ME 200 L6: Energy Rate Balance, Transient Operation, Cyclic Repetitive Operation, Cycle Analysis, Efficiency & Coefficient of Performance Spring 2014 MWF 1030-1120 AM J. P. Gore gore@purdue.edu Gatewood Wing 3166, 765 494 0061 Office Hours: MWF 1130-1230 TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edu

Closed System Energy Balance Energy is an extensive property that includes the internal energy, the kinetic energy and the gravitational potential energy. For closed systems, energy is transferred in and out across the system boundary by two means only: by work and by heat. Energy is conserved. This is the first law of thermodynamics.

Closed System Transient Energy Balance The time rate form of the closed system energy balance is (Eq. 2.37) The rate form expressed in words is rate of change of energy in the system at time t net rate of transfer in by heat net rate of transfer out by work Just as in calculus, separate variables and integrate

Change in Energy of a System The changes in energy of a system from state 1 to state 2 consist of internal, kinetic and potential energy changes. (Eq. 2.27a) DE = DU + DKE + DPE (Eq. 2.27b) Energy at state 1 or state 2 or any other state is defined in reference to a standard state. Definition of energy at all states must have identical standard base state. Changes in the energy of a system between states, defined with identical standard state have significance.

Home Work Problem TV Music speakers Door locked Imagine a party at a college location as sketched below. Bob goes to the refrigerator door to get a soda… Music speakers Well insulated party room A/C Vent   TV Electrical supply cable   Refrigerator (fridge) door open Door locked

Example 1 An electric generator coupled to a windmill produces an average power of 15 kW. The power is used to charge a storage battery. Heat transfer from the battery to the surroundings occurs at a constant rate of 1.8 kW. For 8 h of operation, determine the total amount of energy stored in the battery, in kJ. Find: ΔE in kJ? System Given W = -15 kW Q = -1.8 kW Δt = 8 h Assumptions The battery is a closed system. The work and heat transfer rates are constant. storage battery Q = ?1.8 kW Δt = 8 h W = ?15 kW Basic Equation 6

Example 2 An electric motor draws a current of 10 amp with a voltage of 110 V. The output shaft develops a torque of 10.2 N-m and a rotational speed of 1000 RPM. For operation at steady state, determine for the motor, each in kW. the electric power required. the power developed by the output shaft. the rate of heat transfer. Find Welectric in kW? Wshaft in kW? Q in kW? Sketch Given I = 10 amp V = 110 V τ = 10.2 N-m ω = 1000 RPM Assumptions The motor is a closed system. The system is at steady state. Basic Equations motor τ = 10.2 N-m ω = 1000 RPM I = 10 amp V = 110 V + - 7 7

Example 2 Given Find Sketch Basic Equations Solution I = 10 amp V = 110 V τ = 10.2 N-m ω = 1000 RPM Find Welectric in kW? Wshaft in kW? Q in kW? Sketch Basic Equations Solution motor τ = 10.2 N-m ω = 1000 RPM I = 10 amp V = 110 V + - 8 8 8

Example 3 A gas within a piston-cylinder assembly (undergoes a thermodynamic cycle consisting of) three processes: Process 1-2: Constant volume, V = 0.028 m3, U2 – U1 = 26.4 kJ. Process 2-3: Expansion with pV = constant, U3 = U2. Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ. There are no significant changes in kinetic or potential energy. Sketch the cycle on a p-V diagram. Calculate the net work for the cycle, in kJ. Calculate the heat transfer for process 2-3, in kJ. Calculate the heat transfer for process 3-1, in kJ. 9

Example 3 Assumptions Find System Basic Equations Given The gas is the closed system. For the system, ΔKE = ΔPE = 0. Volume change is the only work mode. Basic Equations Find p-V diagram Wnet = ? in kJ Q23 = ? in kJ Q31 = ? in kJ System Given 1-2: V = 0.028 m3, U2 – U1 = 26.4 kJ 2-3: pV = constant, U3 = U2 3-1: p = 1.4 bar, W31 = -10.5 kJ gas 10

Example 3 Solution 1-2: Constant Volume Heat Addition 2-3: Isothermal Expansion, Heat added to maintain T in spite of Expansion. 3-1: Constant Pressure Heat Rejection and “exhaust,” leading to volume reduction work is put into the system 2 P, atm 3 1 V, m3 11

Example 3 12

Cycle Analysis, Efficiency and Coefficient of Performance When a working substance returns to the original state in a cyclic manner while accepting and rejecting heat from two reservoirs and delivering net work in the process, we have an engine cycle. When a working substance returns to the original state in a cyclic manner while accepting heat from a low temperature reservoir and delivering heat to a high temperature reservoir we have a refrigerator or a heat pump cycle. If the cold reservoir substance is the useful substance then it is a refrigerator if the hot reservoir contains the useful substance then we have a heat pump.

Return to Example 3 1-2: Constant Volume Heat addition 2-3: Isothermal Expansion, Heat added to maintain T in spite of Expansion. 3-1: Constant Pressure Heat Rejection and “exhaust,” leading to volume reduction work is put into the system 2 P, atm 1 3 V, m3 First Law of Thermodynamics or Conservation of Energy is satisfied. For this cycle 1-2 and 2-3 are the heat addition processes and the customer Pays for the fuel that leads to this heat. 14

Flip the Engine to make it a Heating/Cooling Device 3-1: Constant Volume Heat Rejection 2-3: Isothermal Compression, Heat removed to maintain T in spite of Compression. 1-2: Constant Pressure Heat Extraction from cold space leading to expansion of working substance. 3 P, atm 1 2 V, m3 First Law of Thermodynamics or Conservation of Energy is satisfied. 15