5.4 DESIGN OF PRE-PURIFICATION UNIT Design of Pre-cooler (Brazed Aluminum Plate Fin Heat Exchanger) The hot air comes into the precooler at temperature of 390°C and is expected to be cooled to 8°C....

This is the question that i want to know from where the volume flow rate of N2 is come thats value is equal to 0.0000449m3/s5.4 DESIGN OF PRE-PURIFICATION UNIT<br>Design of Pre-cooler (Brazed Aluminum Plate Fin Heat Exchanger)<br>The hot air comes into the precooler at temperature of 390°C and is expected to be<br>cooled to 8°C. the cooled fluid, that is, the coolant used in this design is cold<br>nitrogen at temperature of -195°C.<br>Assumptions;<br>* Heat loss by hot air equals heat gained by cold nitrogen.<br>Given that; øhin = 390°C, Øhout = 8°C, Øcin = -195°C<br>ṁ, = 131.4165 Kgmol<br>hr<br>1hr<br>= 0.0365 Kgmol/s<br>3600s<br>Cpair = 1005J/KgK, CpN,= 1040J/KgK,<br>Density of nitrogen = 1251Kg/m³<br>Volume flow rate of N2 = 0.0000449m³/s<br>Recall that, mass flow rate (m) = density (p) x volume flow rate (v)<br>Hence, m. = 1251 × 0.0000449 = 0.0562Kg/s<br>%3D<br>Qn = m, × Cpair* (Øhin - Øhout)..<br>.3<br>Qu = 0.0365 × 1005 × (390-8) = 14012.715J/s<br>But Qh = Qc<br>Q. = 3668.25 = m, x CPN,× (Øcout - Øcin)<br>14012.715 = 0.0562 × 1040 x (Øcout

Extracted text: 5.4 DESIGN OF PRE-PURIFICATION UNIT Design of Pre-cooler (Brazed Aluminum Plate Fin Heat Exchanger) The hot air comes into the precooler at temperature of 390°C and is expected to be cooled to 8°C. the cooled fluid, that is, the coolant used in this design is cold nitrogen at temperature of -195°C. Assumptions; * Heat loss by hot air equals heat gained by cold nitrogen. Given that; øhin = 390°C, Øhout = 8°C, Øcin = -195°C ṁ, = 131.4165 Kgmol hr 1hr = 0.0365 Kgmol/s 3600s Cpair = 1005J/KgK, CpN,= 1040J/KgK, Density of nitrogen = 1251Kg/m³ Volume flow rate of N2 = 0.0000449m³/s Recall that, mass flow rate (m) = density (p) x volume flow rate (v) Hence, m. = 1251 × 0.0000449 = 0.0562Kg/s %3D Qn = m, × Cpair* (Øhin - Øhout).. .3 Qu = 0.0365 × 1005 × (390-8) = 14012.715J/s But Qh = Qc Q. = 3668.25 = m, x CPN,× (Øcout - Øcin) 14012.715 = 0.0562 × 1040 x (Øcout" - 195) 3668.25 = 239.7467 = Øcout + 195 58.448 Øcout = 44.75°C Øhin = 390°C Øcout = 44.75°C Øhout = 8°C :
Ivi<br>Material Balance for Pre- Purification Unit (PPU)<br>Pre-Cooler<br>Absorber<br>In<br>Out<br>Air<br>Air<br>TABLE 4.2: SUMMARY OF MATERIAL BALANCE ON THE COMPRESSOR<br>Components Amount Mol Wt<br>INPUT<br>OUTPUT<br>(%)<br>(Kg/mol) mol(kmol) mass (kg) mol(Kmol) mass (kg)<br>Nitrogen<br>102.6465 2972.643<br>102.6465 2972.643<br>78<br>28<br>Oxygen<br>21<br>32<br>27.6356 800,3271<br>27.6356 800.3271<br>Argon<br>40<br>1.3159<br>38.1108<br>1.3159<br>38.1108<br>131.598 3811.081<br>131.598 3811.081<br>Total<br>100<br>Ivii<br>Material balance of the w<br>

Extracted text: Ivi Material Balance for Pre- Purification Unit (PPU) Pre-Cooler Absorber In Out Air Air TABLE 4.2: SUMMARY OF MATERIAL BALANCE ON THE COMPRESSOR Components Amount Mol Wt INPUT OUTPUT (%) (Kg/mol) mol(kmol) mass (kg) mol(Kmol) mass (kg) Nitrogen 102.6465 2972.643 102.6465 2972.643 78 28 Oxygen 21 32 27.6356 800,3271 27.6356 800.3271 Argon 40 1.3159 38.1108 1.3159 38.1108 131.598 3811.081 131.598 3811.081 Total 100 Ivii Material balance of the w
Jun 07, 2022
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