Question

A horizontal pipe is running full of water. At a certain point A, it tapers from $60cm$ diameter to $20cm$ diameter at B. The pressure difference between A and B is $100cm$ of water column.The rate of flow of water through the pipe is.

• A. $1.4\times {10}^{5}c{m}^3{s}^{-1}$
• B. $1.4\times {10}^5{m}^3{s}^{-1}$
• C. $0.14\times {10}^{5}c{m}^3{s}^{-1}$
• D. $1.4\times {10}^{5}m{m}^3{s}^{-1}$

Answer 1

The correct option is A $1.4\times {10}^{5}c{m}^3{s}^{-1}$

$P_A-P_B=fgh={10}^{4}N/m^2$

Now,

Applying bernoullis theorem across A and B,

$P_A-P_B=\frac{1}{2}\lambda \left(v^2(1-{\left(\frac{10}{30}\right)}^4)\right)$

$\Rightarrow \frac{2^4}{0.98765\times {10}^3}=v^2\Rightarrow v^2=20.25$$\Rightarrow v=4.5m/s$ at the smaller end

So,

Rate of flow $=AV=4.5\times 100\times \pi \times {\left(10\right)}^{2}c{m}^3/s$

$=1.413\times {10}^{5}c{m}^3/s$