In: Physics

# What does the top pressure gauge in the figure read?

What does the top pressure gauge in the figure read?

## Solutions

##### Expert Solution

Concepts and reason

The concept required to solve this problem is Bernoulli's equation. Initially, convert the pressure into Pascal. Then, write Bernoulli's equation. Finally, substitute the values of the pressure, density, velocities, and the height given in the question into Bernoulli's equation to calculate the pressure at the top.

Fundamentals

The expression of the Bernoulli's equation is, $$P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2}$$

Here, $$P_{1}$$ is the pressure at the bottom position, $$\rho$$ is the density, $$v_{1}$$ is the speed of the liquid at the bottom, $$\mathrm{g}$$ is the acceleration due to gravity, $$h_{1}$$ is the height at the bottom, $$P_{2}$$ is the pressure at the top position, $$v_{2}$$ is the speed of the liquid at the top, and $$h_{2}$$ is the height at the top.

The pressure is,

\begin{aligned} P_{1} &=200 \mathrm{kPa}\left(\frac{10^{3} \mathrm{~Pa}}{1 \mathrm{kPa}}\right) \\ &=200 \times 10^{3} \mathrm{~Pa} \end{aligned}

The height of the tube at the bottom position is, $$h_{1}=0 \mathrm{~m}$$

The pressure at the bottom position can be written in units of Pa from kPa using the unit conversion.

The expression of the Bernoulli's equation is, $$P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2}$$

Substitute $$0 \mathrm{~m}$$ for $$h_{1}$$ and rearrange the expression for $$P_{2}$$.

$$\begin{array}{c} P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g(0)=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2} \\ P_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2} \\ P_{2}=P_{1}+\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)-\rho g h_{2} \end{array}$$

Substitute $$200 \times 10^{3} \mathrm{~Pa}$$ for $$P_{1}, 900 \mathrm{~kg} / \mathrm{m}^{3}$$ for $$\rho, 2 \mathrm{~m} / \mathrm{s}$$ for $$v_{1}, 9.8 \mathrm{~m} / \mathrm{s}^{2}$$ for $$\mathrm{g}, 3.0 \mathrm{~m} / \mathrm{s}$$ for $$v_{2}, 10 \mathrm{~m}$$ for $$h_{2}$$

$$\begin{array}{c} P_{2}=200 \times 10^{3} \mathrm{~Pa}+\frac{1}{2}\left(900 \mathrm{~kg} / \mathrm{m}^{3}\right)\left((2 \mathrm{~m} / \mathrm{s})^{2}-(3 \mathrm{~m} / \mathrm{s})^{2}\right)-\left(900 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(10 \mathrm{~m}) \\ =200 \times 10^{3} \mathrm{~Pa}-2250 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}^{2}-88200 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}^{2} \\ =1.10 \times 10^{5} \mathrm{~Pa} \\ P_{2}=110 \mathrm{kPa} \end{array}$$

The pressure at the top position is $$110 \mathrm{kPa}$$.

Bernoulli's equation is based on the concept of conservation of energy. The expression of Bernoulli's equation in terms of the energy per unit volume is:

$$P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2}$$

The pressure at the top position is 110kPa.