Introductory Chemical Engineering Thermodynamics Unit I. Earth, - - PowerPoint PPT Presentation
Introductory Chemical Engineering Thermodynamics Unit I. Earth, Air, Fire, and Water Chapter 1: Introduction By J.R. Elliott and C.T. Lira A theory is the more impressive the greater the simplicity of its premises is, the more different
Chapter 1: Introduction By J.R. Elliott and C.T. Lira
Elliott and Lira: Chapter 1 - Introduction Slide 1
Elliott and Lira: Chapter 1 - Introduction Slide 2 Introduction to potential energy ∞ r < σ up = -ε(σ/r)6 r ≥ σ where ε ≡ depth of the attractive well at r = σ. The Sutherland Potential
r/ sigma
0 . 5 1 1 . 5 0 . 5 1 1 . 5 2 2 . 5 3
r/sigma u*(r)
0 . 5 1 1 . 5 0 . 5 1 1 . 5 2 2 . 5 3 r/ sigma
0 . 5 1 1 . 5 2 0 . 5 1 1 . 5 2 2 . 5 3
The Square-Well Potential The Lennard-Jones Potential
Elliott and Lira: Chapter 1 - Introduction Slide 3
Temperature is a numerical scale which primarily provides a measure of the kinetic energy of a system of molecules. Classically, T = <Σmvi
2/3nR>t where <...>t denotes an average over time.
Energy is a term which applies to many aspects of the system. Its formal definition is in terms of the capability to perform work, but we cannont say how much work until after the study of thermodynamics. Energy may take the form of heat, kinetic energy, or potential energy and it may refer to energy of a macroscopic or a molecular scale. These are detailed in the next section. Heat is energy in transit between the source from which the energy is coming and a sink toward which the energy is going. (e.g. pizza in the oven) Density is a measure of the mass per unit volume and may be expressed on a molar basis
Pressure is the force exerted per unit area. We will be concerned primarily with the pressure exerted by the molecules of fluids upon the walls of their containers. For our purposes, the kinetic theory of pressure will be sufficient.
Elliott and Lira: Chapter 1 - Introduction Slide 4 An Ultrasimplified Kinetic Theory of Pressure Suppose we have two hard spherical molecules in a container. Both of the molecules are bouncing back and forth in the x-direction only and they are not coming in contact with each other.
F ma m v t = = ∆ ∆ /
1
∆v v v
f i 1 1 1
= −
, but
v v v
i f 1 1 1
= ≡
A2 A1
<-
How much time elapses between collisions of particle 1 with wall A1? L = length between A1 and A2 ∆t = 2 L / v1 ⇒ F1 = m(2 v1)/(2 L v1 ) = m v1
2 / L
By definition, P1 = F1/A1 = F1/ L 2 = m v1
2/ L 3
For the system of two particles we must sum the forces from each particle. ⇒ Ptot = (m/ L 3 )*( v1
2 + v2 2)
Elliott and Lira: Chapter 1 - Introduction Slide 5
For water at 160°C and 0.12 MPa, find the internal energy. Solution: We are superheated, but volues in steam tables are only available at 150 and 200°C and 0.1 and 0.2 MPa. Interpolating on pressure first because the change in U is less with respect to pressure, then interpolating the results on temperature gives: U(150,0.12) = 2582.9 + [(0.12-0.10)/(0.2-0.1)]*(2577.1-2582.9) = 2581.7 J/g U(200,0.12) = 2658.2 + [(0.12-0.10)/(0.2-0.1)]*(2654.6-2658.2) = 2657.5 J/g U(160,0.12) = 2581.7 + [(160-150)/(200-150)]*(2657.5-2581.7) = 2596.9 J/g T°C \ P(MPa) 0.1 0.12 0.2 150 2582.9 2581.7 2577.1 160 2596.9 200 2658.2 2657.5 2654.6
Elliott and Lira: Chapter 1 - Introduction Slide 6 Example 1.8 Quality Calculations from Constant Volume Cooling
Steam is initially contained in a rigid cylinder at P = 30MPa and V=102.498 cm3/mole. The cylinder is allowed to cool to 300°C. What is the pressure, quality, and overall internal energy of the final mixture? Solution: V=102.498 cm3/mole 10-6(m3/cm3)/(0.018kg/mole)= 0.017487m3/kg 0.017487m3/kg < VsatV (300°C) = 0.0216 ⇒ q < 1.0 and P = 8.6028 MPa 0.017487 = (1-q)0.001404 + q (0.0216 ) ⇒ q = 0.796 (Note the log scale in the figure,) U = q (2563) + (1-q) 1333 = 2312 J/g (but we linearly interpolate on V.)
Elliott and Lira: Chapter 1 - Introduction Slide 7
10 20 30 40 50 1 1.5 2 2.5 3 3.5 4 4.5 5 log10 [V(cm
3/mol)]
Pressure (MPa)
100 200 300 400 500
Temperature (