Practice Question: Series Pipe Network Analysis

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Solution

 

\[Q=2\ ft^3/s\]

Calculate the cross-sectional areas of the two pipes using their diameters:

For the first pipe:

\[A_A= {πD_A² \over4}\]

 

\[A_A= {{(π)}({4in×}{1ft\over12 in})²\over4} = 0.09 ft²\]

For the second pipe:

\[A_B= {πD_B² \over4}\]

\[A_B= {{(π)}({6in×}{1ft\over12 in})²\over4} = 0.2 ft²\]

 

Next, we need to calculate the Reynolds number to determine the friction factor (\(f\)). The Reynolds number is given by:

 

\[Re= {V\over Dv}\]Where:

is the velocity of the fluid (ft/s)

 is the kinematic viscosity of the fluid (ft²/s), and

 is the diameter of the pipe (ft).

The kinematic viscosity of water at room temperature is approximately:

 

\[v=1.1\times {10}^{-5}\frac{ft^2}{s}\]

  

First, let's calculate the velocity of the fluid in each pipe:

\[V_A={Q\over A_A} = {2\over 0.09}=22.22\ ft/s\]

 

\[V_B={Q\over A_B} = {2\over 0.2}=10\ ft/s\]

Now, let's calculate the Reynolds number (Re) for each pipe:

 

\[Re_A={{V_AD_A}\over ν} = {22.2{{(4in×}{1ft\over12 in})}\over 1.1×10-5} = 3673333.33\]

                                                                       

\[Re_B={{V_BD_B}\over ν} = {10{{(6in×}{1ft\over12 in})²}\over 1.1×10-5} = 454545.45\]

 

For simplicity, let's take a common value of \(f=0.02\) for both pipes.

 

Now, we can calculate the pressure drop (\(h_f\)) in each pipe using the Darcy-Weisbach equation:

 

For the first pipe:

\[h_{f_A}= {{fL_AV_A^2} \over 2gD_A}\]

 

\[h_{f_A}={{0.02\times500\times22.22^2} \over{{2\times}}32.2 \times0.33} = 230\ ft\]

For the second pipe:

\[h_{f_B}= {{fL_BV_B^2} \over 2gD_B}\]

 

\[h_{f_B}={{0.02\times600\times10^2} \over{{2\times}}32.2 \times0.5} = 37.27\ ft\]

 

Finally, we can add the pressure drops in both pipes to get the total pressure drop at the end of the second pipe:

\[h_{f_t}= h_{f_A}+ h_{f_B}\]

\[h_{f_t}= 230\ ft+ 37.27\ ft\]

\[h_{f_t}= 267.27\ ft\]

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