cosmology - Power Density of Dark Energy (W/m³)

In cosmology empty space has an energy density $\rho_{\Lambda}$ of

$$\rho_{\Lambda}=\rho_c \cdot \Omega_{\Lambda}\cdot c^2$$

with $\Omega_{\Lambda}$ beeing the dark energy fraction ($0.683$ according to Planck 2013) and $\rho_c$ beeing todays critical density defined by

$$\rho_c=3 H_0^2/8/\pi/G$$

where $H_0=2.176\cdot 10^{-18}\, \text{s}^{-1}$ is the Hubble constant, and $G=6.674\cdot 10^{-11}\, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$ Newtons constant. This is in units of $\text{Joule}/\text{m}^3$ or $\text{Pascal}$

$$\rho_{\Lambda}=5.2\cdot 10^{-10} \, \text{kg}\, \text{m}^{-1}\, \text{s}^{-2}$$

Now the universe is expanding, and since the volume increases, so does energy. The rate at which space expands is as mentioned above $2.176\cdot10^{-18}\, \text{m}/\text{m}/\text{s}$ which means that every meter grows by $2.176\cdot10^{-18}$ meters every second.

So one cubic meter, $1\, \text{m}^3$, every second gives birth to

$$\Delta{V} = V_2-V_1=6.528\cdot 10^{-18} \, \text{m}^3$$

Where the volume $V_1$ = $r^3$ with $r=1\, \text{m}$, and $V_2=r\cdot(1+H_0\cdot \Delta{t})$ with $\Delta{t}=1\, \text{s}$

When we multiply the new born volume $\Delta{V}$ with the dark energy density $\rho_{\Lambda}$ and divide it by $\Delta{t}$, we get in units of power, $\text{kg}\,\text{m}^2\,\text{s}^{-3}$, the value of

$$3.394\cdot 10^{-27} \, \text{Watt}$$

Is my interpretation that every cubic meter generates a power of $3.394\cdot 10^{-27} \, \text{Watt}$ correct, or is there a flaw in my considerations?

Answer

is there a flaw in my considerations?

For one thing, the universe is not just empty space.

But, anyways... yeah, if you have a volume of constant energy density and you increase that volume while keeping the energy density constant then... yeah, you increase the energy. It's true.