Question

In a horizontal oil pipe line of constant cross sectional area the decrease of pressure between two points \[100{\rm{ km}}\] apart is \[1500{\rm{ Pa}}\]. The loss of energy per unit volume per unit distance is --Joule.
A. \[15\]
B. \[0.015\]
C. \[0.03\]
D. zero

Answer 1

Hint: We will use the basic concept of energy loss and work done when fluid flows through a pipe. Work done is equal to the product of loss of pressure of oil and cross-sectional area of the pipe.

Complete step by step solution:
Given:
The loss of pressure in the horizontal pipeline of constant cross-sectional area is \[\Delta P = 1500{\rm{ Pa}}\].
The length of the pipeline is \[l = 100{\rm{ km}} = {10^5}{\rm{ m}}\].
We have to find the loss of energy per unit volume and per unit distance.
We know that the energy of a body is equal to work done by it.
\[E = W\]……(1)
Here W is the work done.
Work done can be expressed as the product of the change in pressure and cross-sectional area of the pipe.
\[W = \Delta P \cdot A\]
Here A is the cross-sectional area of the pipe.
Substitute \[\Delta P \cdot A\] for W in equation (1).
\[E = \Delta P \cdot A\]
Let us write the expression of energy lost per length and per unit volume.
\[E = \dfrac{{\Delta P \cdot A}}{{V \cdot l}}\]
Substitute \[A \cdot l\] for V as the volume is the product of cross-sectional area and length in the above expression.
$
E = \dfrac{{\Delta P \cdot A}}{{\left( {A \cdot l} \right)l}}\\
 = \dfrac{{\Delta P}}{{{l^2}}}
$
Substitute \[1500{\rm{ Pa}}\] for \[\Delta P\] and \[{10^5}{\rm{ m}}\] for l in the above expression.
\[E = \dfrac{{1500{\rm{ Pa}}}}{{{{\left( {{{10}^5}{\rm{ m}}} \right)}^2}}}\]……(1)
We can also express one Pascal in terms of Newton and meter as below.
\[1{\rm{ Pa}} = {10^5}{{\rm{N}} {\left/
 {\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right.
 } {{{\rm{m}}^2}}}\]
Substitute \[{10^5}{{\rm{N}} {\left/
 {\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right.
 } {{{\rm{m}}^2}}}\] for Pa in equation (1).
$
E = \dfrac{{1500{\rm{ }}{{10}^5}{{\rm{N}} {\left/
 {\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right.
 } {{{\rm{m}}^2}}}}}{{{{\left( {{{10}^5}{\rm{ m}}} \right)}^2}}}\\
 = 0.015{\rm{ N}}
$
The unit of energy is Joule, which is equal to \[{\rm{N}} \cdot {\rm{m}}\] and unit of energy per unit length is N.

Therefore the loss of energy per unit length and per unit volume is \[0.015{\rm{ J}}\]

So, the correct answer is “Option B”.

Note:
It would be an added advantage if we remember the relation between different pressure units to follow a uniform system of units during units’ substitution. Please do not confuse that the final answer is asked in Joule; we can mention it as energy per unit length and volume.