Question

If the elevation in reservoir b is 100m, what must the elevation in reservoir a be if thevolume flow rate through the cast-iron pipe is 0.03m^3/s?

Answer 1

The elevation in reservoir a must be=100.2631 m

How can we calculate the value of the elevation in reservoir

a ?

To calculate the value of the elevation in reservoir a we are using the formula,

$hf= \frac{flQ^{2}}{12.1d^{5}}$

We know, For cast iron the chart has(f₁) 0.0012  from Moody's chart

(f₂)0.016 for cast iron.

Now the formula becomes,

$h_{1}= h_{2}+\frac{Q^{2}}{12.1}(\frac{f_1l_1}{d_1^{5}} +\frac{f_2l_2}{d_2^{5}} )$

we are given,

h₂= the elevation in reservoir b

=100m

Q= the volume flow rate through the cast-iron pipe.

= 0.03 m³/s

Let l₁ = 100 m pipe

l₂ = 150 m pipe

we know,

d₁=12mm

and d₂=15mm

Now we put the known values in the above equation we get,

$h_{1}= h_{2}+\frac{Q^{2}}{12.1}(\frac{f_1l_1}{d_1^{5}} +\frac{f_2l_2}{d_2^{5}} )$

Or,$h_{1}= 100+\frac{(0.03)^{2}}{12.1}(\frac{0.0012\times 100}{(\frac{12}{1000})^{5}} +\frac{0.0016\times 150}{(\frac{15}{1000})^{5}} )$

Or, $h_{1}= 100+\frac{(0.03)^{2}}{12.1}(\frac{0.12}{0.00032} +\frac{0.24}{0.000759} )$

Or, h₁=100+ 0.000074[375+3162.06]

Or, h₁=100.2631m

From the above calculation we can conclude that, The elevation in reservoir a must be= 100.2631m

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