Chapter 4

1 4-29 Chapter 4 Example ---------------------------------------- ---------------------------------------- -- Steam enters the condenser of a vapor power plant at bar with a quality of and condensate exits at bar and 45oC. Cooling water enters the condenser in a separate stream as a liquid at 20oC and exits as a liquid at 35oC with no change in pressure. heat transfer from the outside of the condenser and changes in the kinetic and potential energies of the flowing streams can be ignored. For steady-state operation, determine (a) the ratio of the mass flow rate of cooling water to the mass flow rate of the condensing steam.

2 (b) the rate of energy transfer from the condensing steam to the cooling water, in kJ per kg of steam passing through the condenser. Solution ---------------------------------------- ---------------------------------------- ---------- (a) Determine the ratio of the mass flow rate of cooling water to the mass flow rate of the condensing steam. The rate of heat transfer from the condensing steam to the cooling water is given by Q = hm (h1 - h2) = cm (h4 - h3) In this equation, hm is the mass flow rate of the hot stream (or condensing steam) and cm is the mass flow rate of the cold stream (or cooling water). The specific enthalpies of the inlet and exit streams are listed in Table For stream (2), (3), and (4) the enthalpies are taken as saturated liquid at the listed temperature so that h2 = hf(T2), h3 = hf(T3), and h4 = hf(T4).

3 6 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 173 4-30 Table Steam properties from CATT2 Specific Temp Pressure Enthalpy Quality Phase C MPa kJ/kg 1 2465 Liquid Vapor Mixture 2 45 0 Saturated Liquid 3 20 0 Saturated Liquid 4 35 0 Saturated Liquid Solving for the ratio of the mass flow rate gives chmm = 1243hhhh-- = = (b) Determine the rate of energy transfer from the condensing steam to the cooling water, in kJ per kg of steam passing through the condenser. Q = hm (h1 - h2) hQm = h1 - h2 hQm = 2465 - = kJ/kg Example ---------------------------------------- ---------------------------------------- -- A supply line carries a two-phase liquid-vapor mixture of steam at 300 lbf/in2.

4 A small fraction of the flow in the line is diverted through a throttling calorimeter and exhausted to the atmosphere at lbf/in2. The temperature of the exhaust steam is measured as 250oF. Determine the quality of the steam in the supply line. Solution ---------------------------------------- ---------------------------------------- ---------- A throttling calorimeter is a device used to reduce the pressure of a gas or liquid stream. This can be simply done by means of a partially opened valve or a porous plus as shown in Figure Figure Examples of throttling devices. 7 Moran, M.

5 J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 177 4-31 Throttling may be used as a means of controlling the flow rate (valves and flow regulators), maintaining a constant downstream pressure (pressure regulator), or measuring the flow rate (flow metering orifices). For throttling devices we usually make the following assumptions: * Steady state steady flow (SSSF) * No work or heat transfer * Potential and kinetic energy are negligible relative to other energy terms Table Steam properties from CATT2 Specific Temp Pressure Enthalpy Quality Phase F psia Btu/lbm 1 300 0 Saturated Liquid 1 300 1204 1 Saturated Vapor 2 250 1169 Superheated Vapor Making energy balance between (1) and (2) we obtain h2 - h1 + g(z2 - z1) + ()222112VV- = mQ - mWs For g(z2 - z1) 0, ()222112VV- 0, mQ = 0, and mWs = 0, we have h2 = h1 = (1 - x1)hf1 + x1hg1 = hf1 + x1(hg1 - hf1)

6 The quality of the steam in the supply line is then x1 = 2111fgfhhhh-- = = 4-32 Energy balance on Integrated or Transient System In real life applications, we usually encounter integrated systems consisting of many components discussed in previous sections. Example ---------------------------------------- ---------------------------------------- -- An industrial process discharges 2 105 ft3/min of gaseous combustion products at 400oF, 1 atm. As shown in Figure , a proposed system for utilizing the combustion products combined a heat -recovery steam generator with a turbine.

7 At steady state, combustion products exit the steam generator at 260oF, 1 atm and a separate stream of water enters at 40 psia, 102oF with a mass flow rate of 275 lb/min. At the exit of the turbine, the pressure is 1 psia and the quality is 93%. heat transfer from the outer surfaces of the stream generator and turbine can be ignored, as can the changes in kinetic and potential energies of the flowing streams. There is no significant pressure drop for the water flowing through the steam generator. The combustion products can be modeled as air as an ideal gas.

8 (a) Determine the power developed by the turbine, in Btu/min. (b) Determine the turbine inlet temperature, in oF. Solution ---------------------------------------- ---------------------------------------- ---------- (a) Determine the power developed by the turbine, in Btu/min. Applying the steady state energy balance on the system consisting of the steam generator and the turbine we obtain with negligible changes in kinetic and potential energies 2m h2 - 1m h1 + 5m h5 - 3m h3 = Q - sW 8 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg.

9 180 4-33 Since the gas and water streams do not mix and heat transfer is negligible, we have 1m = 2m , 3m = 5m , and sW = 1m (h1 - h2) + 3m (h3 - h5) The mass flow rate 1m is given by 1m = ()11 AVv, where the specific volume v1 can be obtained from ideal gas law v1 = 11 RTp = ()11/RMTp v1 = oo1545 ft lbf(860 R) psia 221 ft144 in = lb/ft3 The mass flow rate is then 1m = 533210 ft / lb/ft = lb/min The air properties are listed in Table and steam properties are listed in Table Table Air properties from CATT2 Specific Enthalpy Temp Pressure (Mass) F psia Btu/lbm 1 400 2 260 Table Steam properties from CATT2 Temp Pressure Enthalpy Quality Phase F psia Btu/lbm 3 102 40 Compressed Liquid 5 1 1033 Liquid Vapor Mixture The power developed by the turbine is 4-34 sW = 1m (h1 - h2) + 3m (h3 - h5) sW = ( lb/min)( - ) Btu/lb + (275 lb/min)( - 1033) Btu/lb sW = 49,980 Btu/min (a) Determine the turbine inlet temperature, in oF.

10 We need to know 2 properties at (4). Since pressure drop is negligible, p4 = 40 psia, the enthalpy at (4) can be determine from the energy balance around the steam generator: 0 = 1m (h1 - h2) + 3m (h3 - h4) h4 = h3 + 13mm (h1 - h2) h4 = Btu/lb + ( - ) Btu/lb = Btu/lb The inlet temperature at the turbine inlet is from Table Table Steam properties from CATT2 Specific Temp Pressure Enthalpy Quality Phase F psia Btu/lbm 4 40 1215 Superheated Vapor 4-35 Example ---------------------------------------- ---------------------------------------- -- A tank having a volume of m3 initially contains water as a two-phase liquid-vapor mixture at 260oC and a quality of Saturated water vapor at 260oC is slowly withdrawn through a pressure-regulating valve at the top of the tank as energy is transferred by heat