The rate at which heat is generated inside a chromatographycolumn from friction of flowing liquid is power (watts, $\mathbf{W}\mathbf{=}\mathbf{J}\mathbf{/}\mathbf{s}$)= volume flow raterole="math" localid="1656474760665" $(m^3/s)\times$pressure drop (pascals, role="math" localid="1656474828044" $\mathbf{Pa}\mathbf{=}\mathbf{kg}\mathbf{/}\mathbf{\left[}\mathbf{m}\boxed{\mathbf{?}}{\mathbf{s}}^{\mathbf{2}}\mathbf{\right]}$).

(a) Explain the analogy between heat generated in a chromatography column and heat generated in an electric circuit (power = current$\mathbf{\times }$voltage).

(b) At what rate (watts = J/s) is heat generated for a flow of 1 mL/minwith a pressure difference of3500 bar between the inlet and outlet?

You will need to convert mL/min to${\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{s}$.Also 1 bar$\mathbf{=}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{Pa}$.

Part (a)

The heat generated in an electric circuit is equal to the product of current and voltage where the current is the rate of flow of charge and the voltage is the potential difference driving charge.

The heat generated in a chromatography column is equal to the product of volume flow rate and pressure drop where the volume flow rate is the rate of flow of liquid and the pressure is pressure difference driving liquid.

Part (b)

The power can be calculated using the formula:

power = volume flow rate$\boxed{?}$pressure drop

First, we need to convert 1mL/min to${\mathrm{m}}^3/\mathrm{s}$ :

$1\mathrm{mL}/\min .\frac{{10}^{-6}{\mathrm{m}}^3/\mathrm{mL}}{60\mathrm{s}/\min }=1.67·{10}^{-8}{\mathrm{m}}^3/\mathrm{s}$

Then we need to convert 3500 bars to Pa:

$3500\mathrm{bar}·{10}^5\mathrm{Pa}/\mathrm{bar}=3.5·{10}^8\mathrm{Pa}$

Now we put values in formula

power$=1.67·{10}^{-8}{\mathrm{m}}^3/\mathrm{s}·3.5·{10}^8\mathrm{Pa}$

powerlocalid="1656475464467" $=5.845\mathrm{W}$

Here, the final result is

(a) See the answer is below step 1

(b) power$=5.845\mathrm{W}$