Example 5

Types of Fluid Flow Problems     

Example 5- Determining the Head Loss in a Water pipe

Water at 15°C ( ρ = 999 kg/m^3 and μ = 1.138 × 〖10〗^(-3 )kg/m.s ) is flowing steadily in a 5-cm-diameter horizontal pipe made of stainless steel at a rate of 6 L/s ( Fig. 8-32 ) . Determine the pressure drop, the head loss, and the required pumping power input for flow over a 60-m-long section of the pipe.

SOLUTION : 

The flow rate through a specified water pipe is given. The pressure drop, the head loss, and the pumping power requirements are to be determined.

Assumptions : 

1. The flow is steady and incompressible. 

2. The entrance effects are negligible, and thus the flow is fully developed.

3. The pipe involves no components such as bends, valves, and connectors.

4. The piping section involves no work devices such as a pump or a turbine

Properties :

The density and dynamic  viscosity of water are given to be ρ = 999 kg/m^3 and μ = 1.138 × 〖10〗^(-3 )kg/m.s, respectively.

Analysis :

We recognize this as a problem of the first type, since flow rate,pipe length, and pipe diameter are known. First we calculate the average velocity and the Reynolds number to determine the flow regime :

V = ὐ/A_C  = ὐ/(〖πD〗^2/4) = 〖0.006 m〗^3/(〖π(0.05m)〗^2/4) = 3.06 m/s

Re = ρVD/μ = (〖999 kg/m〗^3 )(3.06 m/s)( 0.05m)/(1.138 ×〖10〗^(-3)  kg/(m.s))=134,300

Re >4000 , the flow is turbulent, The relative roughness of the pipe is estimated using table 8-2

ε/D = (0.002 mm)/(50 mm) = 0.000040

determine  f  from the Colebrook equation on which the Moody chart is based :

1/√f = -2.0 log ((ε/D)/3.7+2.51/(Re√f)) →  1/√f = -2.0 log (0.000040/3.7 + 2.51/(134,300 √f))

f = 0.0172 , the pressure drop ( equivalent to pressure loss in this case), head loss, and the required power input become 

∆P= ∆P_(L )=f L/D  〖ρV〗^2/2=0.0172  60m/0.05m (( 999 kg/m^3 )(3.06 m/s^2 ))/2((1 N)/(1 kg.m/s^2 ))

                                    = 96,540 N/m^2 = 96.5 kPa

h_L= (∆P_L)/ρg = f L/D  V^2/2g = 0.0172 60m/0.05m  (( 3.06 m/s^(2 ) ))/(2( 9.81 m/s^2 )) = 9.85 m

w_(pump= ὐ∆P)   

           = ( 0.006 m^3/s)(96,540  N/m2)((1 W)/(1 N.m/s)) = 579 W