Thermodynamics

Chapter 5 | 225Properties We take the density of water to be 1000 kg/m 3 1 kg/L.Analysis (a) Noting that 10 gal of water are discharged in 50 s, the volumeand mass flow rates of water areV # V 10 gal¢t 50 s a 3.7854 L b 0.757 L/s1 galm # rV # 11 kg>L210.757 L>s2 0.757 kg/s(b) The cross-sectional area of the nozzle exit isA e pr e 2 p 10.4 cm2 2 0.5027 cm 2 0.5027 10 4 m 2The volume flow rate through the hose and the nozzle is constant. Then theaverage velocity of water at the nozzle exit becomesV e V# 0.757 L/s 1 m3A e 0.5027 10 4 m2a 1000 L b 15.1 m /sDiscussion It can be shown that the average velocity in the hose is 2.4 m/s.Therefore, the nozzle increases the water velocity by over six times.EXAMPLE 5–2Discharge of Water from a TankA 4-ft-high, 3-ft-diameter cylindrical water tank whose top is open to theatmosphere is initially filled with water. Now the discharge plug near the bottomof the tank is pulled out, and a water jet whose diameter is 0.5 instreams out (Fig. 5–10). The average velocity of the jet is given byV 12gh, where h is the height of water in the tank measured from thecenter of the hole (a variable) and g is the gravitational acceleration. Determinehow long it will take for the water level in the tank to drop to 2 ft fromthe bottom.h 0AirWaterSolution The plug near the bottom of a water tank is pulled out. The timeit takes for half of the water in the tank to empty is to be determined.Assumptions 1 Water is an incompressible substance. 2 The distancebetween the bottom of the tank and the center of the hole is negligible comparedto the total water height. 3 The gravitational acceleration is 32.2 ft/s 2 .Analysis We take the volume occupied by water as the control volume. Thesize of the control volume decreases in this case as the water level drops,and thus this is a variable control volume. (We could also treat this as afixed control volume that consists of the interior volume of the tank by disregardingthe air that replaces the space vacated by the water.) This is obviouslyan unsteady-flow problem since the properties (such as the amount ofmass) within the control volume change with time.The conservation of mass relation for a control volume undergoing anyprocess is given in the rate form ash h 2D jet0 D tankFIGURE 5–10Schematic for Example 5–2.m # in m # out dm CVdtDuring this process no mass enters the control volume (ṁ in 0), and themass flow rate of discharged water can be expressed as(1)m # out (rVA) out r22ghA jet(2)