Head loss ratio, parallel versus series

Method. Use $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$, i.e. $h_f = rQ^2$ with $r \propto D^{-5}$ for the same $L$ and $f$. Let $D_2 = D_1/2$.

Parallel arrangement (head loss $H_1$)

Both pipes have the same head loss $H_1$, so $Q_i = \sqrt{H_1/r_i}$, which gives $Q_i \propto D_i^{5/2}$:

$$Q = Q_1' + Q_2' = \sqrt{\frac{\pi^2gH_1}{8fL}}\left(D_1^{5/2} + D_2^{5/2}\right) \;\Rightarrow\; H_1 = \frac{8fLQ^2}{\pi^2g\left(D_1^{5/2} + D_2^{5/2}\right)^2}$$

With $D_2 = D_1/2$: $D_2^{5/2} = (0.5)^{2.5}D_1^{5/2} = 0.17678\,D_1^{5/2}$, so $\left(D_1^{5/2} + D_2^{5/2}\right)^2 = (1 + 0.17678)^2D_1^5 = 1.3848D_1^5$:

$$H_1 = \frac{8fLQ^2}{\pi^2g}\cdot\frac{1}{1.3848D_1^5}$$

Series arrangement (head loss $H_2$)

The same $Q$ passes through both pipes and the losses add:

$$H_2 = \frac{8fLQ^2}{\pi^2g}\left(\frac{1}{D_1^5} + \frac{1}{D_2^5}\right) = \frac{8fLQ^2}{\pi^2g}\cdot\frac{1 + 32}{D_1^5} = \frac{8fLQ^2}{\pi^2g}\cdot\frac{33}{D_1^5}$$

Ratio

$$\frac{H_1}{H_2} = \frac{1/1.3848}{33} = 0.02188 \approx \frac{1}{45.7}$$

Answer: $H_1 : H_2 = 1 : 45.7$, i.e. $H_1/H_2 \approx 0.0219$ (the parallel arrangement loses only about 2% of the head lost in series).