= U-tube differential manometer
{c}
{title2=$\Delta p=p_L-p_R=\rho g(z_R-z_L)$}
A <U-tube differential manometer> contains a connected column of <liquid> with one <gas> connection above each <liquid meniscus>. Write $p_L,p_R$ for the left and right <pressures>, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant <mass density> $\rho$ and downward <gravitational acceleration> of magnitude $g$; neglect the <gas> column's <mass density> and differences in <capillary pressure>. The signed vertical <length> difference $\Delta h=z_R-z_L$ then gives
$$
\Delta p=p_L-p_R=\rho g\Delta h.
$$
Larger left <pressure> means a lower left <liquid meniscus>: $p_L>p_R$ implies $z_L<z_R$. Equal <pressures> give equal meniscus heights.
\Image[/manometer-u-tube-signed-height.png]
{title=A <U-tube differential manometer>: higher left <pressure> lowers the left <liquid meniscus>}
{description=Original schematic of two same-height <pressure tappings> connected to a resting <water> column. The constructed example uses $z_L=-30\,\mathrm{mm}$ and $z_R=30\,\mathrm{mm}$, so the vertical difference is $60\,\mathrm{mm}$. Dashed lines mark both meniscus heights and the equal-pressure datum; the dotted bottom line identifies two equal-height <points> within the connected <liquid>. The arrow in the pipe indicates <fluid flow>; the <manometer> <liquid> has no continuing flow at equilibrium.}
The balance follows directly from <hydrostatic pressure>. At a common height $z_0$ below both surfaces, the two <points> in the connected homogeneous <liquid> have equal <pressure>:
$$
p_L+\rho g(z_L-z_0)=p_R+\rho g(z_R-z_0).
$$
Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from <hydrostatic equilibrium>, $\nabla p=-\rho g\mathbf e_z$: a horizontal displacement changes no <pressure>, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected <liquid>. Equal height alone is insufficient for unrelated vessels or for <fluid flow> with significant <acceleration>.
For an original constructed example, choose <water> of stipulated <mass density> $\rho=1000\,\mathrm{kg\,m^{-3}}$, <gravitational acceleration> $g=9.81\,\mathrm{m\,s^{-2}}$, and vertical <length> difference $\Delta h=0.060\,\mathrm m$. Then
$$
\Delta p=1000\times9.81\times0.060=588.6\,\mathrm{Pa}.
$$
The <SI> <pressure> unit is the <pascal (unit)>: the product has units $\mathrm{kg\,m^{-3}}\times\mathrm{m\,s^{-2}}\times\mathrm m=\mathrm{kg\,m^{-1}\,s^{-2}}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\,\mathrm{Pa}$ and reverses which <liquid meniscus> is lower. In equal-bore arms, <volume conservation> makes each surface move $30\,\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred <pressure>. With unequal arm <areas> $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_La=A_Rb$, while the vertical <length> difference is still $a+b$.
If the right port is open to the local atmosphere, $p_L-p_R$ is the left <gauge pressure>. Only with that reference and known <atmospheric pressure> can one obtain <absolute pressure> from
$$
p_{L,\mathrm{abs}}=p_{\mathrm{atmosphere}}+p_{L,\mathrm{gauge}}.
$$
For an explicitly stipulated <atmospheric pressure> of $100000\,\mathrm{Pa}$, the example gives $100588.6\,\mathrm{Pa}$ <absolute pressure>. If the right port is connected to another vessel or a <pressure tapping>, the same reading is a differential <pressure> and supplies no <absolute pressure> without knowing that reference.<University Physics Volume 1 by William Moebs and others>[§14.2, Gauge Pressure vs. Absolute Pressure, equation 14.11; Manometers]{ref}
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