Question

consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pressure drop will \((a)\) double, \((b)\) more than double, (c) less than double, \((d)\) reduce by half, or \((e)\) remain constant.

Short answer

a) double b) halve c) remain the same d) quadruple

Step-by-step solution

Hagen-Poiseuille Equation

The Hagen-Poiseuille equation describes the pressure drop (∆P) in a pipe due to viscous flow, given as: \( ∆P = \dfrac{8 \times µ \times L \times Q}{π \times R^4} \) where ∆P is the pressure drop, µ is the dynamic viscosity of the fluid, L is the length of the pipe, Q is the volumetric flow rate, and R is the radius of the pipe. All other variables remain constant in this exercise, except for the length of the pipe.

Initial Pressure Drop

Let's first find the pressure drop in the pipe of the initial length L. Using the Hagen-Poiseuille equation: \( ∆P_1 = \dfrac{8 \times µ \times L \times Q}{π \times R^4} \)

Doubled Pipe Length

Now, let's find the pressure drop in the pipe of length 2L. Using the same Hagen-Poiseuille equation: \( ∆P_2 = \dfrac{8 \times µ \times (2L) \times Q}{π \times R^4} \)

Compare Initial and Final Pressure Drops

To find how the pressure drop changes, we can divide the pressure drop of the doubled length pipe by that of the initial length pipe: \( \dfrac{∆P_2}{∆P_1} = \dfrac{\dfrac{8 \times µ \times (2L) \times Q}{π \times R^4}}{\dfrac{8 \times µ \times L \times Q}{π \times R^4}} \) Simplify the equation: \( \dfrac{∆P_2}{∆P_1} = \dfrac{2L}{L} = 2 \)

Conclusion

According to our analysis, the pressure drop in the doubled length pipe (∆P₂) will be double of the pressure drop in the initial length pipe (∆P₁). Therefore, the correct option is \((a)\) double.