MEEN 315, SECTION 503 PRINCIPLES OF THERMODYNAMICS Spring 2013 Homework #7 DUE March 19, 2013 N.B.: The enhancement lecture “Example_SS Non Work Device” may be helpful on this homework. It is located on the elearning website LecturesEnhancement LecturesExample_SS Non Work Device. 1. Carbon dioxide enters a steady-state, steady-flow heater at 300 kPa and 15°C, and exits at 275 kPa and 1200°C, as shown below. Changes in kinetic and potential energies are negligible. Calculate the required heat transfer per kilogram of carbon dioxide flowing through the heater. 2. An automotive radiator has glycerine at 200 F enter and return at 130 F, as shown below. Air flows in at 68 F and leaves at 77 F. If the radiator should transfer 33 hp, what is the mass flow rate of the glycerine and what is the volume flow rate of air in at 15 psia? NOTE: You can calculate the change in enthalpy as ߂݄ ൌ ܥ௟௜௤௨௜ௗ߂ܶ. 3. Power plants, such as those that produce our electricity, are thermodynamic systems composed of a collection of steady-state devices (or processes). This collection of processes typically operates on a cycle. As we will learn later, we often call such cycles “heat engine cycles.” The figure below is a schematic example of a steam power plant. The pump pressurizes liquid water and feeds the pressurized water to the boiler (in our figure below, the boiler is a collection of two processes called the “economizer” and the “steam generator”). The boiler’s main function is to vaporize the pressurized liquid water. Thus, exiting the boiler is often high pressure superheated vapor with a high amount of enthalpy (or energy content). Most power plants use combustion of oil, coal, or natural gas within the boiler to provide the high energy into the water. A less number of power plants use nuclear fuel (i.e., a nuclear power plant) to provide the high energy into the water. The high energy could even be provided by the sun, as in solar energy (although this is often not a highly efficient use of solar energy). The turbine is usually the next device in a power plant, which expands the high pressure superheated water vapor to near saturated water conditions. In doing so, the turbine transfers energy within the superheated vapor (enthalpy) into work. In other words, the high-energy steam passing through the turbine causes its shaft to spin,which we can then extract work from. The most common use of this work is to produce electricity. That is, the turbine shaft is commonly coupled to an electrical generator which when spun (by the turbine) generates electricity. After the useful energy is extracted from the water in the turbine, the water passes through a heat exchanger called a “condenser.” The condenser removes the remaining amount of energy in the water so that the water returns to its initial state entering the pump (i.e., liquid water). Often times, the energy removed from the water in the condenser is “rejected” to a nearby cooling source, such as a river or a cooling pond. The following data are for our simple steam power plant shown in the figure above. State 1 2 3 4 5 6 7 P (psia) 900 890 860 830 800 1.5 1.4 T (F) 115 350 920 900 110 h (Btu/lbm) 85.3 323 1468 1456 1029 78 State 6 has a quality, x6 = 0.92 and a velocity of 600 ft/s. The rate of steam flow is 200,000 lbm/h, with 400-hp input to the pump. Piping diameters are 8 in. from the steam generator to the turbine and 3 in. from the condenser to the steam generator. Determine the power output of the turbine and the heat transfer rate in the condenser. 4. Air enters an adiabatic nozzle at 45 psia and 940 F with low velocity and exits at 650 ft/s. If the isentropic efficiency of the nozzle is 85 percent, determine the exit temperature and pressure of the air. 5. Refrigeration systems, such as those we use to cool our homes (air conditioners) and food (refrigerators / freezers) operate on what is called a “refrigeration cycle” or a “heat pump cycle”. There is a difference in objective between a refrigerator and a heat pump, but they functionally operate the same (we’ll talk about this later on). The way a refrigeration system works is as follows (use the figure below as a guide): 1) A refrigerant (e.g., R-134a) enters a compressor as a slightly super-heated vapor (i.e., just above saturated vapor point) and is compressed to a higher pressure and temperature. It then goes through a condenser, which removes thermal energy from the R-134a and rejects it to a high temperature environment (which is at a lower temperature than the R-134a through the condenser). The R-134a then enters an expansion valve, which is like a throttle, which substantially decreases its pressure. The R-134a is now a very cold saturated substance (two-phase); it then absorbs thermal energy from a low temperature environment (which is at a higher temperature than the R-134a) through the evaporator. Suppose R-134a enters an expansion valve (i.e. a throttle) as a saturated liquid at 800 kPa. The expansion valve expands the substance to a pressure of 120 kPa. What is the exit temperature of the R-134a from the expansion valve? N.B., Do not use the temperatures shown in the diagram below. Calculate the change in entropy through the expansion valve and the entropy generation for the process. Why is entropy generation positive (i.e., not equal to zero)?Carbon dioxide enters a steady-state. sleady-flow heater at 300 kPa and 15°C. and exits al275 kPa and 1200°C. as shown below. Changes in kinetic and polential energies are negligible. Calculate the required heat transfer per kilogram ofcarbon dioxide flowing through the heater. ~, ~mJlP-t ___ ...._ .J' ~----~~ c.'S c.. (;).t (,~ ps) ~I r , J -h-k i.. : ?i ;r 300 kP.:... 1/'6 IZ.;;; 1'$ v <:...-S.fz,t.J.e e: Pc-;:: ~1-f:' k.~ prvc.e,s: ~- !, ~I-e Lo.4<r 1'(.. .:;: I~OD"'C g ::-h~-h.: 1::. TCf !::-<;t 6~~ -I.:: ') afl~ ::: O. 'a¥(P ~ - ((). '8'«" ~('I)()() -,'S"') /'00::;. S I ~An automotive radiator ha~ glycerine al 200 F enter and return al 130 F, as shown below. Air flows in at 68 F and leaves at 77 F. If the radiator should transfer 33 hp, what is the mass flow rate of the glycerine and what is the volume flow rate of air in at 15 psia? NOTE: You can calculate the change in enthalpy as llh C/iquidflT . .. s.Jv-.k. \: I, ;:: ~00 F .i ~.J.u.-k::J..: -r-::-130 t=:' · , , I =: It-rIl~~Ol ·: ¢ -"V>.... £.<..o.Q('f-.,Q.,,, t