How long to fill 10.2 m^3 tank from pump delivering 0.75 hL of water per minute?
- Thread starter Ando839
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- Nov 24, 2012
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Hello, and welcome to FMH! 
We know:
[MATH]1\text{ hL}=100\text{ L}[/MATH]
[MATH]1\text{ L}=1000\text{ cm}^3[/MATH]
[MATH]1\text{ m}=100\text{ cm}[/MATH]
Using this, can you convert from [MATH]\text{hL}[/MATH] to [MATH]\text{m}^3[/MATH] ?
We know:
[MATH]1\text{ hL}=100\text{ L}[/MATH]
[MATH]1\text{ L}=1000\text{ cm}^3[/MATH]
[MATH]1\text{ m}=100\text{ cm}[/MATH]
Using this, can you convert from [MATH]\text{hL}[/MATH] to [MATH]\text{m}^3[/MATH] ?
- Joined
- Nov 24, 2012
- Messages
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To follow up:
[MATH]x\frac{\text{hL}}{\text{min}}=x\frac{\text{hL}}{\text{min}}\cdot\frac{100\text{ L}}{1\text{ hL}}\cdot\frac{1000\text{ cm}^3}{1\text{ L}}\cdot\left(\frac{1\text{ m}}{100\text{ cm}}\right)^3=\frac{1}{10}x\frac{\text{m}^3}{\text{min}}[/MATH]
And so:
[MATH]0.75\frac{\text{hL}}{\text{min}}=0.075\frac{\text{m}^3}{\text{min}}[/MATH]
Thus, the time \(t\) required to fill the tank of volume \(V\) at the rate \(r\):
[MATH]t=\frac{V}{r}=\frac{10.2\text{ m}^3}{0.075\dfrac{\text{m}^3}{\text{min}}}=136\text{ min}=2\text{ hr}\,16\text{ min}[/MATH]
[MATH]x\frac{\text{hL}}{\text{min}}=x\frac{\text{hL}}{\text{min}}\cdot\frac{100\text{ L}}{1\text{ hL}}\cdot\frac{1000\text{ cm}^3}{1\text{ L}}\cdot\left(\frac{1\text{ m}}{100\text{ cm}}\right)^3=\frac{1}{10}x\frac{\text{m}^3}{\text{min}}[/MATH]
And so:
[MATH]0.75\frac{\text{hL}}{\text{min}}=0.075\frac{\text{m}^3}{\text{min}}[/MATH]
Thus, the time \(t\) required to fill the tank of volume \(V\) at the rate \(r\):
[MATH]t=\frac{V}{r}=\frac{10.2\text{ m}^3}{0.075\dfrac{\text{m}^3}{\text{min}}}=136\text{ min}=2\text{ hr}\,16\text{ min}[/MATH]