Bernoulli's Equation — Question 2

Question 2

Climbing and narrowing both draw on pressure energy

A steady ideal water flow rises $2.0\,\mathrm{m}$ from section 1 to section 2 while area decreases from $A_1=8.0\,\mathrm{cm^2}$ to $A_2=4.0\,\mathrm{cm^2}$. At section 1, speed is $2.0\,\mathrm{m/s}$ and gauge pressure is $80\,\mathrm{kPa}$. Both gauge references are the same. Use $\rho=1000\,\mathrm{kg/m^3}$ and $g=10\,\mathrm{m/s^2}$; neglect losses.

1. Find speed and gauge pressure at section 2.

2. If the same area change instead ended $2.0\,\mathrm{m}$ lower, find the new pressure. Does greater speed always mean lower pressure between arbitrary points?

Question 2 - Solution

Climbing and narrowing both draw on pressure energy

(a) Derivation and calculation

$$\boxed{v_2=(A_1/A_2)v_1=(8/4)(2.0)=4.00\,\mathrm{m/s}}.$$$$p_2=p_1+\tfrac12\rho(v_1^2-v_2^2)-\rho g(z_2-z_1).$$$$\boxed{p_{2,g}=80000+\tfrac12(1000)(4-16)-1000(10)(2)=54000\,\mathrm{Pa}}.$$

(b) Result and interpretation

For the same area change with a 2 m descent,$$\boxed{p_{2,g}=80000-6000+20000=94000\,\mathrm{Pa}}.$$The pressure rises even while speed increases because gravitational energy supplies both increases. The usual inverse pressure-speed comparison requires equal elevations and the same Bernoulli constant.

Physical check

The 4 m difference between the two outlet elevations changes pressure by $\rho g(4)=40\,\mathrm{kPa}$.