# A non-viscous liquid of constant density: Bernoulli's Equation

**Problem:**
A non-viscous liquid of constant density 1000 kg/m^{3} flows in streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points P and Q at heights of 2 m and 5 m are respectively 4×10^{-3} m^{2} and 8×10^{-3} m^{2}. The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q.
(IIT JEE 1997)

**Answer:** The work done per unit volume by the pressure force is 29025 J/m^{3} and that by the gravity force is -29400 J/m^{}.

**Solution:**
Let $A_1=4\times{10}^{-3}\,\mathrm{m^2}$, $v_1=1$ m/s, $h_1=2$ m, and $p_1$ be the pressure at the end P. Corresponding parameters at the end Q are $A_2=8\times{10}{-3}\,\mathrm{m^2}$, $v_2$, $h_2=5$ m, and $p_2$. The continuity equation between P and Q gives
\begin{align}
v_2=A_1 v_1/A_2=0.5\,\mathrm{m/s}. \nonumber
\end{align}
Bernoulli's equation between P and Q
\begin{align}
&p_1+\tfrac{1}{2}\rho v_1^2+\rho g h_1=p_2+\tfrac{1}{2}\rho v_2^2+\rho g h_2,
\end{align}
gives
\begin{align}
&p_1-p_2=\rho g(h_2-h_1)+\tfrac{1}{2}\rho (v_2^2-v_1^2).\nonumber
\end{align}
The work done per unit volume by the pressure as the fluid flows from the end P to the end Q is
\begin{align}
W_p&=p_1-p_2=\rho g(h_2-h_1)+\tfrac{1}{2}\rho (v_2^2-v_1^2)\nonumber\\
&=(1000)(9.8)(5-2)+\tfrac{1}{2}(1000)(0.5^2-1^2)\nonumber\\
&=29025\,\mathrm{J/m^3}.\nonumber
\end{align}
The work done per unit volume by the gravity is
\begin{align}
W_h=\rho g(h_1-h_2)=-29400\,\mathrm{J/m^3}.\nonumber
\end{align}

## Problem from IIT JEE 1994

**Problem:**
A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where cross-sectional area is 10 cm^{2}, the water velocity is 1 m/s and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm^{2}, is________Pa. (Density of water 1000 kg/m^{3}.)

**Solution:**
The continuity equation, $A_1v_1=A_2v_2$, gives
\begin{align}
v_2&={A_1 v_1}/{A_2}={(10)(1)}/{5}\\
&=2\,\mathrm{m/s}.\nonumber
\end{align}
Bernoulli's equation,
\begin{align}
p_1+\tfrac{1}{2}\rho v_1^2+\rho gh_1=p_2+\tfrac{1}{2}\rho v_2^2+\rho gh_2,\nonumber
\end{align}
with $h_1=h_2$ gives
\begin{align}
p_2&=p_1+\tfrac{1}{2}\rho(v_1^2-v_2^2)\nonumber\\
&=2000+\tfrac{1}{2}(1000) (1^2-2^2)\\
&=500\,\mathrm{Pa}.\nonumber
\end{align}

## More on Bernoulli's Equation

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