Question

Air flows through this tube at a rate of 1200cm3/s.
Assume that air is an ideal fluid. What is the height h
of mercury in the right side of the U-tube? Answer in
cm
1.28 kg m'. Pr = 13.6x10' kg m​

Answer 1

what the heck ...........

serious

Answer 2

Answer:

The height h of mercury in the right side of the U-tube is 4.37 cm.

Explanation:

According to Bernoulli's Principle, the sum of kinetic energy, potential energy, and pressure energy per unit mass stays constant throughout the flow of an ideal, incompressible fluid. Investigate the Bernoulli Principle, which states that the speed of a fluid (in this case, air) dictates the amount of pressure it can generate.

The Bernoulli's principle is equivalent to the principle of conservation of energy, which states that the total amount of energy in an isolated system remains constant.

Given

The volume flow rate $Q=1200 \mathrm{~cm}^3 / \mathrm{s}$

Diameter at the broader end $D_1=2.0 \mathrm{~cm}$

Diameter at the narower end $D_2=4.0 \mathrm{~mm}=0.4 \mathrm{~cm}$

Step 1: Using Bernoulli's theorem:

$\begin{aligned}& \frac{1}{2} \rho v_1^2+P_1=\frac{1}{2} \rho v_2^2+P_2 \\& P_1-P_2=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2\end{aligned}$

Step 2: Using equation of continuity:

$& Q=\pi\left(\frac{D_1^2}{2}\right)^2 v_1=\pi\left(\frac{D_2^2}{2}\right)^2 v_2=1200 \mathrm{~cm}^3 / \mathrm{s}$

$& v_2=\frac{4 \times 1200 \mathrm{~cm}^3 / \mathrm{s}}{\pi \times(0.4 \mathrm{~cm})^2}=9550 \mathrm{~cm} / \mathrm{s}=95.5 \mathrm{~m} / \mathrm{s}$

$& v_1=\frac{4 \times 1200 \mathrm{~cm}^3 / \mathrm{s}}{\pi \times(2 \mathrm{~cm})^2}=382 \mathrm{~cm} / \mathrm{s}=3.82 \mathrm{~m} / \mathrm{s}$

$& P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)$

$& P_1-P_2=\frac{1}{2} \times 1.28 \mathrm{~kg} / \mathrm{m}^3 \times\left((95.5 \mathrm{~m} / \mathrm{s})^2-(3.82 \mathrm{~m} / \mathrm{s})^2\right)$

$& P_1-P_2=5827.6 \mathrm{~N} / \mathrm{m}^2$

Step 3: Now:

$\begin{aligned}& \rho_m g h=5827.6 \mathrm{~N} / \mathrm{m}^2 \\& h=\frac{5827.6 \mathrm{~N} / \mathrm{m}^2}{\rho_m g} \\& h=\frac{5827.6 \mathrm{~N} / \mathrm{m}^2}{13600 \mathrm{~kg} / \mathrm{m}^3 \times 9.8 \mathrm{~m} / \mathrm{s}^2}=0.0437 \mathrm{~m} \\& h=4.37 \mathrm{~cm}\end{aligned}$

Hence the height h of mercury in the right side of the U-tube is 4.37 cm.

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