## 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

The 'Blue Check Mark' means that this solution was answered by an expert.

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:

$\begin{gathered}\frac{p_A}{\gamma_B}-1.626+4.29+5.00=\frac{207 \mathrm{kN} / \mathrm{m}^2}{12.4 \mathrm{kN} / \mathrm{m}^3}=16.69 \mathrm{~m} \\\frac{p_A}{\gamma_B}=9.03 \mathrm{~m}, \quad p_A=9.03(12.4)=112.0 \mathrm{kN} / \mathrm{m}^2=112.0 \mathrm{kPa} \quad \text { ANS }\end{gathered}$

$\frac{p_A}{\gamma_B}-2.4 \frac{8.4}{12.4}+0.4 \frac{13.56(9.81)}{12.4}+5.0=\frac{p_B}{\gamma_B}$

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