Different Values for Different Calculations of Specific Weight of Water in lbf/ft^3

Question

I have been trying to wrap my head around this, but I really can't figure it out. I don't understand why the value for the specific weight of water differs when I directly convert it from $\frac{N}{m^3}$ to $\frac{lb_f}{ft^3}$, and when I compute it using the formula for specific weight $\gamma=\rho g$

Converting it from $\frac{N}{m^3}$:

$$9810\frac{N}{m^3}\times(\frac{1m}{3.28ft})^3\times\frac{0.22lb_f}{1N}=62.4\frac{lb_f}{ft^3}$$

Using the formula $\gamma=\rho g$

$$\gamma=\rho g=(62.4\frac{lb}{ft^3})(32.2\frac{ft}{s})$$ $$\gamma=2009.28\frac{lb_f}{ft^3}$$

What is the cause of this? Is there something I'm missing? Something I'm misunderstanding?

Answer 1

The simple mistake is the conversion from the matrix units to the imperial units has produced the weight density of water, $\gamma_w = \dfrac{62.4 lb_f}{ft^3}$. Note that $lb_f$ has already included the gravitational constant $g$, thus, unless you convert it back to the mass density $\rho = \dfrac{\gamma}{g}$, there is no need to apply the gravitational constant again.

Answer 2

From: Water - Density, Specific Weight and Thermal Expansion Coefficients

In the Imperial system the mass unit is the slug [sl], and is derived from the pound-force by defining it as the mass that will accelerate at $1 ft/s^2$ when a $1 lb_f$ acts upon it.

The density of water is $1.940 sl/ft^3$ at 39 °F (4 °C), and the specific weight in Imperial units is:

\begin{align} γ &= \rho\ g \\ &= 1.940 sl/ft^3 \times 32.174 ft/s^2 \\ &= 1.940 lb_f/(ft/s^2\times ft^3) \times 32.174 ft/s^2 \\ &= 62.4 lb_f/ft^3\end{align}