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EGR 334 Thermodynamics Chapter 4: Section 6-8

EGR 334 Thermodynamics Chapter 4: Section 6-8. Lecture 16: Control Volume Applications: Day 1. Quiz Today?. Today’s main concepts:. Be able to set up mass and energy balance models for Turbines Pumps Compressors Boilers Heat Exchangers Nozzles Diffusers Throttle.

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EGR 334 Thermodynamics Chapter 4: Section 6-8

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  1. EGR 334 ThermodynamicsChapter 4: Section 6-8 Lecture 16: Control Volume Applications: Day 1 Quiz Today?

  2. Today’s main concepts: • Be able to set up mass and energy balance models for Turbines Pumps Compressors Boilers Heat Exchangers Nozzles Diffusers Throttle Reading Assignment: • Read Chapter 4, Sections 10-12 Homework Assignment: Problems from Chap 4: 36, 43, 52, 66

  3. Review: For a Control Volume: Mass Rate Balance: Energy Rate Balance:

  4. Modeling applications with Control Volumes: Many important applications involve one inlet, one exit control volumes at steady state. Today a number of these useful models will be developed using the one inlet, one outlet, steady state forms of the mass balance and energy balance given below. Mass Rate Balance: 1 path, steady state Energy Rate Balance: 1 path, steady state

  5. Control Volume Applications: Boiler Nozzles Diffuser Turbine Pump Compressor Heat Exchanger Throttling Valve

  6. Common Modeling assumptions: Application models generally make use of simplifying assumptions to reduce the complexity of the Energy Balance. By removing terms that do not apply to a particular application or whose impact on the application is generally only minor the models take on simplified, useful, and easy to use forms. Assumption: If: a) Outer surface of CV is well insulated… b) Small change of elevation… c) No mechanical mechanisms present… d) CV maintains same shape and volume… e) No electrical effects act on CV… f) Inlet and Outlet have same physical size… g) Outer surface of CV is small… h) Small ΔT between CV and environment… i) flow passes through CV in short time… j) inlet size much larger than outlet size…

  7. Nozzles and Diffusers • Nozzle: a flow passage of varying cross-sectional area in which the velocity of a gas or liquid increases in the direction of flow. • Diffuser: a flow passage of varying cross-sectional area in which the velocity of a gas or liquid decreases in the direction of flow.

  8. Sec 4.6: Nozzles and Diffusers Nozzles and Diffusers are used to change the speed of the mass flow through the control volume. Ve Vi Ve Vi • For continuous flow, changing the size of the cross section alters the speed of the flow. • if • continuous • if • incompressible •  •  • What common assumptions may be used to simplify the energy balance?

  9. Sec 4.6: Nozzles and Diffusers • Typical Energy Balance simplifications, • Steady State • Horizontal Section • (or very short vertical) • No pump/turbines • Even though there is no insulation, the V is high so there may be little heat transfer. • Therefore,

  10. Sec 4.6: Nozzles and Diffusers Example: (4.34) Air with a mass flow rate of 5 lb/s enters a horizontal nozzle operating at steady state at 800°R, 50 psi and a velocity of 10 ft/s. At the exit, the temperature is 570°R and the velocity is 1510 ft/s. Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing. . m = 5 lbm/s • 50 psi • 800°R • 10 ft/s • 570°R • 1510 ft/s • Ain = ?, Q = ?

  11. Sec 4.6: Nozzles and Diffusers Example: (4.34) Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing. mass balance • 50 psi • 800°R • 10 ft/s • 570°R • 1510 ft/s • Ain = ?, Q = ? continuity ideal gas

  12. Sec 4.6: Nozzles and Diffusers Example: (4.34) Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing. Energy balance at SS reduced using assumptions: • 50 psi • 800°R • 10ft/s • 570°R • 1510ft/s from Table A-22E

  13. Turbines • Turbine: a device in which power is developed as a result of a gas or liquid passing through a set of blades attached to a shaft free to rotate.

  14. Sec 4.7: Turbines A turbine is a device that develops power from a gas or liquid passing through a set of blades which are attached to a shaft free to rotate. • Use Mass and Energy Balances still hold:

  15. Sec 4.7: Turbines • Typical Energy Balance simplifications, • Steady State • Horizontal Section • (or very short vertical) • V  0 • Even though there is no insulation, the V is high so there is no heat transfer. • . • min • Therefore, • WCV • . • mexit • Then,

  16. Sec 4.7: Turbines Example: (4.50) Steam enters the first stage of a turbine at 40 bar and 500 °C with a volumetric flow rate of 90 m3/min. Steam exits the turbine at 20 bar and 400°C. The steam is then reheated at constant pressure to 500 °C before entering the second stage turbine. Steam leaves the second stage as saturated vapor at 0.6 bar. For operation at steady state, and ignoring stray heat transfer and KE and PE effects, determine the • Mass flow rate of steam, in kg/hr • Total power produced by both turbines, in kW • The rate of heat transfer to the steam flowing through the reheater, in kW. • 40 bar • 500°C • 90 m3/min • Sat. vapor • 0.6 bar • WCV,1 • WCV,2 • Assumptions • KE=  PE=0 • QTurbine = 0 • 20 bar • 400°C • 20 bar • 500°C • Reheater

  17. Sec 4.7: Turbines Example: (4.50) • Mass flow rate of steam, in kg/hr • Total power produced by both turbines, in kW • The rate of heat transfer to the steam flowing through the reheater, in kW. • 40 bar • 500°C • 90 m3/min • Sat’d vapor • 0.6 bar • Assumptions • KE=  PE=0 • QTurbine = 0 • 20 bar • 400°C • 20 bar • 500°C 1) Identify state properties given in problem statement. • WCV,1 • WCV,2 2) Find intensive properties from Table A-4 • Reheater

  18. Sec 4.7: Turbines Example: (4.50) • Mass flow rate of steam, in kg/hr • Total power produced by both turbines, in kW • The rate of heat transfer to the steam flowing through the reheater, in kW. To find mass flow rate from volumetric flow rate:

  19. Sec 4.7: Turbines Example: (4.50) • Mass flow rate of steam, in kg/hr • Total power produced by both turbines, in kW • The rate of heat transfer to the steam flowing through the reheater, in kW. To find the power produced in both Turbines:

  20. Sec 4.7: Turbines Example: (4.50) • Mass flow rate of steam, in kg/hr • Total power produced by both turbines, in kW • The rate of heat transfer to the steam flowing through the reheater, in kW. Heat transferred in the reheater…starting with energy balance Simplified with assumptions: 

  21. Compressors and Pumps • Compressors and Pumps: devices in which work is done on the substance flowing through them to change the state of the substance, typically to increase the pressure and/or elevation. • Compressor: substance is gas • Pump: substance is liquid

  22. Sec 4.8: Compressors and Pumps Compressors and Pumps: Device where work is used to increase pressure and/or elevation of the flow substance. • min • Pump Model:used for liquids • Compressor Model: • used for gases . • WCV • mexit . . • WCV .

  23. Sec 4.8: Compressors and Pumps • Typical Energy Balance simplifications, • Steady State • Horizontal Section • (or very short vertical) • v0 • Even though there is no insulation, the V is high so there is no heat transfer. • min . • Therefore, • WCV • mexit . • Then, .

  24. Sec 4.8: Compressors and Pumps • Example: (4.60) Air is compressed at steady state from 1 bar, 300 K, to 6 bar with a mass flow rate of 4 kg/s. Each unit of mass passing from the inlet to the exit undergoes a process described by pV1.27= constant. Heat transfer occurs at a rate of 46.95 kJ/kg of air flowing to the cooling water circulating in a water jacket enclosing the compressor. If ΔKE and ΔPE of the air are negligible, calculate the compressor power in kW. • 1 bar • 300 K • 4 kg/s • Assumptions • Steady State • KE= PE=0 • pV1.27= constant • Ideal gas • QCV /m= 46.95 kJ/kg • 6 bar • WCV . • Table A-22 (Ideal Gas Properties of Air) • h is independent of p

  25. Sec 4.8: Compressors and Pumps • 1 bar • 300 K • 4 kg/s Example: (4.60) • 6 bar • QCV = -46.95 kJ/kg From ideal gas equation and polytropic eq. Rearranging: • WCV Combining . 

  26. Sec 4.8: Compressors and Pumps Example: (4.60) • 1 bar • 300 K • 4 kg/s • 6 bar • QCV = -46.95 kJ/kg Therefore the exit temperature is (Since T2 and p2 are now known, h2 may be found on table A22) The energy balance can then find the pump work • WCV .

  27. end of Lecture 16 Slides

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