Convert a volume flow rate of 3.0 gal/min to m3/s. 6.2 Convert 459 gal/min to m3/s. 6.27 A small metering pump delivers 11.4 gal of a water treatment chemical per 24 h. Express this flow rate in ft3/s. 6.28 A small metering pump delivers 19.5 mL/min of water to dilute a waste stream. Express this flow rate in m3/s. 6.30 Oil for a hydraulic system (sg = 0.90) is flowing at 2.35 x 10-3 m3/s. Calculate the weight flow rate and mass flow rate. 6.36 If a pump removes 1.65 gal/min of water from a tank, how long will it take to empty the tank if it contains 7425 lb of water? The specific weight and density of water needed in problem #30 are 9.81 kN/m3 and 1000 kg/m3, respectively.
The Correct Answer and Explanation is:
Let’s go through each problem step-by-step with correct unit conversions and explanations. Then I’ll provide a 300-word explanation at the end.
6.2 Convert 3.0 gal/min to m³/s
Conversion factors:
3.0 gal/min=3.0×3.78541×10−360=1.89×10−4 m3/s3.0 \, \text{gal/min} = 3.0 \times \frac{3.78541 \times 10^{-3}}{60} = 1.89 \times 10^{-4} \, \text{m}^3/\text{s}
6.2 (again?) Convert 459 gal/min to m³/s
459 gal/min=459×3.78541×10−360=2.89×10−2 m3/s459 \, \text{gal/min} = 459 \times \frac{3.78541 \times 10^{-3}}{60} = 2.89 \times 10^{-2} \, \text{m}^3/\text{s}
6.27 Convert 11.4 gal/24h to ft³/s
Conversions:
- 1 gallon = 0.133681 ft³
11.4×0.13368186400=1.76×10−5 ft3/s\frac{11.4 \times 0.133681}{86400} = 1.76 \times 10^{-5} \, \text{ft}^3/\text{s}
6.28 Convert 19.5 mL/min to m³/s
Conversions:
19.5×1×10−660=3.25×10−7 m3/s19.5 \times \frac{1 \times 10^{-6}}{60} = 3.25 \times 10^{-7} \, \text{m}^3/\text{s}
6.30 Oil flow rate: 2.35 × 10⁻³ m³/s, sg = 0.90
- Density of water = 1000 kg/m³ → Oil density = 0.90 × 1000 = 900 kg/m³
Mass flow rate: m˙=900 kg/m3×2.35×10−3=2.115 kg/s\dot{m} = 900 \, \text{kg/m}^3 \times 2.35 \times 10^{-3} = 2.115 \, \text{kg/s}
Weight flow rate: W˙=8.829 kN/m3×2.35×10−3=0.02075 kN/s\dot{W} = 8.829 \, \text{kN/m}^3 \times 2.35 \times 10^{-3} = 0.02075 \, \text{kN/s}
6.36 Pump removes 1.65 gal/min of water from a tank with 7425 lb
Convert gal/min to lb/min:
- 1 gallon of water ≈ 8.34 lb
1.65 gal/min×8.34 lb/gal=13.761 lb/min1.65 \, \text{gal/min} \times 8.34 \, \text{lb/gal} = 13.761 \, \text{lb/min}
Now divide total mass by flow rate: 742513.761=539.4 minutes≈8.99 hours\frac{7425}{13.761} = 539.4 \, \text{minutes} \approx 8.99 \, \text{hours}
300-Word Explanation
In engineering, converting between various units of flow—whether it’s volume, mass, or weight flow rate—is crucial for understanding system behavior and ensuring compatibility between equipment.
For volumetric flow rates, gallons per minute (gal/min) must often be converted into cubic meters per second (m³/s) for SI unit consistency. Since 1 gallon equals 3.78541 × 10⁻³ m³ and 1 minute equals 60 seconds, we divide accordingly to get flow rates like 1.89 × 10⁻⁴ m³/s for a 3.0 gal/min flow.
In problem 6.27, converting daily gallon use into cubic feet per second requires attention to both volume (gallons to ft³) and time (24 hours = 86,400 seconds). Accurate conversions like 1.76 × 10⁻⁵ ft³/s help estimate chemical dosing properly.
For mass and weight flow rates (problem 6.30), we apply the specific gravity (sg) to find density, and multiply it by volumetric flow to get the mass flow rate (kg/s). Weight flow rate (kN/s) uses specific weight (N/m³), which is the product of specific gravity and gravitational acceleration (9.81 m/s²).
Finally, problem 6.36 combines mass and volumetric flow rates. Since the pump removes a volume flow (1.65 gal/min), we convert that to mass using the density of water (~8.34 lb/gal). The total mass of water (7425 lb) divided by flow rate (13.761 lb/min) gives the time to empty the tank—about 539 minutes or 9 hours.
Understanding and converting these units accurately ensures engineers make correct system designs and avoid operational errors.