Question

Steam at 280°C and 6.00 bar absolute is expanded through a nozzle to 220°C and 3.00 bar. Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. The specific enthalpy of steam is 3020 kJ/kg at 280°C and 6.00 bar and 2906 kJ/kg at 220°C and 3.00 bar. Use the open system energy balance to calculate the exit speed of the steam.

Answer 1

The speed of the steam is 477.4 m/s.

Step-by-step explanation:

Given that,

Temperature

`T_(1)= 280^(\circ)C`

`T_(2)=220^(\circ)C`

Pressure

`P_(1)= 6.00\ bar`

`P_(2)=3.00\ bar`

Specific enthalpy of steam at 280°C = 3020 kJ/kg

Specific enthalpy of steam at 220°C = 2906 kJ/kg

We need to calculate the speed of the steam

Using balance equation

`\Delta H+\Delta E_(k)+\Delta E_(p)=Q-W_(s)`

`\Delta E_(k)=-\Delta H`

Here,
`\Delta E_(p)=Q=W_(s)=0`

`(1)/(2)mv^2=m(H_(out)-H_(in))`

`v^2=2(H_(out)-H_(in))`

Put the value into the formula

`v^2=2*(3020-2906)*10^(3)`

`v=\sqrt{2*(3020-2906)*10^(3)}`

`v=477.4\ m/s`

Hence,The speed of the steam is 477.4 m/s.