Thermodynamics

EXAMPLE 5–9Mixing of Hot and Cold Waters in a ShowerConsider an ordinary shower where hot water at 140°F is mixed with coldwater at 50°F. If it is desired that a steady stream of warm water at 110°Fbe supplied, determine the ratio of the mass flow rates of the hot to coldwater. Assume the heat losses from the mixing chamber to be negligible andthe mixing to take place at a pressure of 20 psia.T 1 = 140°Fm·1Chapter 5 | 241MixingchamberP = 20 psiaSolution In a shower, cold water is mixed with hot water at a specifiedtemperature. For a specified mixture temperature, the ratio of the mass flowrates of the hot to cold water is to be determined.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0 and E CV 0. 2 The kinetic andpotential energies are negligible, ke pe 0. 3 Heat losses from the systemare negligible and thus Q . 0. 4 There is no work interaction involved.Analysis We take the mixing chamber as the system (Fig. 5–33). This is acontrol volume since mass crosses the system boundary during the process.We observe that there are two inlets and one exit.Under the stated assumptions and observations, the mass and energy balancesfor this steady-flow system can be expressed in the rate form as follows:Mass balance:m # in m # out dm system >dt 0¡ 0 (steady)T 2 = 50°F T 3 = 110°Fm·m·23FIGURE 5–33Schematic for Example 5–9.Energy balance:m # in m # out S m # 1 m # 2 m # 3E # in E # out dE system >dt 0¡ 0 (steady)⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭Rate of net energy transferby heat, work, and massE # in E # outRate of change in internal, kinetic,potential, etc., energiesCombining the mass and energy balances,Dividing this equation by ṁ 2 yieldswhere y ṁ 1 /ṁ 2 is the desired mass flow rate ratio.The saturation temperature of water at 20 psia is 227.92°F. Since the temperaturesof all three streams are below this value (T T sat ), the water in allthree streams exists as a compressed liquid (Fig. 5–34). A compressed liquidcan be approximated as a saturated liquid at the given temperature. Thus,Solving for y and substituting yieldsm # 1h 1 m # 2h 2 m # 3h 3 1since Q # 0, W # 0, ke pe 02m # 1h 1 m # 2h 2 (m # 1 m # 2)h 3yh 1 h 2 (y 1)h 3h 1 h f @ 140°F 107.99 Btu/lbmh 2 h f @ 50°F 18.07 Btu/lbmh 3 h f @ 110°F 78.02 Btu/lbmy h 3 h 2 78.02 18.07h 1 h 3 107.99 78.02 2.0Discussion Note that the mass flow rate of the hot water must be twice themass flow rate of the cold water for the mixture to leave at 110°F.TT satCompressedliquid statesP = const.FIGURE 5–34A substance exists as a compressedliquid at temperatures below thesaturation temperatures at the givenpressure.v