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Chapter 2: Review Lecture 05: Chapter 2 Review Quiz Today? Main Concepts of Chap 2. • Work done by compressed gas system: • Heat Transfer W pdV ▫ Conduction ▫ Convection ▫ Radiation • 1st law of Thermodynamics: E Q in W out • Power cycle, Refrigeration cycle, Heat Pump cycle • Thermal Efficiency and Coefficient of Performance. Reading Assignment: • Read Chap 3: Sections 1-5 Homework Assignment: From Chap 2: 46, 59, 82,97 3 Recall Chapter 2 concepts: 1st Law of Thermodynamics: E system P E K E U Q in W o u t system Qin ΔE for most Thermodynamics applications: U Q in W o u t Wout ΔKE = ΔPE = 0 then 4 Work done by Gas System W > 0 : Expansion of Gas W < 0 : Compression of Gas W V f PdV Vi = Area under the p-V graph over process Run Animation 5 Heat Transfer Conduction: Q A where A is area κ is thermal conductivity dT/dx is temperature gradient dT dx Convection: Q h c A Tb T f Radiation: Q A T 4 b where A is area hc is the convection coefficient Tb -Tf is the difference between the body and free steam fluid temp. where Tb is absolute surface temperature ε is emissivity of the surface σ is Stefan-Boltzmann constant A is surface area Thermodynamic Cycle: E Q cycle W cycle = 0 P S1 Clockwise around the cycle: Work is done by the system. S2 Power Cycle: W cycle > 0 Counter clockwise around the cycle: Work is done on the system. S4 S3 Refrigeration Cycle: W cycle < 0 v Sec 2.6: Energy Analysis of Cycles Cycle Models: Power cycle: 7 Effectiveness R esult Effort Refrigeration cycle: Heat Pump Cycle Q in W cycle Q out W cycle Q in Q out W cycle Q in Q out W cycle Q in Q out W cycle Q in Q in Q out Q in Q in Q out Q in W cycle Q out Q out Q in 8 Concept Questions: True or False: a) In principle, expansion work can be evaluated using ∫pdV for both actual and quasi-equilibrium expansion processes. False b) The change in the internal energy of a system between two states is the change in the total energy of the system between the two states less the change of the system’s kinetic and gravitation potential energies between these states. True c) The change in gravitational potential energy of a 2 lb mass whose elevation decreases by 40 ft where g = 32.2 ft/s2 is -2576 ft-lbf . True d) The rate of heat transfer from a hot baked potato to the ambient air is greater with forced convection than natural convection. True 9 Example 3 (Problem 2.91): A heat pump maintains a dwelling at 68oF. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55oF well water at a rate of 500 BTU/min. a) Determine the COP. b) Evaluating electricity at $0.10 per kW-hr, determine the cost of electricity in a month when the heat pump operates for 300 hr. 10 11 Example (2.70): A gas is contained in a vertical piston-cylinder assembly by a piston weighing 1000 lbf and having a face area of 12 in2. The atmosphere exerts a pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy to the gas in the amount of 5 BTU as the elevation Patm=14.7 psi of the piston increases by 2 ft. The piston and cylinder are poor thermal conductors and friction h = 2 ft can be neglected. Determine the change in internal Apiston = 12 in2 energy of the gas, in BTU, assuming it is the only Wpiston = 1000 lbf significant internal energy change of any component present. Welec= - 5 BTU 12 13 Example (2.83): A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ (a) the value of electricity generated per year (b) the waste heat returned to the environment (c) the annual fuel cost (d) Is operation profitable? Qin Fuel Wout =100 MW Air Qout 14 15 Example Problem (2.63) A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar, V2 = 0.02 m3 in a process during which the relation between pressure and volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal energy of the gas increase by 50 kJ/kg during the process, deter the heat transfer in kJ. KE and PE changes are negligible. 16 17 End of Lecture 05: • Solutions to example problems follow 18 Example 3 (Problem 2.91) Page 1 of 2: A heat pump maintains a dwelling at 68oF. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55oF well water at a rate of 500 BTU/min. a) Determine the COP. b) Evaluating electricity at $0.10 per kW-hr, determine the cost of electricity in a month when the heat pump operates for 300 hr. Principle: COP for heat pump (written in terms of power) Q in W in Q out Q out W in W in 5 h p where: and Q in 5 0 0 B tu / m in therefore: Q out W in Q in W in W in Q in W in 1 500 btu / m in 1hp 60 m in 5 hp 2545 btu / h 1hr 1 3.36 19 Example 3 (Problem 2.91)…page 2 of 2: A heat pump maintains a dwelling at 68oF. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55oF well water at a rate of 500 BTU/min. a) Determine the COP. b) Evaluating electricity at $0.10 per kW-hr, determine the cost of electricity in a month when the heat pump operate for 300 hr. Principle: Cost = Cost of energy * Power * time where: W in 5 h p $ / E $0.10 / kW hr t 300 hr therefore: C$ $ E n erg y W in t $ 0 .1 0 kW h r 5hp 300hr 1 kW 1 .3 4 1h p $ 1 1 1 .8 6 20 Example (2.70): Page 1 of 4 A gas is contained in a vertical piston-cylinder assembly by a piston weighing 1000 lbf and having a face area of 12 in2. The atmosphere exerts a pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy to the gas in the amount of 5 BTU as the elevation of the piston increases by 2 ft. The piston and Patm=14.7 psi cylinder are poor thermal conductors and friction can be neglected. Determine the change in internal h = 2 ft energy of the gas, in BTU, assuming it is the only Apiston = 12 in2 significant internal energy change of any component Wpiston = 1000 lbf present. Solution: Apply the 1st law of thermodynamics PE K E U Q in W out U Q in W o u t P E K E Welec= - 5 BTU 21 Example Problem (2.70) …page 2 of 4 where mg = 1000 lbf A = 12 in2 Δh = 2 ft Welec_in = 5 BTU Because of the statement “poor thermal conductors”, it can be assumed that this is an adiabatic process (Q = 0) and we will also assume that the process occurs as a slow quasi-equilibrium process in which case the kinetic energy terms will also be small (ΔKE = 0). Finally, since the piston floats on the contained gas, the outside atmospheric pressure maintains a constant pressure on the cylinder…so this is a constant pressure process (isobaric) therefore: KE 0 Q in 0 P E m g h (1 0 0 0 lb f )( 2 ft ) 2 0 0 0 ft lb f W elect 5 B T U (neg. since its put into the system) V2 W PV V1 p d V p (V 2 V1 ) (for constant pressure) 22 Example Problem (2.70) …page 3 of 4 Ftop=patm A For equilibrium: W=1000lbf F 0 Fb o tto m Fto p W 0 p A p a tm A W 0 p p a tm W Fbottom=p A 1 4 .7 lb f / in 2 A 1 0 0 0 lb f 1 2 in 2 9 8 .0 3 p si and the increase in Volume: V 2 V1 A h V 2 V1 1 2 in ( 2 ft ) 2 1 2 in 2 8 8 in 3 1 ft therefore the work done by the gas was positive work by the system V2 W PV V1 p d V p (V 2 V1 ) (9 8 .2 lb f / in )( 2 8 8 in ) 2 3 1 ft 1 2 in 2 3 5 7 lb f ft 23 Example Problem (2.70) …page 4 of 4 Returning to the 1st law: U Q in W out PE K E U 0 (W P V W elec ) P E 0 U ( 2 3 5 7 ft lb f 5 B T U 7 7 8 ft lb f 1B T U ) ( 2 0 0 0 ft lb f ) U 4 6 7 ft lb f U 4 6 7 ft lb f 1B T U 7 7 8 ft lb f 0 .6 0 B T U 24 Example (2.70): A gas is contained in a vertical piston-cylinder assembly by a piston weighing 1000 lbf and having a face area of 12 in2. The atmosphere exerts a pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy to the gas in the amount of 5 BTU as the elevation Patm=14.7 psi of the piston increases by 2 ft. The piston and cylinder are poor thermal conductors and friction h = 2 ft can be neglected. Determine the change in internal Apiston = 12 in2 energy of the gas, in BTU, assuming it is the only Wpiston = 1000 lbf significant internal energy change of any component present. Welec= - 5 BTU 25 Example (2.83): Page 1 of 3 A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ (a) the value of electricity generated per year (b) the waste heat returned to the environment (c) the annual fuel cost (d) Is operation profitable? Qin Fuel Wout =100 MW Air Qout 26 Example (2.83): Page 2 of 3 A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ (a) the value of electricity generated per year and R evenue 100 M W 10 kW 8000 hr $0.08 $64 10 M W year kW hr year 3 (b) next find the heat generated 1 0 0 M W W out W out Q in Q in 0 .4 0 6 250 M W and the heat returned to the environment Q in W out Q out Q out Q in W out (250 100)M W 150 M W 27 Example (2.83): page 3 of 3 A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ (c) Cost of fuel? C ost fuel $32.4 10 $4.50 G J s 8000 hr 3600 s 250 M W 3 year G J 10 M W year hr (d) Is operation profitable? Profit = Revenue – Costs Profit $ 64 10 year 6 $ 32 . 4 10 year 6 $ 31 . 6 10 6 year So, this could be profitable, but the calculation ignore other costs such as capital and labor. 6 28 Example Problem (2.63) page 1 of 4 A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar, V2 = 0.02 m3 in a process during which the relation between pressure and volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal energy of the gas increase by 50 kJ/kg during the process, deter the heat transfer in kJ. KE and PE changes are negligible. Solution: starting with the 1st Law of Thermodynamics P E K E U Q in W o u t where: ΔKE=0 ΔPE = 0 p1 = 2 bar also: p2 = 8 bar ΔU/m = 50 kJ/kg V1 = ? m = 0.2 kg V2 = 0.02 m3 pV1.3 = constant therefore: 1 .3 p1V1 p 2V 2 1 .3 p V1 2 p1 1 / 1 .3 8 bar V2 2 bar 1 / 1 .3 0 .0 2 m 0 .0 5 8 1 m 3 3 29 Example Problem (2.63)…page 2 of 4 A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar, V2 = 0.02 m3 in a process during which the relation between pressure and volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal energy of the gas increase by 50 kJ/kg during the process, deter the heat transfer in kJ. KE and PE changes are negligible. Solution continued: also:p V p 2V 2 1 .3 1 .3 (8 b a r )(0 .0 2 m ) 3 p 0 .0 4 9 5 / V 1 .3 1 .3 0 .0 4 9 5 b a r m (0 .0 4 9 5 b a r m 3 .9 )V 3 .9 1 .3 therefore: 1 .3 p1V1 p 2V 2 1 .3 p V1 2 p1 1 / 1 .3 8 bar V2 2 bar 1 / 1 .3 0 .0 2 m 0 .0 5 8 1 m 3 3 30 Example Problem (2.63)…page 3 of 4 so work done is: V2 W V1 V2 PdV ((0 .0 4 9 5 b a r m 3 .9 0.0495 bar m 3.9 )V 1 .3 )dV 0 .0 4 9 5 b a r m 0 .3 V1 0 .3 V1 3.9 3.224 m 0.9 1.177 m 0.9 0.3 V (0.02 m 3 ) 0.3 (0.581 m 3 ) 0.3 0.3 0.0495 bar m V2 3 .9 0.0495 bar m 3.9 3.224 m 0.9 1.177 m 0.9 0.3 0.338 bar m 3 100 kN / m 1 bar 2 1kJ 1kN m 33.8 kJ 31 Example Problem (2.63)…page 4 of 4 Internal Energy is given as U ( U / m )( m ) (5 0 kJ / kg )(0 .2 kg ) 1 0 kJ Finally back at the 1st Law: PE K E U Q in W out gives Q in P E K E U W o u t 0 0 (1 0 kJ ) ( 3 3 .7 8 kJ ) 2 3 .7 8 kJ