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##### Chapter 7 E n e r g y a n d E n e r g y B a l a n c e

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**Chapter 7EnergyandEnergyBalance**By :Miss Noorulanajwa Diyana Yaacob noorulnajwa@unimap.edu.my**Content**• Components of total energy, energy balance on a closed and open system, energy or power conversion to other units • First law of thermodynamics • Enthalpies and kinetic energy of a body of mass moving • Enthalpy data and steam tables**Forms of Energy**Three component of total energy of a system • Kinetic energy (Ek) • energy due to the translational motion of the system as a whole relative to some frame of reference (usually the earth’s surface) or to rotation of the system about some axis. • Potential energy (Ep) • energy due to the position of the system in a potential field (such as a gravitational or electromagnetic field). • Internal energy (U) • all energy possessed by a system other than kinetic and potential energy; or • Energy due to translation, rotation, vibration & electromagnetic interactions of the molecules, atom and subatomic particle within the system.**Transfer of Energy**• In closed system (i.e. no mass is transferred across the system boundaries while the process is taking place), energy may be transferred between such a system and its surroundings in two ways as heat or work. • Heat • Energy that flows as a result of temperature difference between a system and its surroundings. • The direction of flow is always from a higher temperature to a low one. • Heat is defined as positive when its transferred to the system from the surroundings. • Work • energy that flows in response to any driving force other a temperature difference, such as a force, a torque or a voltage • Work is defined as positive when it is done by the system on the surroundings.**First Law of Thermodynamics**• Law of conservation of energy, which state that energy can neither be created nor destroyed. • General form of first law of thermodynamics Inlet Energy + Heat - Outlet Energy – Work = Accumulation Inlet energy and outlet energy is summation/total of all energy such as potential, kinetic and internal energy**Kinetic Energy Equation (Ek)**• Kinetic energy, Ek (J) of an object of mass m (kg) moving with velocity u (m/s) relative to the surface of the earth is • If the fluid enters a system with a mass flow rate (kg/s) and uniform velocity u (m/s), the rate at which kinetic energy (J/s) is transported into the system is**Potential Energy Equation (Ep)**• Gravitational potential energy, Ep • if the fluid enters a system with a mass flow rate (kg/s) and an elevation z relative to the potential energy reference plane. • Normally we are interested in the change of potential energy during energy balance calculation**ERT104/4**ppkas@UniMAP Let’s do it Water flows into a process unit through a 2 cm ID pipe at a rate of 2 m3/h. Calculate the kinetic energy transport in this stream in unit J/s. Solution: Ek= 0.870 N.m/s= 0.870 J/s**Let’s try**• Crude oil is pumped at a rate of 15.0 kg/s from a point 220 meters below the earth’s surface to a point 20 meters above ground level. Calculate the attendant rate of increase of potential energy.**Energy Balances on Closed System**• Closed system • no mass is transferred across the system boundaries while the process is taking place • Energy balance Final System Energy – Initial System Energy = Net Energy Transferred to the System Initial energy system = Ui + Eki + Epi Final energy system = Uf + Ekf + Epf Net energy transfer = Q-W (Uf-Ui) + (Ekf-Eki) + (Epf-Epi) = Q-W**Energy Balances on Closed System**• When applying energy balance equation to a given process, the following point must be aware; • The internal energy of a system depends almost entirely on the chemical composition, state of aggregation (solid, liquid, or gas), and temperature of the system materials. If no temperature changes, phase changes, or chemical reactionsoccur in a closed system and ifpressure changes are lessthan a few atmospheres, then∆U ≈ 0. • If a system is not accelerating, then ∆Ek = 0. If a system is not rising or falling, then ∆Ep = 0. • If a system and its surroundings are at the same temperature or the system is perfectly insulated, then Q = 0. The process is then termed adiabatic. • Work done on or by a closed system is accomplished by movement of the system boundary against a resisting force or the passage of an electrical current or radiation across the system boundary. If there no moving parts or electrical current at the system boundary, then W = 0.**Let’s try**• A gas is contained in a cylinder fitted with a movable piston. The initial gas temperature is 25°C. The cylinder is placed in boiling water with the piston held in a fixed position. Heat in the amount of 2.00 kcal is transferred to the gas, which equilibrates at 100°C. The piston is then released and the gas does 100J of work in moving the piston to its new equilibrium position. The final gas temperature is 100°C. • Write the energy balance equation for each of the two stages of this process, and each case solve for the unknown energy term in equation. Consider the gas in the cylinder to be the system, neglect the change in potential energy, and assume the gas behaves ideally. Express all energies in joules.**Energy Balances on Open System**• In open system, mass is transferred across the system boundaries while the process is taking place. • Therefore work must be done on open system to push mass in and work is done on the surrounding by mass that emerges from the systems. • Both work terms must be include in the energy balance for open system • The net work done by an open system and - shaft work rate of work done by the process fluid on a moving part within the system such as a pump rotor. - flow work rate of work done by the fluid at the system outlet minus rate of work done by the fluid at the system inlet.**Energy Balances on Open System**• ^ symbol is used to denote the specific property (property divided by mass or by mole) such as specific internal energy (Û kJ/kg), specific volume ( m3/kg) and so on. • One important property for energy balance on open system is specific enthalpy (Ĥ kJ/kg). • Sometimes, universal gas law constant can be used as a conversion factor to evaluate specific enthalpy.**Let’s try**• The specific internal energy of helium at 300K and 1 atm is 3800 J/mol, and the specific molar volume at the same temperature and pressure is 24.63 L/mol. Calculate the specific enthalpy of helium at this temperature and pressure, and the rate at which enthalpy is transported by a stream of helium at 300K and 1 atm with a molar flow rate of 250 kmol/h**Let’s try**• Five hundreds kilograms per hour of steam drives a turbine. The steam enters that turbine at 44 atm and 450°C at a linear velocity of 60 m/s and leaves at a point 5m below the turbine inlet at atmospheric pressure and a velocity of 360 m/s. The turbine delivers shaft work at a rate of 70 Kw, and the heat loss from the turbine is estimated to be 104 kcal/h. Calculate the specific enthalpy change associated with the process**Reference States and State Properties**• It is not possible to know the absolute value of Û and Ĥ for a process material, but we can determine the change in ΔÛ and change in ΔĤ corresponding to a specific change of state (temperature, pressure, phase). • A convenient way to tabulate ΔÛ and ΔĤ is to choose a temperature, pressure and state of aggregation (i.e. phase) as a reference state. • Since Ĥ cannot be known absolute, for convenience we may assign a value Ĥo=0 to be reference state. Then ΔĤ1= Ĥ1-0; ΔĤ2= Ĥ2-0 and so on. • Some enthalpy tables give the reference states on which the listed values of Ĥ are based and others do not. • However, we do not have to know the reference state to calculate ΔĤ for the transition from one state to another state since the value are based on the same reference in the table. • Ĥ and Û can be said as state properties • Property whose change of value in any process depend only on it initial and final states and do not depend on the path take to reach the state.**Class Discussion**Example 7.5-1**Steam Table**Class Discussion Example 7.5-2**Class Discussion**Example 7.5-3**Energy Balance Tips**• When labeling flowchart, write down together the temperature, pressure and state of aggregation of the process material. • Normally (depend on the process description) for chemical process unit; shaft work, kinetic and potential energy change tend to be negligible compared to heat flows, internal energy and enthalpy changes. • Then simplified energy balance become Closed System: Open System: