1. ME 200 L18: ME 200 L18:Conservation Laws: Heat Exchangers HW 7 Posted Due in One Week: Kim See’s Office ME Gatewood Wing Room 2172 https://engineering.purdue.edu/ME200/ ThermoMentor © Program Launched Spring 2014 MWF 1030-1120 AM I. Sircar for J. P. Gore (No Office Hours Today) gore@purdue.edu Gatewood Wing 3166, 765 494 0061 Office Hours: MWF 1130-1230 TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edurkapaku@purdue.eduhan193@purdue.edu

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

3. 3 Rotating Machinery A turbine is a steady-flow device used to produce mechanical work (W) by reducing the internal &/or kinetic &/or potential energy of the working fluid. For gas turbines, the fluid drives rotating blades while the υ increases from inlet to exit as the working fluid expands (or the p drops). 3

4. 4 Steam Turbine Example 4 6 Kg/s of steam at an inlet velocity of 75 m/s enter a turbine stage at 3 MPa and 400 o C and exit at a velocity of 125 m/s at a pressure of 2 Mpa at 360 o C. Find: (a) the power developed by this steam turbine stage, (b) the percentage change in the steam density across the turbine and (c) the percentage change in the flow area. The heat loss through the casing is 33 kW.

5. Air Compressors September 17th, 2010ME 2005 A pump is a steady-flow device that consumes shaft work from rotating blades that compress the fluid. Compressors are used for gas systems, pumps for liquids so operating assumptions are similar.

6. 6 Example At steady state, a well-insulated compressor takes in air at 60 ºF, 14.2 psi, with a volumetric flow rate of 1200 ft 3 /min, and compresses it to 500 ºF, 120 psi. Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in hp, and the volumetric flow rate at the exit, in ft 3 /min. 6

7. 7 Example Find –W cv = ? in hp –A 2 V 2 = ? in ft 3 /min System (air flowing through compressor) Assumptions The control volume is at steady state; the flow is steady Q, Δke, and Δpe are negligible. T he air is an ideal gas. Basic Equations air P 1 = 14.2 psi T 1 = 60 ºF A 1 V 1 = 1200 ft 3 /min P 2 = 120 psi T 2 = 500 ºF compressor 12 7

8. 8 Example Solution 8

9. 9 Example 9 From Table A-22E

10. 10 Example 10

11. 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.

12. ► 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

13. 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 3 4 1 2 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=256.86 kJ/kg. Table A-8 h 3 =hf 3 +x 3 h fg3 = 58.04+(.16)(195.6) =89.34 kJ/kg.

14. 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 3 4 1 2 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

15. 15 Summary Control volume energy and mass conservation equations that we learned are applicable to many practical energy devices and equipment. and their propertiesImportant learning comes from application of appropriate assumptions, considering the appropriate working substances and their properties in the proper range of operation to estimate different energy quantities. 15