Latent Heat: Hidden Energy In Phase Transitions

Exploring Phase Transitions: Change from Solid Phase to Liquid Phase 3.1. Phase Transition Phase Transition occurs from solid phase to liquid phase, from liquid phase to gas phase, and vice versa. In this transition process, there is a change in enthalpy (and internal energy) which occurs at a fixed temperature called latent heat. The enthalpy change in the same phase is called sensible heat. Types of latent heat: a) Heat of melting (H eat of F usion) (∆H fusion ) Occurs on phase change from solid to liquid b) Heat of Solidification (H eat of Solidifi cation) Occurs on phase change from liquid to solid c) Heat of vaporization (H eat of Vapori zation) (∆H )v Occurs on phase change from liquid to gas d) Heat of Condensation (H eat of Condentation) Occurs on phase change from gas to liquid The overall enthalpy change of pure components is shown in figure 3.1 Figure 3.1 Enthalpy change of pure components as a function of temperature The overall enthalpy change of each is shown in equation 3.1 T fus ion H H (T ) H (Tref ) Cp,sol id T ref dT T vaporizatoi n p ,liquid T dT H vaporizatoi n onTvap C H fus iononTfusion T fus ion C p,vapor (3.1) dT T vap Equations for estimating Heat of Vaporization (H eat of Vaporization) 1. Clapeyron Equation dT dp* HV dp* or T (V g - Vl ) d(ln T ) HV Vg (3.2) - Vl where: p* = vapor pressure T = Absolute temperature H v = Molar heat of vaporization at temperature T V i = molar volume of gas (g)and liquid (l) Figure 3.2 shows that the slope (dp / d(lnT)) can be used to estimate the specific heat of vaporization. Figure 3.2 H v is directly connected to the sloping line on the graph Equation 3.2 can be simplified with the following assumptions: 1. V l is negligible when compared to Vg V g RT/p* 2. The ideal gas law can be applied to So that equation 3.3 is obtained d ln p* If dp* H V dT p* RT 2 (3.3) H v is assumed to be constant over a certain temperature range then the equation 3.3 is integrated i n t o equation 3.4, which is called Clausius-Clapeyron equation HV log 10 p1 * p2 * 1 1 2.303R T2 (3.4) T1 2. Chen Equation 3.978(T / T ) 3.938 1.555ln p HV RT bc b c ) 1.07 (T b /Tc (3.5) where : Tc = Critical pressure Tb = Normal boiling point pc = Critical pressure 3. Riedel equation HV 1,093 RT c T (ln p 1) b c T (0.930 (T / T ) c b (3.6) c 4. Watson equation H V2 H V1 where : 1 0.38 1 T Tr2 (3.7) r1 H v1 = Heat of vaporization of pure liquid at temperature T1 H v 2 = Heat of vaporization of pure liquid at temperature T 2 The Watson equation is used to determine the heat of vaporization at a given temperature based on the known heat of vaporization at the normal boiling point (at 1 atm). Example 3.1 Calculation of heat of vaporization using Watson Equation Calculate the heat of vaporization of water at 600 K if the known properties of water at 100 o C are heat of vaporization = 2256.1 kJ/kg, Tc of water = 647.4 K. Completion: Using equation 3.7 600 H V ,600 H V ,373 1 1 0.38 647.4 373 0.551 647.4 H V ,600 H V ,373 0.551 1244 kJ/kg Example 3.2 Comparison of heat of vaporization estimation calculations Use a) Clausius-Clapeyron equation, b) Chen equation, c) Riedel equation to estimate the heat of vaporization of acetone at normal boiling point and compared to the experimental value of 30.2 kJ/g mol listed in Appendix C of Himmelblau 7th edition. Completion: Completion steps 1) Base 1 g mol 2) Look up properties of acetone in Appendix D of Himmelblau's book 7th edition Normal boiling point: 392.2 K Tc : 508.0 K pc : 47.0 atm 3) Calculating some variable values needed in the estimation equation T b /Tc = 329.2/508.0 = 0.648 H= 34.715 - 0.524 = 34.191 kJ/kg mole Example 3.5 Using a steam table to calculate enthalpy change Steam is cooled from 640o F at 92 psia to 480o F at 52 psia. How much is H in Btu/lb? Settlement Figure C3.1 Changes H that occurs in the problem is shown from part A to part B as shown in figure C3.1 Interpolation is performed twice for points A and B: 1. H interpolated between pressures at a fixed temperature 2. H interpolated between temperatures at fixed pressure (which has been interpolated) In the steam table, after interpolating H at 600o F at 92 psia is 1328.6, here is an example of interpolation 2/5 (1328.7 - 1328.4) = 0.4 (0.3) = 0.12 At p = 92 psia and T = 600o F, H = 1328.7 - 0.12 = 1328.6 The enthalpy change that occurs in the problem is H =1272.8 - 1348.4 = -75.6 Btu/lb Example 3.6 Use of steam table when phase change is involved to calculate the final state of water. Four kg of water at 27 o C and 200 kPa is heated at constant pressure until the volume of water becomes 1000 times greater than the initial volume. What is the final temperature of the water? Completion: The effect of pressure on the volume of water (liquid) is ignored The specific volume of water at 300 K = 0.001004 m 3 /kg, so the final specific volume = 0.001004 (1000) = 1.004 m /kg. 3 At 200 kPa, interpolation was done between 400 K (V = 0.9024 m3 ) and 450 K (V = 1.025 m3 ) where the range is the range between the values of 1.004 and 1.025. m3 /kg. So the T value can be determined by linear interpolation as follows (1.025 0.9024) m3 / kg 0.9024 m3 / kg T 430 (450 400) K 41 (T 400) K 1.004 m3 / kg 2557.5 441 K Figure 3.5 Pressure-enthalpy diagram of n-butane, the dotted line is the volumespecific line, the line with a temperature of o F is the line at a fixed temperature. Example 3.7 Use of n-butane pressure-enthalpy diagram to calculate enthalp change between two states Count H , (saturated) Completion: V , T at 1 lb n-butane saturated vapor from 2 atm to 20 atm H 0 (4.5) Equation 4.4 is the enthalpy balance, used to model the mixing of two fluids with different temperatures. 3. Open system and steady-state flow For steady-state ∆E = 0, so heat transfer ∆Q = 0, hence : W EP EK (4.6) Some energy balance models: a) Isothermal (dT = 0), process at constant temperature b) Isobaric (dP = 0), process at constant pressure c) Isometric or isochoric (dV = 0), process at constant volume d) Adiabatic (Q = 0), no heat transfer occurs Example 4.1 Simplification of the General Energy Balance Equation Figure C4.1 shows the process in segments distinguished by numbered constraints. List assumptions to simplify equation 4.1 for segments 1-5, 4-5, 3-4, 3-5, and 1-3 Figure C4.1 Completion: Analysis for each segment: 1-5 : -) EP = 0 (no height difference) -) Possibilities -) EK=0 E=0 (steady-state process) � Q The result W H H EP 4-5 : -) Q W 0 -) -) 3-4 : EK E=0 The result -) Q -) -) 3-5 : 0 W EK 0 E=0 H The result -) Q -) -) -) 1-3 : 0 0 W 0 EP EK 0 E=0 The result -) -) -) EP H 0 EP EK 0 0 E=0 � The result Q W H Problem! 1. A valve in the laboratory gas cylinder shown in Figure S3.1 is opened so that the gas N2 flows out. If the system is the cylinder, then simplify the generalized energy balance equation by setting all possible assumptions. Figure S4.1 2. Gas flows steadily (valve open) from high pressure at position no.1 to low pressure at position no.2 as shown in figure S3.2 through the turbine producing work. Simplify the generalized energy balance equation for the system consisting of both the cylinder and the turbine by eliminating the parts that may be omitted. Figure S4.2 Energy Balance Problem Solving Strategy 1. Select a system, understand its boundary, determine whether it is closed or open. Write down the decisions made, then change them if they do not provide a good solution. Example 4.6 Energy Balance Application for Water Pumping Water was pumped from the bottom of the 15 ft deep well at a rate of 200 gal/hr to the storage tank to keep the water level in the tank 165 ft above ground. To prevent freezing in winter, a small heater was added at 30000 Btu/hr to heat the water during the process. water transfer to the storage tank. Heat loss of the whole system at a constant rate of 25000 Btu/hr What is the water temperature when it enters the storage tank, assuming the water temperature in the well is 35 F.o A 2 hp pump is used to pump water. 55% of the pump's power becomes work while the rest becomes heat released into the atmosphere. Solution: Stage 1, 2, 3, and 4 Assumed open and steady-state system, consisting of wells, pipes, pumps, and storage tanks as shown in figure C4.6. Figure C4.6 Stage 5 Base: 1 hour of operation Stage 6 Mass balance: 200 gal of water in and 200 gal of water out in 1 hour. Energy balance: � (m( H EK EP ) E Q W -) E=0 (steady-state process) -) m1 = m2 = m -) Possibilities EK = 0 due to assumption v1 = v2 = 0 � 0 So Valu e (m( H EK EP) Q W H at the top of the tank is unknown, but can be calculated from the balance sheet energy If you set the referent temperature as 35 o F (input temperature) then the enthalpy at input = 0, because the temperature difference (Tmasuk - 35) = (35 - 35) = 0 so H the value can be calculated as H2 - 0. Exit temperature can be determined based on property calculations H if C assuming constant T2 H mH m C p dT o mC T 35 F (T - 35) 2 p Stage 8 and 9 Total mass of pumped water 200 gal 8.33lb hr 1 gal 1666 lb Change in potential energy EP m EP m g h 1666 lbm 32.2 ft 180 ft s2 (s2 ) (lb)f 32.2 (ft) (lbm) 1 Btu 778( ft ) (lb f ) 385.4 Btu The heat loss in the system is 25000 Btu when a heater of 30000 Btu is placed in the system, so the net heat exchange: Q = 30000 - 25000 = 5000 Btu Rate of work on water by pump p W 2 hp 0.55 33000 (ft) (lb f) 60 min (min) (hp) Btu hr Q W 5000 778( ft ) (lb f ) H 2800 H 2800 Btu/hr EP H 385 7415 Btu Since the temperature range is considered to be small, the heat capacity of water (liquid) is assumed to be constant and equal to 1.0 Btu/(lb) ( o F) thus 7415 H mC p dT 1666 (1.0) (T2 T - 35) 4.5o F (temperature rise), so T o 39.5 F 2 Problem! 1. In a refinery, the condenser is set to cool 1000 lb/hr of benzene entering at 1 atm pressure, 200 o F, and exiting 171 o F. Assumed heat loss to the environment is negligible. How much (in pounds) cooling water is required per hour if the cooling water enters at 80o F and exits at 100 o F? 2. In a steady-state process, 10 g mol/s O2 at 100o C and 10 g mol/s nitrogen at 150o C are mixed in a vessel that has a heat loss to the environment of 109(T - 25) J/s where T is the temperature of the mixture in o C. Calculate the exit gas temperature in o C. Use the following heat capacity: O2 N 2 :C p 6.15 3.1x10 :C p 6.5 1.25 x10 Where T in K, C p 3 3 T T in cal/(g mol) (K) 3. An exhaust fan in a constant-area, well-insulated duct delivers air with an exit velocity o f 1.5 m/s at pressure difference of 6 cm H2 O. The inlet and outlet temperatures are 21.1o C and 22.8o C. The duct area is 0.60 cm3 . Determine the power required by the fan.