ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

Steady-Flow Devices ● Heat Exchangers Heat exchangers are devices that transfer energy between fluids at different temperatures. For heat exchangers, Classification ► Open Feedwater Heaters (Mixing Chambers) ► Closed Feedwater Heaters

Example 1 Feedwater Heater: Inlet 1 T1 = 200 ºC, p1 = 700 kPa, Exit sat. liquid, p3 = 700 kPa. Find Inlet 2 Inlet 1 Exit

Example 1 (continued)

Example 1 (continued) h2 ~ [hf + vf (p – psat)]2 Inlet 1: Superheated vapor Table A-6, h1 = 2844.8 kJ/kg Inlet 2: compressed liquid h2 ~ [hf + vf (p – psat)]2 Table A-4, 40 ºC psat = 7.384 kPa vf = 0.001008 m /kg hf = 167.57 kJ/kg = 167.57 + 0.001008 (700 – 7.384) = 168.27 kJ/kg Exit: saturated liquid Table A-5, h3 = 697.22 kJ/kg

Example 1 (continued)

Steady-Flow Devices ● Throttling Devices Throttling devices are flow-restricting devices (valves, porous plugs) used to reduce the pressure of a gas or a liquid. Unlike turbines, they produce a pressure drop without producing any work. The pressure drop in the fluid is often accompanied by a change in temperature. The temperature change depends on the Joule-Thomson coefficient.

Steady-Flow Devices hi = he Ti = Te ● Throttling Devices Joule-Thomson Coefficient µ < 0, T increases µ = 0, T no change µ > 0, T decreases For an ideal throttling process, hi = he For ideal gases, Ti = Te

Example 2 A throttling calorimeter is a device for determining the quality of a two-phase liquid-vapor mixture. Inlet 1 sat. mixture, p1 = 2 MPa, Exit T2 = 150 ºC, p2 = 100 kPa, Find the quality of water at the inlet. Sat. mixture p1 = 2 MPa T2 = 150 ºC p2 = 100 kPa

Example 2 (continued) hi = he Assuming an ideal throttling process, Inlet saturated mixture at 2 MPa Table A-5 hf = 908.79 kJ/kg hg = 2799.5 kJ/kg Exit superheated vapor Table A-6 he = 2776.4 kJ/kg hi = hf + xi (hg – hf) = 0.988

Uniform-Flow Process Examples for uniform-flow (unsteady-flow) processes: ► Charge or discharging a vessel, ► Inflate tires or balloons, ► Cooking with a regular or pressure cooker, ● Any instant during the process, the state of the control volume is uniform. ● The fluid properties may differ from one inlet (or exit) to another, but the fluid flow at an inlet (exit) is uniform and steady.

Uniform-Flow Process Conservation of Mass Conservation of Energy + (m2u2 – m1u1)CV

Example 3 A rigid, insulated tank which is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300 ºC. Now the valve is opened and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa at which point the valve is closed. Find the temperature of the steam inside the tank.

Example 3 (continued) p1 = 1 MPa T1 = 300 ºC

Example 3 (continued) m1 = 0, initially evacuated me = 0, no outlet Charging process Uniform-Flow Process m1 = 0, initially evacuated me = 0, no outlet m2 = mi QCV – WCV = Σ mehe - Σ mihi + (m2u2 – m1u1 )CV insulated rigid no outlet initially evacuated mihi = m2u2 hi = u2

Example 3 (continued) There is an increase in temperature of 156.2 ºC Inlet: superheated vapor at 1MPa and 300 ºC Table A-6, hi = 3051.2 kJ/kg Final state: p2 = 1MPa and u2 = 3051.2 kJ/kg Table A-6, T2 = 456.2 ºC There is an increase in temperature of 156.2 ºC due to the conversion of flow work to sensible internal energy.

Example 4 A 0.2-m3 rigid tank equipped with a pressure regulator contained steam at 2 MPa and 300 ºC. When heated, the regulator keeps the steam pressure constant by letting out some steam. But, the temperature inside rises. Find the amount of heat transfer when the temperature inside the tank reaches 500 ºC.

Example 4 (continued) Q Discharging Process Uniform-Flow Process p1 = 2 MPa T1 = 300 ºC Q

Example 4 (continued) me = (m1 – m2)CV No inlet, one outlet Initial state: p1 = 2MPa and T1 = 300 ºC Table A-6, v1 = 0.1255 m3/kg u1 = 2772.6 kJ/kg h1 = 3023.5 kJ/kg Final state: p2 = 2MPa and T2 = 500 ºC Table A-6, v2 = 0.1757 m3/kg u2 = 3116.2 kJ/kg h2 = 3467.6 kJ/kg

Example 4 (continued) me = (m1 – m2)CV = 1.594 – 1.138 = 0.456 kg QCV – WCV = Σ mehe - Σ mihi + (m2u2 – m1u1 )CV rigid no inlet QCV = mehe + (m2u2 – m1u1 )CV

Example 4 (continued) QCV = mehe + (m2u2 – m1u1 )CV = (0.456)(3245.6) + (1.138)(3116.2) - (1.594)(2772.6) = 606.7 kJ

Example 5 A 0.3-m3 rigid tank is filled with saturated liquid water 200 ºC. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank. Heat is transferred to the water such that the temperature in the tank remains constant. Determine the amount of heat transfer by the time one-half of the water (in mass) has been withdrawn.

Example 5 (continued) Q Discharging Process Uniform-Flow Process Sat. liquid T1 = 200 ºC Q

Example 5 (continued) me = (m1 – m2)CV No inlet, one outlet Initial state: Sat. liquid at T1 = 200 ºC Table A-4, v1, u1, h1 Final state: Sat. mixture at T2 = 200 ºC m2 = ½ m1, v2 = 2v1, find x2 Table A-4, v2, u2, h2 Exit: Sat. liquid at T2 = 200 ºC Table A-4, he