The problem asks us to design a pipe network using the equivalent pipe method. The network is a square with inflow at node A and outflow at node C. The flow rate $Q$ is given as 18 l/s. The lengths and diameters of the pipes are as follows: - AB: 2000m, 250mm - BC: 1000m, 150mm - CD: 2000m, 200mm - DA: 1000m, 150mm

1. Problem Description

The problem asks us to design a pipe network using the equivalent pipe method. The network is a square with inflow at node A and outflow at node C. The flow rate $Q$ is given as 18 l/s. The lengths and diameters of the pipes are as follows:

- AB: 2000m, 250mm

- BC: 1000m, 150mm

- CD: 2000m, 200mm

- DA: 1000m, 150mm

2. Solution Steps

Since the problem asks for the design using the equivalent pipe method, it implies that we need to find an equivalent pipe that can represent the whole network.

First, we make the assumption that the flow is divided into two parallel paths: ABC and ADC.

Let us assume that the flow rate in the path ABC is $Q_1$ and in path ADC is $Q_2$. Then $Q = Q_1 + Q_2 = 18 l/s = 0.018 m^3/s$.

We will use the Hazen-Williams equation to calculate the head loss.

$h_f = \frac{10.67 * L * Q^{1.85}}{C^{1.85} * D^{4.87}}$

where

$h_f$ is the head loss

$L$ is the length of the pipe

$Q$ is the flow rate

$C$ is the Hazen-Williams coefficient (assume C=100 for all pipes)

$D$ is the diameter of the pipe

Path ABC:

$h_{f,AB} = \frac{10.67 * 2000 * Q_1^{1.85}}{100^{1.85} * (0.25)^{4.87}}$

$h_{f,BC} = \frac{10.67 * 1000 * Q_1^{1.85}}{100^{1.85} * (0.15)^{4.87}}$

$h_{f,ABC} = h_{f,AB} + h_{f,BC} = \frac{10.67 * Q_1^{1.85}}{100^{1.85}} * (\frac{2000}{0.25^{4.87}} + \frac{1000}{0.15^{4.87}}) = \frac{10.67 * Q_1^{1.85}}{100^{1.85}} * (2000 * 47.58 + 1000 * 413.61) = \frac{10.67 * Q_1^{1.85}}{100^{1.85}} * (95160 + 413610) = \frac{10.67 * Q_1^{1.85}}{100^{1.85}} * 508770 = 0.001067 * Q_1^{1.85} * 508770 = 542822.69 * Q_1^{1.85}$

Path ADC:

$h_{f,AD} = \frac{10.67 * 1000 * Q_2^{1.85}}{100^{1.85} * (0.15)^{4.87}}$

$h_{f,DC} = \frac{10.67 * 2000 * Q_2^{1.85}}{100^{1.85} * (0.2)^{4.87}}$

$h_{f,ADC} = h_{f,AD} + h_{f,DC} = \frac{10.67 * Q_2^{1.85}}{100^{1.85}} * (\frac{1000}{0.15^{4.87}} + \frac{2000}{0.2^{4.87}}) = \frac{10.67 * Q_2^{1.85}}{100^{1.85}} * (1000 * 413.61 + 2000 * 15.26) = \frac{10.67 * Q_2^{1.85}}{100^{1.85}} * (413610 + 30520) = \frac{10.67 * Q_2^{1.85}}{100^{1.85}} * 444130 = 0.001067 * Q_2^{1.85} * 444130 = 473886.71 * Q_2^{1.85}$

Since the head loss is the same along both paths, we have:

$h_{f,ABC} = h_{f,ADC}$

$542822.69 * Q_1^{1.85} = 473886.71 * Q_2^{1.85}$

$(\frac{Q_1}{Q_2})^{1.85} = \frac{473886.71}{542822.69} = 0.873$

$\frac{Q_1}{Q_2} = (0.873)^{\frac{1}{1.85}} = 0.929$

$Q_1 = 0.929 * Q_2$

Since $Q_1 + Q_2 = 0.018$, we have $0.929 * Q_2 + Q_2 = 0.018$, so $1.929 * Q_2 = 0.018$, and $Q_2 = \frac{0.018}{1.929} = 0.00933 m^3/s$.

Then $Q_1 = 0.018 - 0.00933 = 0.00867 m^3/s$.

$h_f = 473886.71 * (0.00933)^{1.85} = 473886.71 * 0.000232 = 109.94 m$

To find the equivalent pipe, we consider a pipe with length equal to the total length of the square which is $2000 + 1000 + 2000 + 1000 = 6000 m$.

We want to find the diameter D such that the head loss is the same as above.

$h_f = \frac{10.67 * L * Q^{1.85}}{C^{1.85} * D^{4.87}}$

$109.94 = \frac{10.67 * 6000 * (0.018)^{1.85}}{100^{1.85} * D^{4.87}}$

$109.94 = \frac{10.67 * 6000 * 0.000363}{851.14 * D^{4.87}}$

$109.94 * 851.14 * D^{4.87} = 23.25$

$D^{4.87} = \frac{23.25}{109.94 * 851.14} = \frac{23.25}{93574.33} = 0.0002485$

$D = (0.0002485)^{\frac{1}{4.87}} = 0.115 m = 115 mm$

3. Final Answer

The equivalent pipe has a length of 6000m and a diameter of 115mm.