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37 Practice Problems

Practice Problems

Flow Fundamentals

  • 1.       What is volumetric flow and what are the common units of measure?
  • 2.       What is mass flow and what are the common units of measure? How is it different from volumetric flow?
  • 3.       How can you calculate volumetric flow if you know the pipe’s cross-sectional area and fluid velocity?
  • 4.       What is the formula for mass flow if you know the volumetric flow and density of the fluid?
  • 5.       Define the internal diameter of a pipe and explain its importance in flow calculations.
  • 6.       How do you calculate the cross-sectional area of a pipe if you know its internal diameter?
  • 7.       Define specific gravity and explain its significance in flow calculations.
  • 8.       Explain how to find the velocity of a fluid if you know the volumetric flow and cross-sectional area.
  • 9.       Define flow rate and provide the formula for calculating it. Include the conversion factor to convert from cfs to gpm.
  • 10.   What is the relationship between velocity and cross-sectional area in flow calculations?
  • 11.   Calculate the density of a fluid with a specific gravity of 0.8 and a density of water at 62.425 lb/ft³.
  • 12.   Calculate the cross-sectional area if the internal diameter of a pipe is 18 inches. Answer must be in square feet
  • 13.   Calculate the mass flow of a fluid with a volumetric flow of 10 cfs and a density of 62.425 lb/ft³. Answer must be in pounds per minute.
  • 14.   A fluid with a volumetric flow of 8 cfs flows through a pipe with a cross-sectional area of 10 ft². Calculate the velocity of the fluid.
  • 15.   Calculate the volumetric flow if the velocity is 2 ft/s and the cross-sectional area is 1.5 ft².
  • 16.   Calculate the flow rate (Q) for a volumetric flow (q) of 5 cfs.
  • 17.   Calculate the mass flow rate for a fluid with a volumetric flow of 15 gpm and a density of 50 lb/ft³. Answer must be pounds per minute.
  • 18.   Calculate the volumetric flow rate (Q) if the cross-sectional area is 0.2 ft² and the velocity is 0.4 ft/s. Answer must be in gallons per minute.
  • 19.   Calculate the mass flow rate of a fluid with an SG of 0.8009612 if the velocity is 3 ft/s and the cross-sectional area is 0.25 ft². Answer must be in pounds per minute.
  • 20.   Given a mass flow rate of 19,432 lb/min of water in a 6-inch pipe, determine the fluid velocity.

Flow Measurement

Practice Problem Answers

  • 1.       Answer: Volumetric flow is the volume of fluid passing through a pipe per unit of time. It is measured in cubic feet per second (cfs) or gallons per minute (gpm).
  • 2.       Answer: Mass flow is the mass of fluid passing through a pipe per unit of time. It is measured in pounds per minute (lb/min) or pounds per second (lb/s). It is different from volumetric flow as it considers the density of the fluid.
  • 3.       Answer: Volumetric Flow (cfs) = Cross-sectional Area (ft²) x Velocity (ft/s)
  • 4.       Answer: Mass Flow (lb/min) = Volumetric Flow (cfs) x Density (lb/ft³)
  • 5.       Answer: The internal diameter is the measure of the inner width of a pipe. It is crucial in flow calculations as it is used to determine the cross-sectional area of the pipe.
  • 6.       Answer: Cross-sectional Area (ft²) = π * (Internal Diameter/2)²
  • 7.       Answer: Specific gravity is the ratio of the density of a fluid to the density of water. SG can be used to calculate mass flow calculations when given.
  • 8.       Answer: Velocity (ft/s) = Volumetric Flow (cfs) / Cross-sectional Area (ft²)
  • 9.       Answer: Flow Rate is the quantity of fluid passing through a point per unit of time. Flow Rate (gpm) = Volumetric Flow (cfs) x 448.831
  • 10.   Answer: Velocity is inversely proportional to cross-sectional area. As one increases, the other decreases, and vice versa.
  • 11.   Answer: Density = Specific Gravity x Density of Water = 0.8 x 62.425 lb/ft³ = 49.94 lb/ft³
  • 12.   Answer: Cross-sectional Area = π * (18/2)² = 254.47 square inches / 144 = 1.767 square feet
  • 13.   Answer: Mass Flow = (10 cfs * 60s/min) x 62.425 lb/ft³ = 37,455 lb/min
  • 14.   Answer: Velocity = 8 cfs / 10 ft² = 0.8 ft/s
  • 15.   Answer: Volumetric Flow = 2 ft/s x 1.5 ft² = 3 cfs
  • 16.   Answer: Flow Rate = 5 cfs x 448.831 = 2244.155 gpm
  • 17.   Answer: Mass Flow = Volumetric Flow x Density = 15 gpm x 8.345lb/gal * (50 lb/ft³/62.425 lb/ft³) = 334.2 lb/min
  • 18.   Answer: Volumetric Flow = Cross-sectional Area x Velocity = 0.2 ft² x 0.4 ft/s = .08 cfs * 448.83 = 35.9 gpm
  • 19.   Answer: Mass Flow = Velocity x Density x Cross-sectional Area = (3 ft/s * 60 s/min) x (0.8009612 * 62.425 lb/ft³) x 0.25 ft² = 2250lb/min
  • 20.   Answer: Convert mass to volumetric. q = (mass / density)(1min/60seconds) = (19,432 lb/min * 1min/60sec) (1ft3 / 62.425lb) = 5.188cfs. OR q = Q / 448.83 = (mass / density)(19432 lb/min * 1 gallon / 8.345 lb) / 448.83 = 5.188cfs. Solve for velocity = q/area = 5.188cfs/0.196ft2  = 26.422ft/s

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