Pipe Flow worksheet

1. Flow rate: 42.955 gpm

1. Find the open area inside 1 1/2 in Type K copper, since that's the space the water fills (inside diameter 1.481 in = 0.037617 m)

   $A$$=\frac{\pi }{4}\times {0.037617}^2$$=0.0011113924\text{\hspace{0.17em}m²}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $8\times 0.3048$$=2.4384\text{\hspace{0.17em}m/s}$

3. Multiply the speed by the area to get the flow

   $2.4384\times 0.0011113924$$=0.0027100193\text{\hspace{0.17em}m³/s}$

4. Show the flow in gpm

   $\frac{0.0027100193}{0.0000630902}$$=42.954681\text{\hspace{0.17em}gpm}$

2. Minimum inside diameter: 0.948 in

1. Change the flow to cubic meters per second so all units match (1 gpm = 0.0000630902 m³/s)

   $11\times 0.0000630902$$=0.00069399\text{\hspace{0.17em}m³/s}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $5\times 0.3048$$=1.524\text{\hspace{0.17em}m/s}$

3. Find the inside diameter (d) that lets this flow (Q) pass at your target speed (V)

   $d$$=\sqrt{\frac{4\times 0.00069399}{\pi \times 1.524}}$$=0.02407908\text{\hspace{0.17em}m}$

4. Show the diameter in inches (1 in = 0.0254 m)

   $\frac{0.02407908}{0.0254}$$=0.947995\text{\hspace{0.17em}in}$

3. Flow rate: 15.669 gpm

1. Find the open area inside 1 1/4 in Type L copper, since that's the space the water fills (inside diameter 1.265 in = 0.032131 m)

   $A$$=\frac{\pi }{4}\times {0.032131}^2$$=0.000810846\text{\hspace{0.17em}m²}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $4\times 0.3048$$=1.2192\text{\hspace{0.17em}m/s}$

3. Multiply the speed by the area to get the flow

   $1.2192\times 0.000810846$$=0.0009885834\text{\hspace{0.17em}m³/s}$

4. Show the flow in gpm

   $\frac{0.0009885834}{0.0000630902}$$=15.669367\text{\hspace{0.17em}gpm}$

4. Velocity: 31.52 ft/s

1. Find the open area inside 3/8 in PEX (CTS, SDR 9), since that's the space the water fills (inside diameter 0.36 in = 0.009144 m)

   $A$$=\frac{\pi }{4}\times {0.009144}^2$$=0.0000656693\text{\hspace{0.17em}m²}$

2. Change the flow to cubic meters per second so all units match (1 gpm = 0.0000630902 m³/s)

   $10\times 0.0000630902$$=0.0006309\text{\hspace{0.17em}m³/s}$

3. Divide the flow by the area to get the water's speed

   $\frac{0.0006309}{0.0000656693}$$=9.60726\text{\hspace{0.17em}m/s}$

4. Show the speed in feet per second (1 ft = 0.3048 m)

   $\frac{9.60726}{0.3048}$$=31.519883\text{\hspace{0.17em}ft/s}$

5. Velocity: 4.802 ft/s

1. Find the open area inside 1 in PEX (CTS, SDR 9), since that's the space the water fills (inside diameter 0.875 in = 0.022225 m)

   $A$$=\frac{\pi }{4}\times {0.022225}^2$$=0.0003879479\text{\hspace{0.17em}m²}$

2. Change the flow to cubic meters per second so all units match (1 gpm = 0.0000630902 m³/s)

   $9\times 0.0000630902$$=0.00056781\text{\hspace{0.17em}m³/s}$

3. Divide the flow by the area to get the water's speed

   $\frac{0.00056781}{0.0003879479}$$=1.463629\text{\hspace{0.17em}m/s}$

4. Show the speed in feet per second (1 ft = 0.3048 m)

   $\frac{1.463629}{0.3048}$$=4.801932\text{\hspace{0.17em}ft/s}$

6. Flow rate: 9.973 gpm

1. Find the open area inside 3/4 in Schedule 40 steel, since that's the space the water fills (inside diameter 0.824 in = 0.02093 m)

   $A$$=\frac{\pi }{4}\times {0.02093}^2$$=0.0003440422\text{\hspace{0.17em}m²}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $6\times 0.3048$$=1.8288\text{\hspace{0.17em}m/s}$

3. Multiply the speed by the area to get the flow

   $1.8288\times 0.0003440422$$=0.0006291844\text{\hspace{0.17em}m³/s}$

4. Show the flow in gpm

   $\frac{0.0006291844}{0.0000630902}$$=9.972776\text{\hspace{0.17em}gpm}$

7. Flow rate: 360.19 gpm

1. Find the open area inside 6 in Schedule 40 steel, since that's the space the water fills (inside diameter 6.065 in = 0.154051 m)

   $A$$=\frac{\pi }{4}\times {0.154051}^2$$=0.0186388419\text{\hspace{0.17em}m²}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $4\times 0.3048$$=1.2192\text{\hspace{0.17em}m/s}$

3. Multiply the speed by the area to get the flow

   $1.2192\times 0.0186388419$$=0.0227244761\text{\hspace{0.17em}m³/s}$

4. Show the flow in gpm

   $\frac{0.0227244761}{0.0000630902}$$=360.190289\text{\hspace{0.17em}gpm}$

8. Minimum inside diameter: 0.738 in

1. Change the flow to cubic meters per second so all units match (1 gpm = 0.0000630902 m³/s)

   $8\times 0.0000630902$$=0.00050472\text{\hspace{0.17em}m³/s}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $6\times 0.3048$$=1.8288\text{\hspace{0.17em}m/s}$

3. Find the inside diameter (d) that lets this flow (Q) pass at your target speed (V)

   $d$$=\sqrt{\frac{4\times 0.00050472}{\pi \times 1.8288}}$$=0.01874554\text{\hspace{0.17em}m}$

4. Show the diameter in inches (1 in = 0.0254 m)

   $\frac{0.01874554}{0.0254}$$=0.738013\text{\hspace{0.17em}in}$

9. Flow rate: 3.017 gpm

1. Find the open area inside 3/4 in Type L copper, since that's the space the water fills (inside diameter 0.785 in = 0.019939 m)

   $A$$=\frac{\pi }{4}\times {0.019939}^2$$=0.0003122458\text{\hspace{0.17em}m²}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $2\times 0.3048$$=0.6096\text{\hspace{0.17em}m/s}$

3. Multiply the speed by the area to get the flow

   $0.6096\times 0.0003122458$$=0.000190345\text{\hspace{0.17em}m³/s}$

4. Show the flow in gpm

   $\frac{0.000190345}{0.0000630902}$$=3.017031\text{\hspace{0.17em}gpm}$

10. Flow rate: 21.55 gpm

1. Find the open area inside 1 in Schedule 40 steel, since that's the space the water fills (inside diameter 1.049 in = 0.026645 m)

   $A$$=\frac{\pi }{4}\times {0.026645}^2$$=0.0005575814\text{\hspace{0.17em}m²}$

2. Change the speed to meters per second to match (1 ft = 0.3048 m)

   $8\times 0.3048$$=2.4384\text{\hspace{0.17em}m/s}$

3. Multiply the speed by the area to get the flow

   $2.4384\times 0.0005575814$$=0.0013596065\text{\hspace{0.17em}m³/s}$

4. Show the flow in gpm

   $\frac{0.0013596065}{0.0000630902}$$=21.550203\text{\hspace{0.17em}gpm}$