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Pipe Volume worksheet

Grades 7–8 · CCSS 8.G.C.9, 7.G.B.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Pipe or tube: Schedule 80 PVC
- Pipe size (the size it's sold as): 1/2 in
- Length of pipe: 70 ft
Find: Water in the pipe
Answer
2.
Problem 2.
- Pipe or tube: Schedule 40 PVC
- Pipe size (the size it's sold as): 3 in
- Length of pipe: 60 ft
Find: Water in the pipe
Answer
3.
Problem 3.
- Pipe or tube: Type L copper
- Pipe size (the size it's sold as): 1 1/4 in
- Length of pipe: 40 ft
Find: Water in the pipe
Answer
4.
Problem 4.
- Pipe or tube: PEX (CTS, SDR 9)
- Pipe size (the size it's sold as): 1 1/2 in
- Length of pipe: 34 ft
Find: Water in the pipe
Answer
5.
Problem 5.
- Pipe or tube: Schedule 40 PVC
- Pipe size (the size it's sold as): 3 in
- Length of pipe: 45 ft
Find: Water in the pipe
Answer
6.
Problem 6.
- Pipe or tube: Schedule 40 steel
- Pipe size (the size it's sold as): 3/4 in
- Length of pipe: 60 ft
Find: Water in the pipe
Answer
7.
Problem 7.
- Pipe or tube: Type L copper
- Pipe size (the size it's sold as): 2 1/2 in
- Length of pipe: 40 ft
Find: Water in the pipe
Answer
8.
Problem 8.
- Pipe or tube: Type L copper
- Pipe size (the size it's sold as): 2 1/2 in
- Length of pipe: 41 ft
Find: Water in the pipe
Answer
9.
Problem 9.
- Pipe or tube: Schedule 40 PVC
- Pipe size (the size it's sold as): 4 in
- Length of pipe: 58 ft
Find: Water in the pipe
Answer
10.
Problem 10.
- Pipe or tube: Type M copper
- Pipe size (the size it's sold as): 2 in
- Length of pipe: 70 ft
Find: Water in the pipe
Answer

Answer key: Pipe Volume worksheet

1. Water in the pipe: 0.79 US gal
1. Find the area of the open circle inside 1/2 in Schedule 80 PVC (A), using its inside diameter d of 0.526 in
  $A$ $= \frac{π}{4} \times {0.526}^{2}$ $= 0.217301 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $70 \times 12$ $= 840 in$
3. Multiply the area by the length to get the volume of water inside
  $0.217301 \times 840$ $= 182.5327 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{182.5327}{231}$ $= 0.790185 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $0.790185 \times 3.785411784$ $= 2.991175 L$
2. Water in the pipe: 22.65 US gal
1. Find the area of the open circle inside 3 in Schedule 40 PVC (A), using its inside diameter d of 3.042 in
  $A$ $= \frac{π}{4} \times {3.042}^{2}$ $= 7.267889 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $60 \times 12$ $= 720 in$
3. Multiply the area by the length to get the volume of water inside
  $7.267889 \times 720$ $= 5,232.8803 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{5,232.8803}{231}$ $= 22.653161 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $22.653161 \times 3.785411784$ $= 85.751544 L$
3. Water in the pipe: 2.61 US gal
1. Find the area of the open circle inside 1 1/4 in Type L copper (A), using its inside diameter d of 1.265 in
  $A$ $= \frac{π}{4} \times {1.265}^{2}$ $= 1.256814 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $40 \times 12$ $= 480 in$
3. Multiply the area by the length to get the volume of water inside
  $1.256814 \times 480$ $= 603.2706 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{603.2706}{231}$ $= 2.611561 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $2.611561 \times 3.785411784$ $= 9.885834 L$
4. Water in the pipe: 2.22 US gal
1. Find the area of the open circle inside 1 1/2 in PEX (CTS, SDR 9) (A), using its inside diameter d of 1.264 in
  $A$ $= \frac{π}{4} \times {1.264}^{2}$ $= 1.254828 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $34 \times 12$ $= 408 in$
3. Multiply the area by the length to get the volume of water inside
  $1.254828 \times 408$ $= 511.9696 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{511.9696}{231}$ $= 2.216319 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $2.216319 \times 3.785411784$ $= 8.389679 L$
5. Water in the pipe: 16.99 US gal
1. Find the area of the open circle inside 3 in Schedule 40 PVC (A), using its inside diameter d of 3.042 in
  $A$ $= \frac{π}{4} \times {3.042}^{2}$ $= 7.267889 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $45 \times 12$ $= 540 in$
3. Multiply the area by the length to get the volume of water inside
  $7.267889 \times 540$ $= 3,924.6602 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{3,924.6602}{231}$ $= 16.989871 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $16.989871 \times 3.785411784$ $= 64.313658 L$
6. Water in the pipe: 1.66 US gal
1. Find the area of the open circle inside 3/4 in Schedule 40 steel (A), using its inside diameter d of 0.824 in
  $A$ $= \frac{π}{4} \times {0.824}^{2}$ $= 0.533267 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $60 \times 12$ $= 720 in$
3. Multiply the area by the length to get the volume of water inside
  $0.533267 \times 720$ $= 383.9519 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{383.9519}{231}$ $= 1.662129 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $1.662129 \times 3.785411784$ $= 6.291844 L$
7. Water in the pipe: 9.92 US gal
1. Find the area of the open circle inside 2 1/2 in Type L copper (A), using its inside diameter d of 2.465 in
  $A$ $= \frac{π}{4} \times {2.465}^{2}$ $= 4.772256 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $40 \times 12$ $= 480 in$
3. Multiply the area by the length to get the volume of water inside
  $4.772256 \times 480$ $= 2,290.6829 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{2,290.6829}{231}$ $= 9.916376 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $9.916376 \times 3.785411784$ $= 37.537567 L$
8. Water in the pipe: 10.16 US gal
1. Find the area of the open circle inside 2 1/2 in Type L copper (A), using its inside diameter d of 2.465 in
  $A$ $= \frac{π}{4} \times {2.465}^{2}$ $= 4.772256 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $41 \times 12$ $= 492 in$
3. Multiply the area by the length to get the volume of water inside
  $4.772256 \times 492$ $= 2,347.9499 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{2,347.9499}{231}$ $= 10.164285 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $10.164285 \times 3.785411784$ $= 38.476006 L$
9. Water in the pipe: 37.82 US gal
1. Find the area of the open circle inside 4 in Schedule 40 PVC (A), using its inside diameter d of 3.998 in
  $A$ $= \frac{π}{4} \times {3.998}^{2}$ $= 12.553807 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $58 \times 12$ $= 696 in$
3. Multiply the area by the length to get the volume of water inside
  $12.553807 \times 696$ $= 8,737.4499 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{8,737.4499}{231}$ $= 37.824459 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $37.824459 \times 3.785411784$ $= 143.181151 L$
10. Water in the pipe: 11.53 US gal
1. Find the area of the open circle inside 2 in Type M copper (A), using its inside diameter d of 2.009 in
  $A$ $= \frac{π}{4} \times {2.009}^{2}$ $= 3.169931 in²$
2. Change the length to inches so it matches the area (1 ft = 12 in)
  $70 \times 12$ $= 840 in$
3. Multiply the area by the length to get the volume of water inside
  $3.169931 \times 840$ $= 2,662.7417 in³$
4. Turn cubic inches into US gallons (1 gal = 231 in³)
  $\frac{2,662.7417}{231}$ $= 11.52702 gal$
5. Turn gallons into litres (1 gal = 3.785411784 L)
  $11.52702 \times 3.785411784$ $= 43.634519 L$