Fluid Flow Rate

This problem set deals with fluid flow rate.

1. The rates associated with the units below may be classified as volume flow or amss flow rates. Identify each by QV or Qm.
1. QV m3/sec
2. Qm grams/min
3. QV liters/sec
4. QV gal/min
5. Qm kg/min
6. QV in3/sec

2. If a leak in a faucet causes a water loss of 120 gallons in two days, what is the volume flow rate per hour?
 Knowns Unknowns Equations/Remarks V = 120 gals t = 2 days = 48 hrs QV = ? in gal/hr QV = V/t QV = V/tQV = (120 gals)/(48 hrs) QV = 2.5 gal/hr Substitute the known values in the equation and calculate the answer.

3. A supertanker carries 1.8 ×106 gallons of crude oil. The oil is unloaded at an offshore discharge buoy in 35 hours. Find the volume flow rate in gallons per minute.
 Knowns Unknowns Equations/Remarks V = 1.8 ×106 gallons t = 35 hrs 1 hr = 60 min QV = ? in gal/min QV = V/t QV = V/tQV = [(1.8 ×106 gal)/(35 hrs)] · (1 hr/60 min) QV = 857 gal/min Substitute in the known values and calculate the answer, converting hours to minutes.

4. A supertanker unloads oil at the rate of 850 gal/min. It takes 25 hours to unload. How many gallons of oil are unloaded in 25 hours?
 Knowns Unknowns Equations/Remarks QV = 850 gal/min t = 25 hrs 1 hr = 60 min V = ? QV = V/t QV = V/tV = QV · t = (850 gal/min)(25 hrs)(60 min/1 hr) V = 1.28 ×106 gal Rearrange the equation to solve for the unknown, then substitute in the known values and calculate the answer.

5. A supertanker unloads 2 ×106 gallons of crude oil at 800 gal/min. How long does it take to unload the tanker, in hours?
 Knowns Unknowns Equations/Remarks V = 2 ×106 gal QV = 800 gal/min 1 hr = 60 min t = ? in hrs QV = V/t QV = V/tt = V/QV = [(2 ×106 gal)/(800 gal/min)] · (1 hr/60 min) t = 41.7 hr Rearrange the equation to solve for the unknown, then substitute in the known values and calculate the answer.

6. A hydraulic control pump moves 3.6 ×103 liters of fluid through a closed system in a 10 hour period. Find the volume flow rate in liters per minute.
 Knowns Unknowns Equations/Remarks V = 3.6 ×103 liters t = 10 hr 1 hr = 60 min QV = ? in liters/min QV = V/t QV = V/tQV = [(3.6 ×103 liters)/(10 hr)] · (1 hr/60 min) QV = 6 liters/min Substitute in the known values and calculate the answer, converting hours to minutes.

7. A propane tank used to supply propane gas to a boiler will empty 250 kg of fuel into the boiler in 8 hours. Find the mass flow rate of the gas in kg/hr.
 Knowns Unknowns Equations/Remarks m = 250 kg t = 8 hrs Qm = ? Qm = m/t Qm = m/tQm = (250 kg)/(8 hrs) Qm = 31.3 kg/hr Substitute the known values in the equation and calculate the answer.

8. An air conditioning system is designed to completely exchange the air in a room every 8 minutes. The room has dimensions of 6m × 6m × 2.5m. Dry air at 20°C (68°F) has a mass of 1.2 kg/m3. What mass flow rate does the air conditioner provide?
 Knowns Unknowns Equations/Remarks L = 6 m W = 6 m H = 2.5 m r  =  1.2 kg/m3 t = 8 mins Qm = ? Qm = m/t r = m/V V = L·W·H (volume of a box) V = L·W·H = (6 m)(6 m)(2.5 m) = 90 m3 r = m/V m = r · V = (1.2 kg/m3)(90 m3) = 108 kg Qm = m/t =  (108 kg)/(8 min) Qm = 13.5 kg/min Find the volume of the room with the volume of a box equation.Use the volume and the density of the air to find the mass of the air in the room. Then use the mass flow rate equation to find the mass flow rate.

9. An air compressor is rated at 210 cfm ("cubic feet per minute," or ft3/min), free air. "Free air" means air at atmospheric pressure. Find the weight of air taken into the compressor while operating for one hour at its rated value. One cubic foot of air weighs 7.5 ×10-2 lb.
 Knowns Unknowns Equations/Remarks QV = 210 ft3/min t = 1 hr rw = 7.5 ×10-2 lb/ft3 1 hr = 60 min w = ? rw = w/V QV = V/t QV = V/t V = QV · t = (210 ft3/min)(1 hr)(60 min/1 hr) = 12,600 ft3 rw = w/V w = rw · V = (7.5 ×10-2 lb/ft3)(12600 ft3) w = 945 lb Rearrange the volume flow rate equation to find the volume taken into the compressor in one hour.Then rearrange the weight density equation to find the weight of the air taken into the compressor in one hour.

10. A tank is filled with 480 ft3 of water in 1.50 hours. Find the volume flow rate.
 Knowns Unknowns Equations/Remarks V = 480 ft3 t = 1.50 hr QV = ? QV = V/t QV = V/t = (480 ft3)/(1.50 hr)QV = 320 ft3/hr Substitute the known values in the equation and calculate the answer.

11. 200 kg of air flows through a hose in 8.50 minutes. Find the mass flow rate.
 Knowns Unknowns Equations/Remarks m = 200 kg t = 8.50 min Qm = ? Qm = m/t Qm = m/t = (200 kg)/(8.50 min)Qm = 23.5 kg/min Substitute the known values in the equation and calculate the answer.

12. 263 cm3 of water flows through a pipe in 86.5 seconds. Find:
1. the volume flow rate, and
2. the mass flow rate.

 Knowns Unknowns Equations/Remarks V = 263 cm3 t = 86.5 sec r = 1 g/cm3 A. QV = ? B. Qm = ? QV = V/t Qm = rQV (Equation II-11) A. QV = V/t = (263 cm3)/(86.5 sec)     QV = 3.04 cm3/sec B. Qm = rQV = (1 g/cm3)(3.04 cm3/sec)     Qm = 3.04 g/sec Substitute the known values in the equation and calculate the answer.Use Equation II-11 to convert the volume flow rate to mass flow rate. Substitute in the known values and calculate the answer.

13. Water flows through a 3 ft wide channel with a water depth of 18 inches. The average flow speed is 4.0 ft per sec. Find the volume flow rate.
 Knowns Unknowns Equations/Remarks W = 3 ft H = 18 in v = 4.0 ft/sec 1 ft = 12 in QV = ? QV = v · A A = W · H A = W · H = (3ft)(18 in)(1 ft/12 in) = 4.5 ft2 QV = v · A = (4.0 ft/sec)(4.5 ft2) QV = 18 ft3/sec Find the cross-sectional area of the channel.Then substitute in the flow rate equation and calculate the answer.

14. Water at a depth of 4.5 cm flows through a 26.0 cm wide channel at an average speed of 18.5 cm/s. Find the volume flow rate in
1. cm3/s and
2. m3/hr.

 Knowns Unknowns Equations/Remarks W = 26.0 cm H = 4.5 cm v = 18.5 cm/s 1 m3 = 106 cm3 1 hr = 3600 sec A. QV = ? in cm3/s B. QV = ? in m3/hr QV = v · A A = W · H A. A = W · H = (26.0 cm)(4.5 cm) = 117 cm2     QV = v · A = (18.5 cm/s)(117 cm2)     QV = 2164.5 cm3/s = 2160 cm3/s B. QV = (2164.5 cm3/s)(3600 s/1 hr)(1 m3/106 cm3)     QV = 7.79 m3/hr Find the cross-sectional area of the channel, then substitute in the known values and calculate the answer in cm3/s.Convert seconds to hours and cubic centimeters to cubic meters to find the answer in m3/hr.

15. A pipe supplies liquid to a storage tank at a rate of 2.5 cubic meters per minute. How long will it take to fill a tank with a radius of 2.50 meters and a depth of 6.50 meters?
 Knowns Unknowns Equations/Remarks QV = 2.5 m3/min r = 2.50 m h = 6.50 m t = ? QV = V/t V = pr2h (volume of a cylinder) V = pr2h QV = V/t = (pr2h)/t t = (pr2h)/QV = (p)(2.50 m)2(6.50 m)/(2.5 m3/min) t = 51.1 min Substitute the cylinder volume equation in for V in the volume flow rate equation.Rearrange the equation to solve for t. Substitute in the known values and calculate the answer.

16. What is the mass flow rate in kg/sec through a hose if 20 liters of water flows out of the hose in 10 sec?
 Knowns Unknowns Equations/Remarks V = 20 liters t = 10 s r = 1000 kg/m3 1 m3 = 1000 liters Qm = ? in kg/s Qm = m/t r = m/V r = m/V m = r · VQm = m/t = (r · V)/t Qm = (1000 kg/m3)(20 liters)(1 m3/1000 liters)/(10 s) Qm = 2 kg/s Rearrange the mass density equation to solve for mass, then substitute in for m in the flow rate equation.Substitute in the known values, convert liters to cubic meters, and calculate the answer.

17. If the volume flow rate of water is 0.5 m3/s, what is the mass flow rate in kg/min?
 Knowns Unknowns Equations/Remarks QV =  0.5 m3/s r = 1000 kg/m3 1 min = 60 sec Qm = ? in kg/min Qm = rQV (Equation II-11) Qm = rQV = (1000 kg/m3)(0.5 m3/sec)(60 sec/1 min)Qm = 30,000 kg/min Substitute in the known values, convert seconds to minutes, and calculate the answer.