Water flows through a heat exchanger with a volumetric flow rate of 105 m^3/min.
ID: 500073 • Letter: W
Question
Water flows through a heat exchanger with a volumetric flow rate of 105 m^3/min. Water is fed with a velocity of 22.0 m/s and a pressure of 3 atm. The water leaves the heat exchanger with a velocity of 35.5 m/s and with a pressure of 2 atm. a. What is the density of water throughout the process? Why is the density that value? b. What is the mass flow rate in kg/s? c. Consider only the mass that flows for one second, (i.e. m * 1 second = m) What is the change in the kinetic energy of that particular mass? Consider BEVERAGEX flowing around a beverage plant through a cylindrical pipe with diameter of 3.5 inches. BEVERAGEX is traveling at a velocity of 4.28 m/s. The specific gravity (SG) of BEVERAGEX with respect to water is 1.06. a. Find the volumetric flowrate (V) of BEVERAGEX in m^3/s. b. Find the mass flowrate (m) of BEVERAGEX in kg/s.Explanation / Answer
Water is a liquid and liquid is considered to be incompressible. Incompressible fluids are not sensitive to change in pressrue or temperature. Water density is 1000 kg/m3
mass flow rate= Volumetric flow rate* density = 105m3/min*1000 kg/m3=1.05*105 kg/min
Change in kinetic energy= m*(V22-V12)/2=1.05*105 kg/min(35.52-222)/2=4.1*107 kg/min* m2/s2
2. 12 inches =0.3048m, 3.5 inches = (3.5/12)*0.3048 m=0.089 m,
cross sectional area of pipe = (pi/4)*d2 = (22/28)*(0.089)2 =0.006 m2
volumetric flow rate = veloctity*cross sectional area = 4.28*0.006 m3/s=0.027 m3/s
mass flow rate = volumetric flow rate*density = volumetric flow rate*specific gravity*density of water =0.027*1.06*1000=28.62 kg/s
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