Energy Balances ChE Spring /13. closed: No transfer of mass across the system boundaries.

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1 Energy Balances ChE Spring /13 1 Basic Concepts Definitions: System: Any specified mass of material or piece(s) of equipment we wish to devote our attention to. closed: No transfer of mass across the system boundaries. open: Transfer of mass across the system boundaries is permitted. Surroundings: Anything that is not part of the system. Properites: A characteristic of the material that can be calculated or measured. extensive: A property that is the sum of the values of each of the subsystems comprising the whole system. A property dependent on system size. Example: mass or volume. intensive: A property whose value is not additive, and does not vary with system size. Example: temperature, pressure, density specific: An intensive property that is obtained by dividing an extensive property by the total amount of process material. Example: A liquid where V = 200 cm 3 and the mass = 200 g, then the specific volume ˆV = 1 cm 3 /g. State: A set of properites of the material at a point in time. depends only on the system s intensive properties. The state of a system 1.1 Work The term Work is used widely in casual conversation, but in this context we refer to a specific definition. Work is a form of energy that represents a transfer of energy between the system

2 Energy Balances ChE Spring /13 and surroundings. For a mechanical force: W = state 2 state 1 F dl (1) where F is an external force in the direction of l acting on the system, or a system force acting on the surroundings. The amount of work done by a system on its surroundings (or by the surroundings on the systems) is dependent upon the inital and final states of the system and the path taken to get from state 1 to state 2. For this reason, we say that work is a path function. Path function: A function who s final value is dependent on the path taken to get from the initial to final state. Example: work, heat. State function: A function that is only dependent upon the initial and final states of the system. Example: internal energy, enthalpy Illustration of work as a path function An ideal gas at 300 K and 500 kpa is enclosed in a cylinder capped by a frictionless piston. The gas slowly expands from 0.1 m 3 to 0.2 m 3. Calculate the work done by the gas on the piston if: Path 1: expansion occurs at constant pressure (p = 500 kpa) Path 2: expansion occurs at constant temperature (T = 300 K) Path 1 W = V2 V 1 F A A dl = V2 V 1 p dv (2) V2 W = p dv = p (V 2 V 1 ) (3) V 1 = kpa (0.2 m m 3 ) = 50kJ

3 Energy Balances ChE Spring /13 Remember 1 J = 1 N m. Path 2 V2 V2 nrt W = p dv = V 1 V 1 V = nrt ln(v 2 /V 1 ) dv (4) n = P V RT = ( Pa)(0.1m 3 ) (300K)(8.314 Pa m 3 /mol K) = 20.0 mol (5) W = (20.0mol)(8.314 Pa m 3 /mol K)(300K) ln(0.2/0.1) = 34.57kJ (6) 1.2 Heat Heat (Q): The part of the total energy flow across a system boundary that is caused by a temperature difference between the system and its surroundings. Heat, like work, is a path function. 1.3 Kinetic Energy Kinetic energy (KE): The energy a system possesses because of a velocity difference between it and it s surroundings at rest. KE = 1 2 mv2 (7) where m is the mass of the object and v is the object s velocity relative to the reference state Example: Calculation of Kinetic Energy Water is pumped from a storage tank into a 4.0 cm ID tube at a rate of m 3 /s. What is the kinetic energy of the water? Step 1, determine the linear velocity of the water v = V /A = 0.001m3 /s = m/s (8) π(0.02) 2 m2

4 Energy Balances ChE Spring /13 Step 2, find the mass flow rate of mater ρ = m/v ; ρ V = ṁ; (1000 kg/m 3 )(0.001 m 3 /s) = 1.0 kg/s (9) Step 3, calculate the kinetic energy KE = 1 = 1 2ṁv2 2 (1.0 kg/s)(0.795 m/s)2 = J/s (10) check units 1 N = kg m/s 2 1 J = N m Therefore: (kg m/s 2 )(m/s) = N m/s = J/s 1.4 Potential Energy Potential Energy (PE): The energy a system possesses because of forces exerted on its mass by gravitational or electromagnetic fields with respect to a reference surface. P E = mgh (11) where h is the distance from the reference surface, m is the mass of the object and g is the acceleration due to gravity (9.807 m/s 2 ) Example: Calculation of Potential Energy Water is pumped from one reservoir to a second reservoir 10 m above the water level of the first. What is the increase in the potential energy of the water per kg of water moved (the increase in the specific PE of the water)? P E = mgh = (9.80 m/s 2 )(10m) = 98.0 J/kg (12)

5 Energy Balances ChE Spring / Internal Energy Internal Energy (U): A macroscopic measure of the molecular, atomic and subatomic energies contained within a fluid. Cannot be measured, but is instead calculated from other variables that can be measured, such as pressure, temperature and volume. U is a state function. From the Gibbs phase rule we know for a one componet, one phase system that specifying two intensive variables will definite U. Û is commonly defined in terms of T and ˆV Û = f(t, ˆV ) (13) Taking the derivative yields ( ) ( ) dû = Û Û dt + T ˆV ˆV d ˆV (14) T Note: C v = ( ) Û ; T V ( ) Û ˆV T 0 (15) Therefore, changes in Û can be calculated by integration of equation Enthalpy Û 2 Û1 = T2 T 1 C v dt (16) Enthalpy (H): Enthalpy, like U, is a state function and is defined: H = U + pv (17) for a one component, one phase system: H = f(t, p) (18) Taking the derivative yields: ( ) dĥ = Ĥ dt + T p ( ) Ĥ dp (19) p T

6 Energy Balances ChE Spring /13 Note: C p = ( ) Ĥ ; T p ( ) Ĥ p T 0 (20) Therefore, changes in Ĥ can be calculated by integration of equation 19. Ĥ 2 Ĥ1 = T2 T 1 C p dt (21) Again, we note that the absolute values of both U and H cannot be calculated directly. Only differences in U or H relative to a reference state may be calculated.

7 Energy Balances ChE Spring /13 2 Heat Capacities Definitions: Sensible heat: The heat that must be transfered to raise or lower the temperature of a substance(s). Q = U (closed system) (22) Q = Ḣ (open system) (23) Latent heat: The specific enthalpy change associated with a phase transition (e.g. liquidvapor) of a substance at constant T and P. Heat of vaporization ( Ĥv): The specific enthalpy difference between the liquid and vapor forms of a substance at a given T and P. Heat of fusion ( Ĥm): The specific enthalpy difference between the solid and liquid forms of a substance at a given T and P. NOTE: Both Ĥv and Ĥm are strong functions of T, but only weak functions of P. Knowing the sensible heat and the latent heat, it is possible to determine the change in internal energy ( U, closed system), or the enthalpy ( H, open system). The determination of U for a system that does not undergo a phase transition is straightforward. Û 2 Û1 = T2 This equation is valid under the following conditions: Exact for ideal gases A good approximation for liquids and solids Only valid for a non-ideal gas if V = constant. T 1 C v dt (24)

8 Energy Balances ChE Spring /13 We can write a similar expression for H: Ĥ 2 Ĥ1 = T2 This equation is valid under the following conditions: Exact for ideal gases Only valid for a non-ideal gas if P = constant. For liquids or solids: Ĥ 2 Ĥ1 = ˆV P + T 1 C p dt (25) T2 T 1 C p dt (26) However, the ˆV P term in the above equation is usually negligible compared to the second term. C p and C v are related to each other in the following way: For liquids and solids C p C v Ideal gases C p = C v + R 2.1 Example of a U Calculation from C v. 1 mol of liquid cyclohexane is stored in a covered beaker at 1 atm and 20 C. Calculate the energy input required to heat the cyclohexane to 60 C. For a closed system, the energy balance becomes: Q = n Û (27) U may be calculated from integration of 14. C p C v for liquid cyclohexane, taken from Table B.2 [kj/(mol C)]is: C v = T T T 3 (28)

9 Energy Balances ChE Spring /13 Integration of 28 yields: Û = (T 2 T 1 ) (29) (T2 2 T1 2 ) (T2 3 T1 3 ) (T2 4 T1 4 ) 4 Since Q = n Û, Q =(1.0 mol)(4.582 kj/mol) = kj Û = kj/mol (30) If the fluid undergoes a change of phase (solid to liquid, liquid to vapor, etc), a more complex procedure to calculate the amount of heat required. 2.2 Hypothetical Process Paths Suppose we wish to calculate the heat required to turn solid phenol at 25 C and 1 atm to vapor at 300 C at 3 atm. Ĥ = Ĥ(vapor, 300 C, 1 atm) Ĥ(solid, 25 C, 3 atm) (31) Problem - we don t know Ĥ for each state. Fortunately Ĥ is a state function, so it doesn t matter what path we take to get from point A to point B. Ĥ can be evaluated by constructing a convenient alternate path. The change in the enthalpy for the process is simply the sum of the enthalpy changes for each leg of the hypothetical process path. Ĥ = Ĥ1 + Ĥ2 + Ĥ3 + Ĥ4 + Ĥ5 + Ĥ6 (32)

10 Energy Balances ChE Spring / C, vapor, 1 atm dh6 300 C, vapor, 3 atm dh C, liquid dh , vapor dh3 dh C, solid 42.5 C, liquid dh1 25 C, 1 atm, solid Figure 1: Hypothetical process path for the vaporization of Phenol. Red line denotes actual path, black line denotes the hypothetical process path. 2.3 Example of Ĥ Calculation 100 mol of liquid hexane at 25 C and 7 bar is vaporized and heated to 300 C at constant pressure. Step 1, collect data From table B.1, Ĥv = kj/mol at 69 C. Find T boil (7 bar). Solve for T, T boil = 144 C. log p = ( T ) (33) Since T boil T n boil, we need to construct a hypothetical process path to get from the initial to final states. The change in enthalpy for this process is given by: Ĥ = Ĥ1 + Ĥ2 + Ĥ3 (34) Step 2, determine the Ĥ values for each leg of our hypothetical process. Ĥ1 = ˆV P dt (35)

11 Energy Balances ChE Spring / C, 7 bar, vap dh3 69 C, 1 bar, liq dh2 69 C, 1 bar, vap dh1 25 C, 7 bar, liq Figure 2: Hypothetical process path for the vaporization of Hexane. ρ = g/cm 3, ˆV = 1/ρ = m 3 /kg. 100 mol*86.17 g/mol = 8617 g or kg H 1 = (8.617) ( )( ) ( ) +100(0.2163(69 25)) (36) H 1 = kj/h kj/h (37) H 2 = (100)(28.85) = 2885 kj/h (38) H 3 = T T T 3 dt (39) Sum values of H determined above: H 3 = kj/h (40) H = = 8551 kj/h (41)

12 Energy Balances ChE Spring / Estimation of Latent Heats Trouton s Rule: Ĥv 0.088T b (non polar fluids) where Ĥv = [kj/mol] and T b =[K] T b (water, low MW alcohols) (42) For greater accuracy, one should use Chen s equation Ĥv = T b[0.0331(t b /T c ) log P c ] 1.07 (T b /T c ) where Ĥv = [kj/mol], T b, T c =[K] and P c = [atm]. (43) If Ĥv is known for one temperature, Watson s Correlation can be used to predict Ĥv at a different temperature ( ) Tc T 0.38) 2 Ĥv(T 2 ) = Ĥv(T 1 ) (44) T c T 1 The following relationships can be used to predict heats ofr fusion ( Ĥm) when experimental data is not availible Ĥm T m (metallic elements) T m (inorganic compounds) 0.050T m (organic compounds) (45) 3 Energy Balances on Closed Systems Remember, a closed system is one where there is no transfer of material across the system boundary. In this case, the general balance equation beomes: accumulation = input output (46) or final system energy inital system energy = net energy transfered to system (47)

13 Energy Balances ChE Spring /13 initial system energy final system energy energy transfered U i + KE i + P E i (48) U f + KE f + P E f (49) Q + W (50) Combining the above equations we can write the energy balance for a closed system U + KE + P E = Q + W (51) NOTES: U depends almost entirely on chemical composition, temperature, and phase (gas, solid, liquid) of the system materials. For a closed system with no T, phase changes or chemical reactions, U 0 If the system is not accelerating or decelerating (v = 0), KE = 0. If the system is not rising or falling, P E = 0. If the system and its surroundings are at the same temperature, or the system is perfectly insultated, Q = 0. This kind of process is refered to as adiabatic. If there are no moving parts, electrical currents or radiation at the system boundary, W = 0. 4 Energy Balances for Open Systems In an open system, mass is allowed to cross the system boundaries. The general balance equation in this case becomes simply: input = output and H + KE + P E = Q + W (52)