+0  
 
0
455
5
avatar+7 

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
avatar+94088 
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
 #2
avatar+94088 
0

Do you have any questions lisasondra?

Melody  Jan 4, 2015
 #3
avatar+7 
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
avatar+94088 
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
avatar+94088 
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  1cm3  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 cm2 and the water is 1 cm high.

So that is a volume of 1130973 x 1 cm3  = 1130973 cm3 

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

27 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.