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Steam at 260°C and 7.00 bar absolute is expanded through a nozzle to 200°C and 4.00 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 2974 kJ/kg at 260°C and 7 bar and 2860 kJ/kg at 200°C and 4 bar. Use the open-system energy balance to calculate the exit steam velocity

Respuesta :

Answer:

u = 477 m/s

Explanation:

[tex]\delta H + \delta E_k + \delta E_p = \dot Q - \dot W_s[/tex]

[tex]\delta E_p = \dot Q = \dot W_s = 0[/tex]

[tex]\delta E_k = \delta H [/tex]

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

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

      [tex]= 2 (2974 - 2860) kj/kg \times \frac{10^{3} N m}{1 kg} \times\frac{ 1kg m/s2}[/tex]{1 N}

[tex]u^2 = 2.28\times 10^5 m62/s^2[/tex]

u = 477 m/s