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

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.

{\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.