02 Process Process Variables - Week 1 Day 1 LATEST.pdf -...

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Any operation or series of operations that causes a physical or chemical change in a substance or mixture of substances Process
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Input or feed Output or product An apparatus in which one of the operation that constitute a process is carried out. Each process unit has a set of input and output process stream Process Unit o Distillation Column o Absorption Column o Evaporator Column o Extraction Column o Reactor
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Properties and conditions of the materials that enter and leave each process unit Examples: o Mass, volume and density o Flow rate o Chemical composition o Concentration o Pressure o Temperature Process Variables
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= density of a substance density of a reference substance = / ref Specific gravity, SG Density, = mass , m (kg) volume, V (m 3 ) Specific volume = volume , V (m 3 ) mass, m (kg) Reference substance = water at 4 o C = 1.0 g/cm 3 1000 kg/m 3 62.43 Ib m /ft 3 Mass and Volume SG can also being obtained from Table B.1
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Calculate the density of mercury (specific gravity at 20 o C = 13.546) in Ib m /ft 3 and calculate the volume in ft 3 occupied by 215 g of mercury. Solution: 845.7 Ib m /ft 3   Hg = 13.546 (62.43 Ib m /ft 3 ) = Specific gravity = 13.546 = / ref = 62.43 Ib m /ft 3 0.454 kg 845.7 Ib m V = 215 kg 1 Ib m 1 ft 3 = 0.560 ft 3 Mass, Volume and Density
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A liquid has a specific gravity of 0.50. What is, (a) its density in g/cm 3 and in Ib m /ft 3 ? (b) its specific volume in cm 3 /g? (c) the mass of 3.0 cm 3 of this liquid? (d) the volume occupied by 18 g of this liquid? Exercises
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Rate at which material is transported through a process line Mass flow rate = mass (kg) time (s) Volumetric flow rate = volume (m 3 ) time (s) = m can be used to convert a known volumetric V flow rate of a process stream to the mass flow rate of that stream or vice versa Flow Rate
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Flow rate For flowing system: ? = 𝑉
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The specific gravity of nitrobenzene is 1.20. Calculate the volumetric flow rate in ml/min corresponding to a mass flow rate of 30 Ib m /h nitrobenzene. Solution:   Hg = 1.20 (62.43 Ib m /ft 3 ) = Specific gravity = 1.20 = / ref = 62.43 Ib m /ft 3 Volumetric flow rate = 74.916 Ib m /ft 3 30 Ib m 1 h ft 3 10 6 ml h 60 min 74.916 Ib m 35.3145 ft 3 = 188.99 ml/min Flow Rate
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Atomic weights can be obtained from the periodic table: Molecular weight, MW - sum of all the atomic weights of its atoms Moles and Molecular Weight
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Mole - amount of a species whose mass (g) equivalent to its molecular weight Molecular weight, MW - sum of all the atomic weights of its atoms MW O 2 = 2(16) = 32 1 gmol O 2 = 32 g 1 kmol O 2 = 32 kg 1 Ib-mole O 2 = 32 Ib m If the molecular weight of a substance is MW, then there are MW kg/kmol, MW g/gmol, and MW Ib m /Ib-mole of this substance Moles and Molecular Weight
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Therefore, n (NH 3 ) = 34 kg NH 3 1 kmol NH 3 17 kg NH 3 4 Ib-moles NH 3 17 Ib m NH 3 h 1 Ib-mole NH 3 = 68 Ib m NH 3 /h
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