Question

A 160' horizontal run of 2" schedule 40 steel pipe carries 1000 GPM of water at 150°F through a heat exchanger. In addition to the pipe loss, minor losses include two gate valves, six standard 90 elbows, four 45° elbows, and the heat exchanger (K=12) a) What is the velocity of flow? b) What is the Reynolds number? c) What is the Darcy friction factor? d) What is the head loss due to pipe friction? e) What is the total of minor losses?

Answer 1

1. Answer: a) 116.34 ft/s b) 1.77E(6) c) 0.0115 d) 2246 ft e) 3134 ft Explanation: DATA: from table of properties of water v(60ºF)= 1.21E(-5) ftˆ2/s v(70ºF)= 1.052E(-5) ftˆ2/s from tables of comercial piping Inner diameter of a 2" sch 40 pipe is, 2.067 inch rugosity of a steel pipe is, 0,000005 ft loss coefficient: gate valve = 0.15 90º elbow = 0.3 45º elbow = 0.2 heat exchanger = 12 a)
   `u=(q)/(A)` the area of a circular section is:
   `A=(\pi D^(2))/(4) =(\pi.2.067^(2) )/(4) =3.355inch^(2) *(1 ft^(2) )/(144inch^(2) ) =0.02329 ft^(2)` it is necesary convert the flow from gpm to ftˆ3/s
   `1000gpm*(0,16054ft^(3) /s)/(1gpm) *(1min)/(60s) =2.676(ft^(3) )/(s)` the velocity of the fluid is:
   `u=(q)/(A) =(2.676ft^(3) /s)/(0,02329ft^(2) ) =116.34ft/s` b) to calculate the reynolds number we need to know the viscosity at 65 ºF, so it is necesary to interpolate between 60 and 70 ºF, so :
   `m=(v_(70)-v_(60)  )/(T_(2)-T_(1)  ) =(1,052*10^(-5)-1.21*10^(-5)  )/(70-60) =-1.58*10^(-7) \\v_(65) =m*(65-T_(1))+v_(60)\\v_(65)=-1.58*10^(-7)  *(65-60)+1.21*10^(-5)=1.131*10^(-5)  ft^(2) /s` this number of reynolds indicates that that is a turbulent flow. c) from the last question the fluid is turbulent, so to calculate the friction factor it is necesary to use the colebrook equation. when we use this equation is necesary to iterate and suppose an initial value, for this excercise the initial value is 0.011 and our final value is 0.0115. this is the final value because when you solve the equation you get 0.
   `Re=(u*D)/(v) =(116.34*0.1723)/(1.131*10^(-5) ) = 1.77*10^(6)`
   `R=(e)/(D) =(0.000005)/(0.1723) =2.902*10^(-5) \\(1)/(√(f) ) +2 log((R)/(3.7)+(2.51)/(Re*√(f) )  )=0\\(1)/(√(0.0115) ) +2*log((2.902*10^(-5) )/(3.7)  +(2.51)/(1.77*10^(6) *√(0.0115) ) )=-0.028` almost 0. d)
   `h_(L) =f*(l)/(D) (u^(2) )/(2*g)  =0.0115*(160ft)/(0.1723ft) *((116.34ft/s)^(2) )/(2*32.174ft/s^(2) ) =2246 ft` e) minor losses ∑
   `h_(k)=EK_(L) *(u^(2) )/(2*g) =(2*0.15+6*0.3+4*0.2+12)*((116.34ft/s)^(2))/(2*32.174) =3134 ft`