Chapter 3
Q. 3.SP.4
In Fig. S3.4 liquid A weighs 53.5 \mathrm{lb} / \mathrm{ft}^3\left(8.4 \mathrm{kN} / \mathrm{m}^3\right) and liquid B weighs 78.8 \mathrm{lb} / \mathrm{ft}^3\left(12.4 \mathrm{kN} / \mathrm{m}^3\right) . Manometer liquid M is mercury. If the pressure at B is 30 \mathrm{psi}(207 \mathrm{kPa} ), find the pressure at A . Express all pressure heads in terms of the liquid in bulb B .
Step-by-Step
Verified Solution
Proceeding from A to B :
\frac{p_A}{\gamma_B}-\left(z_a-z_c\right) \frac{\gamma_A}{\gamma_B}+\left(z_a-z_b\right) \frac{\gamma_M}{\gamma_B}+\left(z_b-z_d\right) \frac{\gamma_B}{\gamma_B}=\frac{p_B}{\gamma_B}
BG units:
\begin{gathered}\frac{p_A}{\gamma_B}-8.0 \frac{53.5}{78.8}+1.3 \frac{13.56(62.4)}{78.8}+16.7=\frac{p_B}{\gamma_B} \\\frac{p_A}{\gamma_B}-5.43+13.96+16.7=\frac{30(144)}{78.8}=54.8 \mathrm{ft} \\\frac{p_A}{\gamma_B}=29.6 \mathrm{ft} \quad p_A=29.6 \frac{78.8}{144}=16.19 \mathrm{psi} \end{gathered}
SI units:
\begin{gathered}\frac{p_A}{\gamma_B}-1.626+4.29+5.00=\frac{207 \mathrm{kN} / \mathrm{m}^2}{12.4 \mathrm{kN} / \mathrm{m}^3}=16.69 \mathrm{~m} \\\frac{p_A}{\gamma_B}=9.03 \mathrm{~m}, \quad p_A=9.03(12.4)=112.0 \mathrm{kN} / \mathrm{m}^2=112.0 \mathrm{kPa} \quad \text { ANS }\end{gathered}
\frac{p_A}{\gamma_B}-2.4 \frac{8.4}{12.4}+0.4 \frac{13.56(9.81)}{12.4}+5.0=\frac{p_B}{\gamma_B}