# If the pressure between the cross-sections of a pipe is equal, will $P_1$ and $P_2$ be cancelled out of Bernoulli's Equation?

a). Water is flowing through a pipe of two circular cross- sections area $$A_1$$ and $$A_2$$ lying horizontally as shown in figure below. The pressure difference between the cross-section $$A_1$$ and $$A_2$$ is 7500 Pascals. If the velocity of the water through cross-section $$A_1$$ is $$3.25\ \mathrm{m/s}$$, what would be the velocity of the water through cross-section $$A_2$$?

Using Bernoulli's Equation, I tried rearranging the equation so that it can be used to answer a problem like the one above. $$P_1 + \rho gh_1 + 1/2 \rho v^2_1 = P_1 + \rho gh_2 + 1/2 \rho v^2_2$$ Height is constant. Therefore equation will be: $$P_1 + 1/2 \rho v^2_1 = P_2 + 1/2 \rho v^2_2$$ My question: Since the pressure is constant, will $$P_1$$ and $$P_2$$ be removed from the equation?

• Comments are not for extended discussion; this conversation has been moved to chat. Jun 13, 2016 at 17:14

$P_1$ is not equal to $P_2$. The question says that there is a "pressure difference" between $A_1$ and $A_2$. The height $h$ is assumed to be the same at both ends (the pipe is lying horizontally). The difference in area causes a difference in flow speed (continuity equation) and a difference in pressure (Bernoulli equation).
The illustration suggests that $A_2 > A_1$. Therefore $P_2 > P_1$. However, this leads to an imaginary value for $v_2$, the speed through $A_2$.
However, the illustration could be misleading. The text does not state that $A_2 > A_1$. Assuming instead that $A_1 > A_2$ and therefore $P_1 > P_2$ leads to a real (and realistic) value for $v_2$ :
$$P_1 + \frac{1}{2} \rho v^2_1 = P_2 + \frac{1}{2} \rho v^2_2$$ $$P_1 - P_2 = \frac{1}{2} \rho (v^2_2 - v_1^2)$$ $$7500Pa = \frac{1}{2} (1000kg/m^3)(v^2_2 - (3.25m/s)^2)$$ which gives $$v=5.06m/s$$.