# LTR: Drag in Pipes

You may have noticed that the normalizing "drag pressure" used thus far has been $\frac12 \rho v_\infty^2$, which is often called the dynamic pressure. Another common way to denote "pressure" losses due to drag is to quote them as head losses.

##### DEFINITION:

The head loss is the energy lost per unit weight of the fluid.

Head losses for straight lengths of pipe are directly related to the friction factor and constitute "major losses" in pipe flow:

$\displaystyle{h_L = 2 f_f \frac{L}{D}\frac{v_\infty^2}{g}}$

Head losses for bends and other pipe components are related to an empirical (experimentally measured) factor called the loss coefficient/factor, Ki (see page 175 for values in pipe bends) and constitute "minor losses" in pipe flow:

$\displaystyle{h_L = 2f_f\frac{L_{eq}}{D}\frac{v_\infty^2}{g} = K_i\frac{v_\infty^2}{2g}}$

Another examples of loss coefficient/factor is for pipe constrictions, where:

$\displaystyle{K_i = 0.45 \left (1- \frac{A_2}{A_1} \right )}$

or pipe enlargements, where:

$\displaystyle{K_i = \left (\frac{A_2}{A_1} - 1\right )^2}$

##### NOTE:

The terms "major" and "minor" losses do not necessarily denote which is larger, simply which is most common!

In order to use head losses to calculate pressure drops, we use the following equation (which we will derive later in the course):

$\displaystyle{\rho g h_L = \frac{\rho \left(v_1^2 - v_2^2 \right )}{2} + (P_1-P_2)}$

##### OUTCOME:

Estimate friction losses in pipes and pipe networks

##### TEST YOURSELF

You are analyzing the flow between the hot water heater and the shower in a ranch-style home (so that elevation changes may be neglected). Calculate the mainline water pressure when the flowrate is 1L/s and the water is flowing through 10 meters of 2cm ID (smooth) pipe, going through 3 90o elbows and you may consider the shower-head equivalent to a wide-open angle valve. (Note: the fluid velocity does not change along the region of interest.)

In order to calculate the mainline pressure, we need to figure out the necessary pressure-drop in the system described. In order to obtain that, we need to calculate the head losses for each of the parts of the pipe network.

Our calculations, therefore, need to include contributions from:

• Straight pipe ("major") losses
• "Minor" losses from the elbows
• "Minor" losses from the shower-head

Head losses for the straight length of pipe is obtained from the friction factor, so we must first calculate fluid velocity so that we can get the Re:

$\displaystyle{Q = V_{avg}A_{cs} = 1L/s = \pi (0.01m)^2 V_{avg}}$

This gives $V_{avg}$ = 3.18m/s, so that:

$\displaystyle{Re = \frac{(0.02m)(3.18m/s)}{5x10^{-7}m^2/s} = 1.27x10^6}$

Since the flow is strongly turbulent, we will assume that $V_{avg} = V_\infty$. Also, we get $f_f$ from the charts to be $f=0.0027$, so that:

$\displaystyle{h_L = 2 f_f \frac{L}{D}\frac{v_\infty^2}{g} = 2(0.0027)\left (\frac{10m}{0.02m}\right ) \left (\frac{(3.18m/s)^2}{9.81m/s^2} \right ) =2.98m}$

Head losses for both the bends and the shower-head are obtained from the loss factor, using:

$\displaystyle{h_L = K_i\frac{v_\infty^2}{2g}}$

Looking up the values of $K_i$, we get $K_{90^oelbow} = 0.7$, while $K_{nozzle} = 3.8$, so that the head losses from these four things (3 elbows and a nozzle) are:

$\displaystyle{h_L = (3*0.7+3.8)\frac{(3.18m/s)^2}{2(9.81m/s^2)} = 3.04m}$

##### NOTE:

The "minor" losses are bigger than the "major"! Also, we could not group them in this way if $v_\infty$ was different for each.

Then, the total head losses are calculated simply by adding these up, whereby we can get the pressure drop from:

$\displaystyle{\rho g \sum h_L = \frac{\rho \left(v_1^2 - v_2^2 \right )}{2} + (P_1-P_2)}$

which in our problem becomes:

$\displaystyle{\rho g \sum h_L = (P_1-P_2)}$

so that:

$\displaystyle{(1000kg/m^3)(9.81m/s^2)(6m) = (P_1-P_2) = 58.9kPa}$

This means that the pressure in the line is about 160kPa (since the pressure at the nozzle end must be atmospheric).