Question

Water at 158C (r 5 999.1 kg/m3 and m 5 1.138 3 1023 kg/m·s) is flowing steadily in a 30-m-long and 5-cm-diameter horizontal pipe made of stainless steel at a rate of 9 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

(a) the pressure drop is $\Delta P_L = 100.185\ kPa$

(b) the head loss is the $h_L =10.22\ m$

(c) the pumping power requirement to overcome this pressure drop $W_p = 901.665\ W$

Explanation:

So the question we are told that the water is at a temperature of 15°C

 And also we are told that the density of the water is $\rho=999.1\ kg/m^3$

 We are also told that the dynamic viscosity of the water is                                        $\mu = 1.138 *10^{-3} kg/m \cdot s$

   From the diagram the length of the pipe is $L=$ 30 m    

  The diameter is given as $D=$ 5 cm $\frac{5}{100} m = 0.05m$

   The volumetric flow rate is given as  $Q=$ 9 L/s $= \frac{9}{1000} m^3/ s = 0.009\ m^3/s$

Now the objective of this solution is to obtain

      i the pressure drop,  ii the head loss iii  the pumping power requirement to overcome this pressure drop

to obtain this we need to get the cross-sectional area of the pipe which will help when looking at the flow analysis

       So mathematically the cross-sectional area is

                               $A_s =\frac{\pi}{4} D^2$

                                    $= \frac{\pi}{4} (5*10^{-2})$

                                   $=1.963*10^{-3} m^2$

Next thing to do is to obtain average flow velocity which is mathematically represented as

                 $v = \frac{Q}{A_s}$

                 $v =\frac{9*10^{-3}}{1.963*10^{-3}}$

                  $=4.585 \ m/s$

 Now to determine the type of flow we have i.e to know whether it a laminar flow , a turbulent flow or an intermediary flow

  We use the Reynolds number

if it is below $4000$ then it is a laminar flow but if it is higher then it is a turbulent flow ,now when it is exactly the value then it is an intermediary flow

     This Reynolds number is mathematically represented as

                           $Re = \frac{\rho vD}{\mu}$

                                $=\frac{(999.1)(4.585)(0.05)}{1.138*10^{-3}}$

                               $=2.0127*10^5$

Now since this number is greater than 4000 the flow is turbulent

So we are going to be analyse the flow using the Colebrook's  equation which is mathematically represented as

                $\frac{1}{\sqrt{f} } =-2.0\ log [\frac{\epsilon/D_h}{3.7} +\frac{2.51}{Re\sqrt{f} } ]$

 Where f is the friction  factor , $\epsilon$ is the surface roughness ,

Now generally the surface roughness for stainless steel is

                                             $\epsilon = 0.002 mm = 2*10^{-6} m$

Now substituting the values into the equation we have

                                   $\frac{1}{\sqrt{f} } =-2.0\ log [\frac{2*10^{-6}/5*10^{-2}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ]$

So solving to obtain f we have

                                  $\frac{1}{\sqrt{f} } =-2.0\ log [\frac{4*10^{-5}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ]$

                                 $= -2.0log[1.0811 *10^{-5} +\frac{1.2421*10^{-5}}{\sqrt{f} } ]$

                              $f = 0.0159$

Generally the pressure drop is mathematically represented as

                                            $\Delta P_L =f\ \frac{L}{D} \ \frac{\rho v^2}{2}$

Now substituting values into the equation

                                           $= 0.0159 [\frac{30}{5*10^{-5}} ][\frac{(999.1)(4.585)}{2} ]$

                                           $=100.185 *10^3 Pa$

                                           $=100.185 kPa$  

Generally the head loss in the pipe is mathematically represented as

                        $h_L =\frac{\Delta P_L}{\rho g}$

                            $= \frac{100.185 *10^3}{(999.1)(9.81)}$

                             $=10.22m$

Generally the power input required to overcome this pressure drop is mathematically represented as

                     $W_p = Q \Delta P_L$

                         $=(9*10^{-3}(100.185*10^3))$

                         $= 901.665\ W$