Bernoulli's Equation — Question 1

Question 1

A Venturi pressure difference determines a flow rate

Water flows steadily through a horizontal Venturi with areas $A_1=6.0\,\mathrm{cm^2}$ and $A_2=2.0\,\mathrm{cm^2}$. The upstream static pressure exceeds throat pressure by $4.0\,\mathrm{kPa}$. Assume incompressible, inviscid flow with uniform section speeds and apply Bernoulli along a connecting streamline. Use $\rho=1000\,\mathrm{kg/m^3}$.

1. Find both speeds and the volume flow rate.

2. If upstream absolute pressure is $105\,\mathrm{kPa}$, find throat absolute pressure. Explain why the pressure drop is not itself an energy loss in this model.

Question 1 - Solution

A Venturi pressure difference determines a flow rate

(a) Derivation and calculation

$$A_1v_1=A_2v_2\Rightarrow v_2=3v_1.$$$$p_1-p_2=\tfrac12\rho(v_2^2-v_1^2)=\tfrac12(1000)(9-1)v_1^2.$$$$4000=4000v_1^2\Rightarrow\boxed{v_1=1.00,\quad v_2=3.00\,\mathrm{m/s}}.$$$$\boxed{Q=A_1v_1=(6.0\times10^{-4})(1.00)=6.00\times10^{-4}\,\mathrm{m^3/s}}.$$

(b) Result and interpretation

$$\boxed{p_{2,\mathrm{abs}}=105-4.0=101\,\mathrm{kPa}}.$$$$\Delta p=-4000\,\mathrm{J/m^3},\quad\Delta(\tfrac12\rho v^2)=+4000\,\mathrm{J/m^3}.$$$$p+\tfrac12\rho v^2=\text{constant}.$$Static pressure is converted into kinetic energy; it is not dissipated in the stated ideal model.

Physical check

A lossless downstream expansion can recover static pressure as the speed decreases.