Fluid Flow Rate

This problem set deals with fluid flow rate.

The rates associated with the units below may be classified as volume flow or amss flow rates. Identify each by Q_V or Q_m.

Q_V m³/sec
Q_m grams/min
Q_V liters/sec
Q_V gal/min
Q_m kg/min
Q_V in³/sec

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

Q_V = ? in gal/hr

Q_V = V/t

Q_V = (120 gals)/(48 hrs)

Q_V = 2.5 gal/hr

Substitute the known values in the equation and calculate the answer.

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

Q_V = ? in gal/min

Q_V = V/t

Q_V = [(1.8 ×10⁶ gal)/(35 ~~hrs~~)] · (1 hr/60 min)

Q_V = 857 gal/min

Substitute in the known values and calculate the answer, converting hours to minutes.

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

Q_V = 850 gal/min
t = 25 hrs
1 hr = 60 min

V = ?

Q_V = V/t

V = Q_V · 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.

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
Q_V = 800 gal/min
1 hr = 60 min

t = ? in hrs

Q_V = V/t

t = V/Q_V = [(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.

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

Q_V = ? in liters/min

Q_V = V/t

Q_V = [(3.6 ×10³ liters)/(10 hr)] · (1 hr/60 min)

Q_V = 6 liters/min

Substitute in the known values and calculate the answer, converting hours to minutes.

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

Q_m = ?

Q_m = m/t

Q_m = (250 kg)/(8 hrs)

Q_m = 31.3 kg/hr

Substitute the known values in the equation and calculate the answer.

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

Q_m = ?

Q_m = 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

Q_m = m/t = (108 kg)/(8 min)
Q_m = 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.

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
Q_V = 210 ft³/min t = 1 hr r_w = 7.5 ×10^-2 lb/ft³ 1 hr = 60 min	w = ?	r_w = w/V Q_V = V/t
Q_V = V/t V = QV · t = (210 ft³/~~min~~)(1 hr)(60 ~~min~~/1 hr) = 12,600 ft³ r_w = w/V w = r_w · 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	Q_V = ?	Q_V = V/t
Q_V = V/t = (480 ft³)/(1.50 hr) Q_V = 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	Q_m = ?	Q_m = m/t
Q_m = m/t = (200 kg)/(8.50 min) Q_m = 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:

the volume flow rate, and
the mass flow rate.

Knowns	Unknowns	Equations/Remarks
V = 263 cm³ t = 86.5 sec r = 1 g/cm³	A. QV = ? B. Qm = ?	Q_V = V/t Q_m = rQ_V (Equation II-11)
A. Q_V = V/t = (263 cm³)/(86.5 sec) Q_V = 3.04 cm³/sec B. Q_m = rQ_V = (1 g/cm³)(3.04 cm³/sec) Q_m = 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 = ?

Q_V = v · A
A = W · H

A = W · H = (3ft)(18 in)(1 ft/12 in) = 4.5 ft²

Q_V = v · A = (4.0 ft/sec)(4.5 ft²)

Q_V = 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

cm³/s and
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. Q_V = ? in cm³/s
B. Q_V = ? in m³/hr

Q_V = v · A
A = W · H

A. A = W · H = (26.0 cm)(4.5 cm) = 117 cm²

Q_V = v · A = (18.5 cm/s)(117 cm²)
Q_V = 2164.5 cm³/s = 2160 cm³/s

B. Q_V = (2164.5 ~~cm3/s~~)(3600 s/1 hr)(1 m³/10⁶ cm³)
Q_V = 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

Q_V = 2.5 m3/min
r = 2.50 m
h = 6.50 m

t = ?

Q_V = V/t
V = pr²h (volume of a cylinder)

V = pr²h

Q_V = V/t = (pr²h)/t

t = (pr²h)/Q_V = (_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	Q_m = ? in kg/s	Q_m = m/t r = m/V
r = m/V m = r · V Q_m = m/t = (r · V)/t Q_m = (1000 kg/m³)(20 ~~liters~~)(1 m³/1000 ~~liters~~)/(10 s) Q_m = 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
Q_V = 0.5 m³/s r = 1000 kg/m³ 1 min = 60 sec	Q_m = ? in kg/min	Q_m = rQ_V (Equation II-11)
Q_m = rQ_V = (1000 kg/m³)(0.5 m³~~/sec~~)(60 ~~sec~~/1 min) Q_m = 30,000 kg/min		Substitute in the known values, convert seconds to minutes, and calculate the answer.