Under ideal conditions, for a drop of 80 kJ/kg enthalpy, what will be the approximate velocity of steam at the outlet of the nozzle if the inlet velocity of the steam is 2 m/s? 

Concept:

To determine the velocity of steam at the outlet of the nozzle under ideal conditions, we use the principles of energy conservation. Specifically, we apply the Steady Flow Energy Equation (SFEE) for a nozzle, which simplifies to the conservation of enthalpy and kinetic energy.

Calculation:

Given:

Enthalpy drop, \( h_1 - h_2 = 80 \, \text{kJ/kg} \)

Inlet velocity of the steam, \( V_1 = 2 \, \text{m/s} \)

Calculation:

Convert the enthalpy drop into compatible units: \( 1 \, \text{kJ/kg} = 1000 \, \text{m}^2/\text{s}^2 \)

\( h_1 - h_2 = 80 \, \text{kJ/kg} = 80000 \, \text{m}^2/\text{s}^2 \)

Using the Steady Flow Energy Equation (SFEE):

\( h_1 - h_2 = \frac{V_2^2}{2} - \frac{V_1^2}{2} \)

Substitute the given values:

\( 80000 = \frac{V_2^2}{2} - \frac{2^2}{2} \)

\( 80000 = \frac{V_2^2}{2} - 2 \)

\( 80000 + 2 = \frac{V_2^2}{2} \)

\( 80002 = \frac{V_2^2}{2} \)

\( V_2^2 = 2 \times 80002 \)

\( V_2^2 = 160004 \)

\( V_2 = \sqrt{160004} \)

\( V_2 \approx 400 \, \text{m/s} \)

The approximate velocity of steam at the outlet of the nozzle is:

\( 400 \, \text{m/s} \)