constriction is a disadvantage. It produces work. This is an External Work process. The device involve complex mechanisms. Normally used or a single phase luids. Regular maintenance and periodic checks are required. 24
Gas Liqueaction Systems System Thermodynamically Ideal System Linde Hampson System Precooled Linde Hampson System Linde Dual Pressure System Claude System Kapitza System Heylandt System Collins System 25
Thermodynamic Ideal System m 1 2 W c Q R The salient eatures o this system are as ollows. All the gas that is compressed, gets liqueied. m W e All the processes are ideal in nature and there are no irreversible pressure drops. Process o compression and expansion are isothermal and isentropic respectively. 26
Thermodynamic Ideal System Q R m 1 2 2 1 W c T W e m g s The initial condition 1 o the gas determines the position o point. 27
Thermodynamic Ideal System Q R m 1 2 It is an open thermodynamic system because the working luid lows across the system. W c Consider a control volume or this system as shown in the igure. m W e 1 st Law o Thermodynamics is applied to analyse the system. The changes in the velocities and datum levels are assumed to be negligible. 28
Thermodynamic Ideal System m 1 2 W c Q R The quantities entering and leaving the system are as given below. IN m 1 @ 1 OUT Q R W e W c W e m 1 @ Using 1 st Law, we get m E in = E out mh + W = Q + W + m h 1 1 c R e 29
Thermodynamic Ideal System m 1 2 W c Q R The work W e produced by the expander is negligible as compared to other terms. mh + W = Q + W + m h 1 1 c R e W e Rearranging the terms, we have ( ) Q W = m h h R c 1 1 m The compression process is assumed to be isothermal. 30
Thermodynamic Ideal System m 1 2 W c Q R W e Hence, rom the Second Law o Thermodynamics, we can write R ( ) Q = mt s s 1 1 2 1 Also, the expansion process is an isentropic process. Thereore, s 2 =s. m By substitution, ( ) ( ) W = mt s s m h h c 1 1 1 1 1 31
Thermodynamic Ideal System m 1 2 W c Q R This work o compression is done on the system. Hence, the value is expressed as a negative quantity. ( ) ( ) W = mt s s m h h c 1 1 1 1 1 m W e Work required per unit mass o the gas compressed is given by Wc T 1 s 1 s h 1 h m 1 ( ) ( ) = 32
Thermodynamic Ideal System Q R m 1 2 Since in an ideal system, mass o gas compressed is same as mass o gas liqueied, m 1 =m. W c Work required per unit mass o the gas liqueied is given by m W e Wc T 1 s 1 s h 1 h m ( ) ( ) = Work required per unit mass o the gas is dependent on the initial condition o the gas. 33
Tutorial 1 Determine the ideal work requirement or liqueaction o nitrogen beginning at 1 bar pressure and 300 K. Step 1 The T s diagram or an ideal thermodynamic cycle is as shown T 2 1 s g 34
Tutorial 1 Step 2 The state properties at dierent points are as given below. p T h s (bar) (K) (J/g) (J/gK) 1 1 300 462 4.42 1 77 29 0.42 T 2 1 g s 35
Step 3 Substitution into the equation. Wi T 1 s 1 s h 1 h m ( ) ( ) Tutorial 1 p T h s (bar) (K) (J/g) (J/gK) 1 1 300 462 4.42 1 77 29 0.42 = = 300(4.42 0.42) (462 29) = 767 J / g T 2 1 s g 36
Tutorial 2 Calculate the ideal work requirement or liqueaction o Helium and Hydrogen beginning at 1 bar pressure and 300 K. Compare the results. Step 1 The T s diagram or an ideal thermodynamic cycle is as shown T 2 1 s g 37
Step 2 Tutorial 2 The state properties or Hydrogen and Helium at dierent points are as given below. p T h s (bar) (K) (J/g) (J/gK) Hydrogen 1 1 300 4190 65 1 20-75 18 Helium 1 1 300 1575 31.5 1 4.2 9.5 3.45 38
Step 3 Substitution into the equation. Tutorial 2 p T h s (bar) (K) (J/g) (J/gK) Hydrogen 1 1 300 4190 65 1 20-75 18 Helium 1 1 300 1575 31.5 1 4.2 9.5 3.4 = Wi T 1 s 1 s h 1 h m H 2 ( ) ( ) = 300(65 18) (4190 + 75) = 9835 J / g He = 300(31.5 3.4) (1575 9.5) = 6864.5 J / g 39
Ideal Work Requirement Gas Normal Boiling Point (K) Ideal Work (kj/kg) Helium 4.21 6819 Hydrogen 20.27 12019 Nitrogen 77.36 768.1 Air 78.8 738.9 Argon 87.28 478.6 Oxygen 90.18 635.6 Ammonia 239.8 359.1 40
Assignment 1. Calculate using the charts, ideal work o liqueaction or Air Oxygen Helium Ammonia 2. Compare the values obtained with the values given in the table 41
Summary For an ideal gas µ JT = 0, but or a real gas, J T coeicient depends upon the relative magnitude o departure rom Joule s Law and Boyle s Law. The gases like nitrogen, air show J T cooling when expanded at room temperature because the T INV is more than room temperature. While the gases like He, Hydrogen and Neon, are to be precooled in order to experience J T eect. 42
Summary In expansion devices like turbo-expanders and expansion engines, the expansion process is isentropic or reversible adiabatic. Coeicient o an isentropic expansion is given by µ s T v = + c T p p The isentropic expansion is always positive or both real and ideal gases. It results in cooling or any initial state, unlike the J T expansion which is dependent on T INV. 43
Summary J T expansion is normally used where phase changes are required, where as isentropic expansion is used or single phase luids. In a thermodynamic ideal system, all the gas that is compressed gets liqueied. The work required per unit mass o the gas compressed and gas liqueied are given by Wc T 1 s 1 s h 1 h m ( ) ( ) = 44
Thank You! 45