A case of conversion - From $lb_m$ to $gpm$

 Imperial units prove to be somtimes hard to use. Here the case of conversion from mass to volumetric flowrate is presented.

As one may imagine a physical property of the fluid is required: the density $\rho$. This can be written as,

$\dot{Q}= Q\, \rho$        Eq. (01)

where $\dot{Q}$ is the mass flow rate, in $lb_m/hr$ for example, and $Q$ is the volumetric flow rate, in $gpm$, for example.

Let us now consider the case of $40,000.00\, lb_m/hr$ flowing liquid water at $190\,^\circ \, C$. In this case, its density would be $\rho=54.70\,lb_m/ft^3$.

Using Eq. (01), it follows,

$Q=\dfrac{\dot{Q}}{\rho}=\dfrac{40,000.00\, lb_m/hr}{54.70\, lb_m/ft^3}$

$Q=924.56\,ft^3/hr=115.27\,gpm$

which is the desired conversion. If you were working with steam, working pressure must be considered to get the proper fluid density.