If an impact sprinkler pumps 9.425 litres per minute over a circular lawn with perimeter of 37.7 metres, and the water is evenly distributed, how long do you need to water, to give the entire lawn 10mm of water?

lisasondra
Jan 4, 2015

#1**0 **

First change everything to ml and cm.

9425ml/min

P=3770cm

10mm=1cm

2pi×r=3770 so

r= 3770/(2pi)

V=pi× r^2 × 1 ml

(1minute / 9425ml) × V

will give the time in minutes.

I am sorry about the presentation. I am on my phone.

Melody
Jan 4, 2015

#3**0 **

Thanks Melody for your help. A bit lost where to go next though with those details. What formula would I use to calculate the time?

Cheers,

Lisa-sondra

lisasondra
Jan 4, 2015

#4**0 **

Sorry I only just saw your reply.

I will go home soon and explain better

It will be easier when I am on my computer. :)

Melody
Jan 4, 2015

#5**0 **

Okay - I said I would be back and here I am :)

If an impact sprinkler pumps 9.425 litres per minute over a circular lawn with perimeter of 37.7 metres, and the water is evenly distributed, how long do you need to water, to give the entire lawn 10mm of water?

First you need to know that a container 1cm^{3} will hold 1ml of water.

So I will change all the units to cm and ml.

the question has become

**If an impact sprinkler pumps 9425 mL per minute over a circular lawn with perimeter of 3770 cm, and the water is evenly distributed, how long do you need to water, to give the entire lawn 1cm of water?**

Now the perimeter is the circumference of the circle which is $$C=2\pi r$$

so

$$\\3770=2\pi r\\

$divide both sides by 2\pi\\

r=\frac{3770}{2\pi}$$

I would prefer to keep r in this form until the end but you may prefer to get a figure for this.

$${\frac{{\mathtt{3\,770}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}} = {\mathtt{600.014\: \!135\: \!456\: \!445\: \!415\: \!8}}$$

So the radius of the cirlce is very close to 600cm (that is 6m which sounds reasonable for a sprinkler)

**radius=600cm**

Now, the area of the circle is $$A=\pi r^2$$

$$\\A=\pi * 600^2 \quad cm^2\\

A=360000\pi\quad cm^2\\$$

$${\mathtt{360\,000}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{1\,130\,973.355\: \!292\: \!325\: \!565\: \!846\: \!6}}$$

$$\\A=1130973\quad cm^2$$

Now think of this like a very shallow circular pond.

The area of the base is 1130973 cm^{2} and the water is 1 cm high.

So that is a volume of 1130973 x 1 cm^{3 }= 1130973 cm^{3}

And a pond this size will hold 1130973 mL

So NOW the question is redued to

**An impact sprinkler pumps 9425 mL per minute. How long will it take to pump out 1130973 mL?**

So we have

$$\frac{9425mL}{minute},\qquad 1130973\;ml,\qquad $and we want to know $\qquad ?\;minutes$$

I have my own pet method of dealing with rates and if you learn it then you can work out really complicated rates questions easily.

I want the answer to be minutes so I want to start with minutes on the top of the fraction.

if the sprinkler pumps 9425ml / minute THEN it is also true to say that in 1 minute it pumps 9425ml.

SO it is valid to turn the fraction upside down

so I have $$\frac{1 \;minute}{9425ml}$$ if I multiply this by 1130973 ml the ml will cancel out leaving only minutes which is the answer that you want.

i.e.

$$\frac{1 \;minute}{9425ml}\times \frac{1130973\;ml}{1}=\frac{11310}{9425}\;minutes$$

$${\frac{{\mathtt{1\,130\,973}}}{{\mathtt{9\,425}}}} = {\mathtt{119.997\: \!135\: \!278\: \!514\: \!588\: \!9}}$$

So it will take 120 minutes which is **2 hours**

Now let me know if you understand. Maybe I can explain individual bits better.

Melody
Jan 4, 2015