Module 05 Lecture 35 : Low temperature process design Key words: Refrigeration system

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1 Refrigeration System Module 05 Lecture 35 : Low temperature process design Key words: Refrigeration system A refrigerator is simply a heat pump where heat is rejected to atmosphere( the sink). Fig.41.3(lecture 41) shows closed cycle system and Fig.41.5(lecture 41) shows the absorption refrigeration system. Since, absorption refrigeration system is more complex and expensive, it is generally, not integrated with processes. The grand composite curve, once again, can give the clue as to when the absorption system might be favored over the compression system. The absorption system requires a large aboveambient heat input. If there is sufficient waste heat from the process below the pinch at a sufficiently high temperature is available, then the absorption system could be run completely on free energy, whereas the competing compression refrigeration system has to run on expensive imported power. Conversely, compression refrigeration will be favoured if there is little waste heat available, or the pinch temperature is close to ambient, or the refrigeration load is required well below ambient, or a CHP system can be installed above the pinch. Refrigeration systems tend to be the most expensive of all site utilities per unit of heat load. The reason for this can easily be understood from the Carnot efficiency. The work required to absorb heat from the below ambient heat source is given by a rearrangement of Eq The upper temperature, T 1 is usually ambient temperature, but as the lower temperature T 2 is below ambient, the value of W rises exponentially and becomes infinite as T 2 approaches to absolute zero. Hence, the power consumed in refrigeration rises sharply as the required refrigeration temperature falls. Further, for running a refrigeration system the T 1 has to be more than ambient temperature to reject heat in to it and similarly the Temperature of evaporator should be lower by an amount T( about 5 C) to absorb heat at T 2 temperature. For a typical refrigeration system with a mechanical efficiency of 50% (most losses occur in the compressor), to remove 2 kw of heat from a process at 0 C already requires 0.44 kw power. For refrigeration system, the energy and cost penalty increase sharply as temperature falls. Thus, any possibility to increase the energy efficiency of a refrigeration system are worth investigation. Since process cooling is carried out by evaporating the liquid refrigerant, these represent on the GCC as a series of constant temperature cold utility levels ( for multi level refrigeration system), and loads and levels are matched in the usual way.

2 BShifted Temp., T C Low temperature process design Module 06 Lecture 35 One way of reducing the power required by the simple cycle in Fig.35.1(a) is to incorporate an economiser, as shown in the figure. The compression and flash expansion are split into two stages, with flash vapour from the first expansion stage is returned to the suction of the highest pressure compressor( Comp. H). In this way, the quantity of vapour flowing through the low pressure compressor(comp. L) of the system is reduced, saving power. T AMBIENT Pinch point C liquid Cooling water Comp. H Vapor E D Vapor+Liquid Comp. L H R 1 2 Vapor + Liquid Vapor Process stream F G 1 2 H R (a) (b) H Fig (a) schematic of the refrigeration system and (b) GCC of process Matching of refrigeration cycles against the grand composite curve is illustrated in Fig.35.1(b). The GCC has a pinch temperature little below the cooling water temperature marked by B on the GCC. Thus the cooling duty below the cooling water inlet temperature should be handled by the refrigeration system. The GCC profile below the pinch is shown by BCDEF and the required cooling load of the GCC is H R which is serviced by a single level refrigeration system. A close analysis of the system shows that for EG section this arrangement is good. For section BC of GCC the supplied driving force is much higher than required. Thus substantial power saving can be achieved if the single level refrigeration system is converted into a two level refrigeration system which adds a refrigeration level for BC section of GCC separately as shown in Fig.35.2.

3 3 Process stream Vapor+Liquid Cooling water liquid Comp. H Vapor 4 H R 1 2 Vapor + Liquid Vapor Process stream T AMBIENT Comp. L B Shifted Temp., T C Pinch point C 3 4 E D F 2 H R (a) (b) Fig (a) Two level refrigeration system and (b) GCC of process G 1 H Shifting load upwards ( load of BC section) in temperature level reduces vapour flow to the compressor(comp. L) servicing the low pressure part of the refrigeration cycle. The net result, however, is a substantial reduction in power consumption. Now is it possible to further reduce the power cost of the refrigeration system by increasing the level of refrigeration in the present system. A close examination of process GCC shows that this can be done by exploiting the pocket( represented by hatch area) in the GCC. The concept is shown in Fig.35.3.

4 BShifted Temp., T C Low temperature process design Module 06 Lecture Vapor+Liquid Process stream liquid Cooling water Vapor T AMBIENT Pinch point C Process stream E D F G 2 1 Vapor+Liquid Vapor 7 Process 8 stream (b) H H R 1 2 Vapor + Liquid Process stream Vapor (a) Fig (a) Two level refrigeration system and (b) GCC of process A further power saving can be obtained if some part of the heat absorbed in 1 2 is rejected at 7 8. This will save energy as the temperature level of 7 8 is below ambient. The rest part of 1 2 can be rejected at ambient temperature. The savings are more than maintaining a low temperature level load at 5 6. However, do this the complexity of the system has gone from Fig.35.1(a) to Fig.35.3(a). The economics of this entirely depends on the refrigeration temperature levels and their loads. To show the temperature lifts used in Fig.35.1 to Fig.35.3, Fig.35.4 is plotted. Ambient Temperature Amount of lift For Fig.35.1(a) For Fig.35.2(a) For Fig.35.3(a) Fig.354 Temperature lift shown for Figs.35.1, 35.2 and 35.3

5 Illustrative Example Problem: Determine the refrigeration requirement of the low temperature distillation process given in Table 35.1 for ΔT min = 5 0 C Table 35.1 Stream data for the illustrative problem Stream Type T S ( 0 C) T T ( 0 C) CP (MW/ K) 1 Hot Hot Hot Cold Cold Cold Cold a) Plot the grand composite curve and determine the temperature and duties of the refrigeration if two levels of refrigeration are to be used. Assume isothermal vaporization and condensation of the refrigerant. b) Calculate the power requirements for the refrigeration for heat rejection to cooling water operating between 15 and 20 0 C approximated by Equation given below: Approximate COP Q W Where, T = Evaporation Temperature, 0 C T = Condensation Temperature, 0 C W = Refrigeration power requirement, kw COP = Coefficient of performance Q C = Cooling duty 0.6 T T T c) Repeat the calculation from Part b using refrigeration power targeting, assuming propylene as the refrigerant and a reciprocating compressor

6 d) Heat rejection from the refrigeration system into the process can be used to reduce the refrigeration power requirements. Calculate the power using above equation. e) Repeat the calculation from Part d using refrigeration power targeting, assuming propylene as the refrigerant and a reciprocating compressor. Solution: The grid diagram of the above problem is shown in Fig.35.5 H CP (MW/ K) Fig.35.5 Grid diagram of the illustrative problem The corresponding problem table cascade and GCC are shown in Fig & Fig respectively. Fig.35.6 GCC of the problem

7 a.) The two levels of refrigeration are shown in Fig with the details given in Table 35.2 Fig.35.7 Problem table cascade

8 1.13 MW 0.9 MW Fig.35.8 Two level refrigeration Table 35.2 Levels T * ( 0 C) T ( 0 C) Q EVAP (MW) Level ( 24.5) 246 ( 27) 1.13 MW Level ( 46.5) 224 ( 49) 0.9 MW b.) For heat rejection to cooling water operating between 15 and 20 0 C, Thus, Now, for level 1 refrigeration, T COND = Condensing Temperature = = 25 0 C = = 298 K T EVAP = Evaporation Temperature = 246 K Q C = 1.13 MW Thus,

9 W MW Similarly, for level 2 refrigeration, T COND = Condensing Temperature = = 25 0 C = = 298 K T EVAP = Evaporation Temperature = 224 K Q C = 0.9 MW Thus, Hence, W MW Power requirement = = MW c.) In this part, we will design cycle 1 ( Refrigeration level 1) to operate between 27 0 C to 25 0 C and cycle 2 (Refrigeration level 2) to operate between 49 0 C to 25 0 C. For cycle 1: Using Antoine equation: Now, for propylene, log Thus, A = B = {When P (mmhg) and T ( 0 C)} C = log Also, So, log Pressure ratio = / = (11.658/ 2.378) =

10 Similarly, For cycle 2: Pressure ratio = Thus, Table 35.3 can be generated as follows: Table 35.3 Pressure (bar) Cycle 1 Cycle 2 Evaporator Condenser Pressure Ratio The pressure ratio is a little high for Cycle 2 to be carried out in a single compression stage. However, single stage compression will be assumed for the sake of comparison between different options. Also, in this example, the heat capacity ratio γ will be assumed to be constant with a value of The next step involves the calculation of mass flow rate of the refrigerant by performing an energy balance around the evaporator. Where, H 2 = Specific enthalpy at the evaporator outlet (sat. vapor enthalpy at the evaporator pressure) H 4 = Specific enthalpy at the condenser outlet (sat. liquid enthalpy at the condenser pressure) Thus, for cycle 1,

11 Hence, Also, So, kg/ m Similarly. for cycle 2, Thus, Table 35.4 can be generated as follows: Table 35.4 Cycle 1 Cycle 2 Q EVAP MW H 2 H 1 kj/ kg m Kg/ s ρ v Kg/ m F in m 3 / s The isentropic compressor efficiency, polytrophic coefficient, outlet temperature and power requirement can now be calculated as shown below for cycle 1 & cycle 2. For cycle 1: The isentropic efficiency of the compressor can be calculated by: η IS = (ln r) (ln r) ln r η IS = (ln ) (ln ) ln = *(1.59^3) * (1.59^2) *(1.59)

12 = (0.1091*4.02) (0.5274*2.5281) + (0.8577*1.59) = = The polytrophic coefficient 'n' can be calculated using ln ln 1 / ln The outlet temperature is given by: The power required for compression : Or, 1 1 / Thus, Table 35.5 can be generated for cycle 1 & cycle 2: Table 35.5 Cycle 1 Cycle 2 Isentropic efficiency (η IS ) Polytrophic coefficient (n) Outlet Temperature (T out, K) Power (W, MW) Therefore, Total power for heat rejection to cooling water = = MW d.) Now, consider that part of level 2 heat can be rejected to the process above the pinch as shown in Fig The heat exchangers shown in the figure may be several exchangers in practice. The rejection load = 0.54 MW as clear from the figure.

13 Now, W= MW Therefore, process cooling by level 2 by this arrangement across the pinch = = MW Balance of cooling demand at level 2 = = MW 0.54 MW 1.13 MW 0.9 MW Fig.35.9 Two level refrigeration with heat rejected above pinch This balance of the cooling demand on Level 2 ( MW) together with the load from Level 1 must be either rejected to the process at a higher temperature above the pinch or to cooling

14 water. The process has a heating demand at 23 C, which means that heat could be rejected at 28 C. However, rejection to the process would add to the complexity of both design and operation. Also, there seems little advantage in such an arrangement since the heat can be rejected to cooling water at 30 C, Therefore, the rest of the rejection heat will be assumed to go to the cooling water. Thus, W MW W MW Therefore, total refrigeration power for part rejection of level 2 to the process = = MW Hence, saving in the refrigeration process due to integration with the background process = = MW e.) The duty on cycle 1 will remain unchanged from part (c). Hence, However, the duty on cycle 2 must be reduced from part (c) to compensate by the duty on cycle 3. For cycle 2: Now, the calculation in part (c) will be adjusted to reduce Q EVAP to MW. Thus, m = kg/ s F in = m 3 / s W = 0.25 MW For cycle 3, the cycle will be designated to operate between 224 k ( 49 0 C) and 9 0 C. Thus, the calculations for cycle 3 are given in Table 35.6 below: Table 35.6 Pressure of evaporator (bar) 0.96 Pressure of condenser (bar) 7.6

15 Pressure ratio Q EVAP (MW) H 2 H 4 (kj/ kg) = m (kg/ s) ρ V (kg/ m 3 ) 2.26 F in (m 3 / s) η IS n T out (K) W (MW) 0.14 Q COND (MW) Now, Total refrigeration power for part rejection of Level 2 to the process from refrigeration targeting Hence, = = The saving in the refrigeration power by integration with the process from refrigeration targeting = = MW References 1. vings/sidor/default.aspx 2. Pongsid Srikhirin, Satha Aphornratana and Supachart Chungpaibulpatana, A review of absorption refrigeration technologies, Renewable and Sustainable Energy Reviews, 5 (2001) Xiao Feng and Thore Berntssont, Critical Cop For An Economically Feasible Industrial Heat Pump Application, Applied Thermal Engineering Vol. 17, No. I. pp , 1997