Outline First law of thermodynamics Closed systems (no mass flow) Device for in-class thought experiments to prove 1 st law Rubber stops GAS Features: Quasi-equlibrium expansion/compression Constant volume option Insulated (adiabatic) Non-QE shaft work Heat addition/rejection ICE Experiments Starting from state 1, consider... Insulated (adiabatic), fix volume, do 8 kj of shaft work on system. Then measure change in energy to be 8 kj so Wsh = E2 E1 = E Insulated, compress gas doing 10 kj of boundary work. Then measure change in energy to be 10 kj Wb = E Initial conclusion: For adiabatic process, amount of work done is equal to change in energy for closed system W = E when Q = 0 1
More experiments Fix volume, no shaft work, add 10 kj of heat. Then measure change in energy to be 10 kj Q = E when W = 0 Add 15 kj of heat and do 6 kj of shaft work. Then measure change in energy to be 21 kj E = Q W Still more experiments From state 1 cause a change to state 2 by several processes Compress gas doing 9 kj of boundary work adiabatically so W = ( 9) = E Do 9 kj shaft work adiabatically so b W = ( 9) = E sh For all adiabatic processes between two specified states of closed system, net work is the same regardless of the nature of the system or details of the process Energy change of system during a process is equal to net work and heat transfer between the system and its surroundings Illustration 2
Summary of experimental findings For closed system (Joule, 1800s): Q W = E Net energy transfer to system as heat and work Net energy increase of system Above is first law of thermodynamics E is property (point function) Q, W are path functions (except when one or the other is zero) Other forms of the 1 st law Q W = E = U + KE + PE 1 2 2 = m( u2 u1) + m ( V2 V1 ) + mg( z2 z1) 2 Q W = U (stationary system) q w = e (per unit mass) δ q δ w = de (differential form) de Qɺ Wɺ = (rate form) dt Q W = 0 (cyclic process) Example A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kj. During the cooling process, the fluid loses 500 kj of heat, and the paddle wheel does 100 kj of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel. Q W = E = U 500kJ + 100kJ = U U = U 800kJ U = 400kJ 2 2 1 2 3
Outline Phase and pure substance Features of phases of matter Phase change processes Property diagrams Phase and Pure substance Phase refers to quantity of matter that has same chemical composition and physical structure throughout e.g. all solid, all liquid, or all vapor (gas) A system can contain more than one phase e.g. system of liquid water and water vapor (steam) contains two phases A pure substance with fixed chemical composition throughout i.e. water, N 2, He, CO 2, etc. A mixture of chemical elements or compounds is still a pure substance as long as mixture is homogeneous i.e. air A mixture of two or more phases is still a pure substance as long as chemical composition of each phase is same i.e. ice and liquid water but not liquid and gaseous air Phases of a pure substance Solid molecules are closely packed and arranged in a 3D pattern (lattice) which is repeated throughout solid e.g. long range order Due to closely packed nature attractive forces between molecules are large and keep them at fixed positions e.g. definite shape and size Molecules maintain their position but oscillate about this equilibrium position and velocity of molecules is related to temperature Higher temperature results in higher velocity and may be high enough to overcome attractive forces and groups of molecules may break away - melting Temperature and molecular motions Molecular interactions 4
Phases (cont.) Liquid chunks of molecules float about each other but w/in each chunk molecules maintain orderly structure e.g. short range order Definite volume but fluid shape Slight increase in distance between molecules over solid (except water where solid is less dense ice floats in your drink) Phases (cont.) Gas molecules far apart w/ no molecular order Molecules move about at random colliding w/ each other and walls of container Shape and volume defined by container At low densities intermolecular forces are very small and collisions are the only mode of interaction i.e. ideal gas Much higher energy levels than liquid/solid Phase change schematic Gas Deposition Sublimation Solid Vaporization Freezing Condensation Melting Liquid ENTHALPY 5
Phase change processes Practical situations involve two phases of a pure substance existing in equilibrium Water exists as mixture of liquid and vapor in boiler and condenser of steam power plant Refrigerant turns from liquid to vapor in freezer Most important cases involve liquid and vapor phases We will illustrate using water... Constant pressure phase change experiment subcooled liquid What happens when we heat water at constant pressure? Consider water at 1 atm, 20 C, e.g. tap water, in a piston-cylinder device It is called compressed or subcooled liquid (CL or SL) because it is not about to vaporize (boil) Compressed suggests above boiling P at give T Subcooled suggests below boiling T at given P Saturated liquid state Add heat, T rises to say 40 C, liquid water expans slightly, specific volume increases, piston moves up slightly P still 1 atm (why?) Water still compressed liquid Add more heat until T=100C Still liquid but any more heat will cause vaporization Liquid about to boil is called saturated liquid (can t hold any more heat) (SL) 6
Saturate liquid-vapor mixture state Add more heat T=constant as energy goes into vaporization process (phase change); breaking bonds Midway we have ½ liquid and ½ vapor; continue to add heat until all liquid is vaporized Saturated liquid-vapor mixture (SLVM) Saturated vapor state Entire cylinder contains vapor Any heat loss, no matter how small, will cause some vapor to condense (phase change from vapor to liquid) Vapor is about to condense Saturated vapor (SV) Superheated vapor state Single-phase (vapor) Add heat T and specific volume increases At say 300 C if we transfer some heat away T will decrease but no condensation Not about to condense Superheated vapor (SHV) 7
T-v diagram of constant pressure phase change process SL SLVM SHV CL SV Saturation temperature and pressure The temperature at which water starts boiling depends on the pressure If pressure is fixed, so is boiling temperature For example, water boils at 100 C at 1 atm If P was 5 atm (add weights to piston) boiling would occur at 151.9 C Saturation temperature and pressure (cont.) At given P, T at which pure substance starts boiling is saturation temperature T sat At given T, pressure for boiling is saturation pressure P sat Example: At P=101.35 kpa, T sat =100 C; at T=100 C, P sat =101.35 kpa T sat = f (P sat ) during phase change; not independent and cannot be used to fix state (see Fig. 2-12) 8
Repeat constant P phase change experiment at different P Add weights to piston to change P Water boils at higher T as P increases Specific volume of SL is larger Specific volume of SV is smaller than at 1 atm Horizontal line connecting SL and SV states shorter Critical point (next slide) Critical point Point at which SL and SV states identical Above critical point there is no sharp difference between liquid and vapor T cr, P cr, v cr = 374.14 C, 22.09 MPa, 0.003155 m3/kg For other substances critical point values tabulated in A-1 Connect the dots... SL/SV states connected by SL/SV lines They meet at critical point forming wet dome T=const. phase change by decreasing piston weight; note downward slope of T=cont. lines 9
Extension to include solid phase Most substances contract during solidification (freezing); except water which expands during freezing Under some conditions all three SLV phases can coexist in equilibrium On P-v or T-v diagrams these triple states form a triple line on a P-T diagram at triple point P-T diagram and triple point P-v-T surfaces Recall that state of a simple compressible substance is fixed by 2 independent, intensive properties e.g. state postulate Recall equation z=z(x,y) represents a surface in space, so we can represent P-v- T relationship as surface in space i.e. T-v space with P as height (see next slide) P-v, T-v, P-T diagrams are projections from this 3D plot 10
P-v-T surfaces Contract on freezing Expand on freezing (like water) Summary Is iced water a pure substance? What is the difference between saturated liquid and compressed liquid? What is the difference saturated vapor and superheated vapor? Is there any difference between the properties of saturated vapor at a given temperature and the vapor of a saturated liquid-vapor mixture at the same temperature? Is it true that water boils at higher temperatures at higher pressures? Explain. What is the difference between the critical point and the triple point? Outline Property tables Property evaluations 11
Property tables Properties are listed in the form of tables i.e. steam tables SI and English unit versions They list property information for each region of interest, such as SHV, CL, and SLVM regions Looking at steam tables you find enthalpy, h, and entropy, s, which we have not discussed yet Entropy later, enthalpy now... H=U+PV (kj) or h=u+pv (kj/kg) P, T, v, u, h, and s data are tabulated Saturated water tables (A-2 T table, A-3 P table) u = u u h = h h fg g f fg g f s = s s fg g f Examples A rigid tank contains 50 kg of saturate liquid water at 90 C. Determine the pressure in the tank and volume in the tank. P = P = 0.7014bar = 70.14 kpa ( A 4) sat @90C υ = υ = f @90C 3 0.001036 m / kg ( A 4) V = mυ = (50 kg)(0.001036 m / kg) = 0.0518m 3 3 Can you confirm these numbers from the tables listed? 12
Examples A piston-cylinder device contains 2ft 3 of saturated water vapor at 50 psia. Determine T of vapor and mass of vapor inside cylinder. T = T = 281.03 F ( A 3 E) sat @50 psia υ = υ = g @50 psia 3 8.518 ft / lbm ( A 3 E) 3 V 2 ft m = = = 0.235lbm 3 υ 8.518 ft / lbm Can you confirm these numbers from the tables listed? Saturated liquid-vapor mixture During vaporization substance exists as part liquid and part vapor i.e. mixture of SL and SV We need to know proportions of each phase in mixture We define quality, x, as ratio of mass of vapor to total mass of mixture m vapor x mtotal = mliq + mvap = m f + mg mtotal x always between 0 and 1, i.e. x=0 for SL and x=1 for SV x can be used as one of two independent, intensive properties to fix state Properties of SL and SV are same in mixture as if alone SLVM properties We normally pretend SLVM is homogeneous mixture V = V + V mυ = m υ + m υ f g t av f f g g ( ) ( ) m = m m mυ = m m υ + m υ f t g t av t g f g g m υ = (1 x) υ + xυ = υ + xυ t av f g f fg x = υ υ / υ av f fg x has no meaning outside of the wet dome f ( ) / y = y + x y x = y y y y y y f fg f fg g (usually drop av subscript) 13
Quality on the wet dome x = 0.5 x = 0 x = 1 ( ) / x = υ υ υ av f fg υ f υ av υg Examples A rigid tank with 10 kg of water at 90 C. If 8 kg of water is in liquid form and rest is in vapor form, determine (a) P in tank, (b) V of tank. OR P = P = 70.14 kpa ( A 2) υ sat @90C = 0.001036 m / kg, υ = 2.361 m / kg ( A 2) 3 3 f @90 C g @90C V = V + V = m υ + m υ f g f f g g = (8 kg)(0.001036 m / kg) + 2(2.361 m / kg) = 4.73m 3 3 3 x = m / m = 2 /10 = 0.2 υ = υ + xυ = g t f fg 3 3 = 0.001036 m / kg + 0.2(2.361 0.001036) m / kg = 0.473 m / kg V = mυ = 10(0.473) = 4.73m 3 3 Superheated vapor Table A-4 Single phase region to right of SV line T, P independent and fix state P<P sat at given T T>T sat at given P v, u, h > SV values at given P or T 14
Example Find internal energy of water at 20 psia and 400 F Tsat @20 psia = 227.96 F ( A 3 E); T > T SHV ; u = 1145.1 Btu / lbm( A 4 E) sat Find temperature of water at P=0.5 MPa and h=2890 kj/kg hg @0.5MPa = 2748.7 kj / kg; h > hg SHV T = 216.4C T, C h, kj / kg 200 2855.4 250 2960.7 Linear interpolation Assumes any two data points connected by straight line (set slopes equal to find missing value) A B 100 5 130 X 200 130 100 200 100 10 x 5 = 10 5 Compressed liquid Not much data due to relative independence of CL properties with pressure They are incompressible i.e. change in pressure barely changes v, T, u, etc. A good approximation if data is unavailable is to treat CL as SL at same T (properties of CL due vary with T) y y f @ T h hf @ T + v f ( P Psat ) P>P sat at given T (CL) and T<T sat at given P (SC) v, u, h < SL values at given P or T 15
Example Determine internal energy of CL water at 80 C and 5 MPa using (a) data from CL table and (b) SL data. Error? P = 47.4 kpa; P = 5MPa > P sat@80c sat CL u = 333.72 kj / kg vs. u u = 334.86 kj / kg (0.34%) f @80C Reference state and values u,h,s cannot be measured directly P,T,v can be measured directly Changes in u, h, s computed using derived relations with P, T, v Note: only changes can be computed not u, h, s, at specific state But fortunately in 1 st and 2 nd law s only changes in these properties is needed We choose a convenient reference state and ste properties to zero at that state For water, u=0, s=0, at 0.01C for SL state; h=u+pv; could have negative values relative to reference; different ref. state for different substances; ref. state cancels out ( ref ) ( ref ) u u u u = u u = u 2 1 2 1 Summary What does the saturated water table represent? What is quality? Does it have any meaning in SHV region? What if h data is needed at T=125 C and the table only lists h data at T=100 and T=150 C? In the absence of CL tables, how is specific volume of CL at given P and T determined? Does reference point selected for properties of a substance have any effect on thermodynamic analysis? Why? 16