ME 200 L19: ME 200 L19:Conservation Laws: Cycles HW 7 Due Wednesday before 4 pm HW 8 Posted Start early Kim See’s Office ME Gatewood Wing Room ThermoMentor © Program Launched Spring 2014 MWF AM J. P. Gore Gatewood Wing 3166, Office Hours: MWF TAs: Robert Kapaku Dong Han

2 Common Steady-flow Energy Devices Nozzles Compressors Heat Exchangers and Mixers Throttles 2 Water, Steam, Gas Turbines Pump Diffusers

Heat Exchangers ► Direct contact: A mixing chamber in which hot and cold streams are mixed directly. ► Tube-within-a-tube counterflow: A gas or liquid stream is separated from another gas or liquid by a wall through which energy is conducted. Heat transfer occurs from the hot stream to the cold stream as the streams flow in opposite directions.

► if there is no stirring shaft or moving boundary. ► ΔKE = ( V i 2 /2-V e 2 /2) negligible unless specified. ► ΔPE = negligible unless specified. ► If Heat transfer with surroundings is negligible. ► Control Volume includes both hot and cold flows. The “heat exchange,” between them is internal! Heat Exchanger Modeling

Example Problem: Heat Exchanger Given: Air and Refrigerant R-22 pass through separate streams through an insulated heat exchanger. Inlet and exit states of each are defined. Find:(a) Mass flow rates, (b) Energy transfer from air to the refrigerant. R-22 Air Assumptions: Flow work only, insulated casing, steady state, steady flow, no leaks. Data: Av 1 = 40m 3 /min,1=27 C=300K, P1= 1.1 bars T2=15 C= 288 K, P2= 1bar P4= 7 bars, T4=15 C, P3=7 bars, x3=0.16 R22 properties: Table A-9, A-8. P=7 bars, Tsat=10.91 C. Therefore, 4 is superheated and h4= kJ/kg. Table A-8 h 3 =hf 3 +x 3 h fg3 = (.16)(195.6) =89.34 kJ/kg.

Example Problem: Heat Exchanger Given: Air and Refrigerant R-22 pass through separate streams through an insulated heat exchanger. Inlet and exit states of each are defined. Find:(a) Mass flow rates, (b) Energy transfer from air to the refrigerant. R-22 Air Assumptions: Flow work only, insulated casing, steady state, steady flow, no leaks. Data: Av 1 = 40m 3 /min,1=27 C=300K, P1= 1.1 bars T2=15 C= 288 K, P2= 1bar P4= 7 bars, T4=15 C, P3=7 bars, x3=0.16 R22 properties: Table A-9, A-8. P=7 bars, Tsat=10.91 C. Therefore, 4 is superheated and h 4 = kJ/kg. Table A-8 h 3 =hf 3 +x 3 h fg3 = (.16)(195.6) =89.34 kJ/kg.

Example Problem: Heat Exchanger Given: Air and Refrigerant R-22 pass through separate streams through an insulated heat exchanger. Inlet and exit states of each are defined. Find:(a) Mass flow rates, (b) Energy transfer from air to the refrigerant. R-22 Air Assumptions: Flow work only, insulated casing, steady state, steady flow, no leaks. Data: Av 1 = 40m 3 /min,1=27 C=300K, P1= 1.1 bars T2=15 C= 288 K, P2= 1bar P4= 7 bars, T4=15 C, P3=7 bars, x3=0.16

► Engineers creatively combine components to achieve some overall objective, subject to constraints such as minimum total cost. This engineering activity is called system integration. System Integration ► The simple vapor compression refrigeration cycle provides an illustration. ► An integrated system that transfers heat from a low T reservoir to a high T reservoir using work is called a Heat Pump or a Refrigerator

Heat Pumps and Refrigerators Heat Pumps and Refrigerators are thermodynamic devices that take heat from a low temperature reservoir and pump it into a high temperature reservoir using cyclic external work input. The only difference between a refrigerator and a heat pump is in the “desired energy result”: –it’s Q C the heat removed from a low temperature reservoir for a refrigerator. –it’s Q H the heat input into a high temperature reservoir for a heat pump. The required energy input is the compressor work in each case. 9

Heat Pumps and Refrigerators The figure of merit is the same, desired energy output/required energy input. It’s termed the coefficient of performance Carnot or Max performance 10

Heat Pumps and Refrigerators 11 Vapor-compression refrigeration cycle 120 kPa, -22 o C, Sat. Vapor 120 kPa, -22 o C, Sat. L+V 800 kPa, 60 o C, SHV 800 kPa, 31 o C, SCL

Refrigerator and Heat Pump Example 12

Heat Engines 13 Steam Power Plant

Heat Engines A generic heat engine can be represented as: Features include: –receiving heat from a high T source. –producing net mechanical work. –rejecting heat to a low T sink. –operating in a cyclic manner. 14

Heat Engines Application of the First Law to our (cyclic) heat engine gives Cyclic operation implies no net change in (h+V 2 /2+gZ) over the cycle hence: 15

Heat Engines Application of the First Law to our (cyclic) heat engine gives Cyclic operation implies no net change in (h+V 2 /2+gZ) over the cycle hence: This is usually written in terms of the net work. 16

Heat Engines Thermal efficiency is the “figure of merit” for heat engines. It’s utility comes from indicating what fraction of the energy added to the system is converted to mechanical work. Thermal efficiency = Energy you get / Energy you pay for 17 Thermal efficiency = Desired energy result / Required energy input

Heat Engines 18

2 nd Law Corollaries for Power Cycles No power cycle can have a thermal efficiency of 100%. Carnot Corollaries –The thermal efficiency of an irreversible power cycle is always less that the thermal efficiency of a reversible power cycle when each operates between the same two thermal reservoirs. –All reversible power cycles operating between the same two thermal reservoirs have the same thermal efficiency. 19