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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.

Respuesta :

Answer:

The speed of the steam is 477.4 m/s.

Explanation:

Given that,

Temperature

[tex]T_{1}= 280^{\circ}C[/tex]

[tex]T_{2}=220^{\circ}C[/tex]

Pressure

[tex]P_{1}= 6.00\ bar[/tex]

[tex]P_{2}=3.00\ bar[/tex]

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

[tex]\Delta H+\Delta E_{k}+\Delta E_{p}=Q-W_{s}[/tex]

[tex]\Delta E_{k}=-\Delta H[/tex]

Here, [tex]\Delta E_{p}=Q=W_{s}=0[/tex]

[tex]\dfrac{1}{2}mv^2=m(H_{out}-H_{in})[/tex]

[tex]v^2=2(H_{out}-H_{in})[/tex]

Put the value into the formula

[tex]v^2=2\times(3020-2906)\times10^{3}[/tex]

[tex]v=\sqrt{2\times(3020-2906)\times10^{3}}[/tex]

[tex]v=477.4\ m/s[/tex]

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

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