Question

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

Answer 1

Answer:

u = 477 m/s

Explanation:

$\delta H + \delta E_k + \delta E_p = \dot Q - \dot W_s$

$\delta E_p = \dot Q = \dot W_s = 0$

$\delta E_k = \delta H$

$\frac{mv^2}{2} = - \dot m (H_{out} - H_{in})$

$u^2 = 2(H_{in} - H_{out})$

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

$u^2 = 2.28\times 10^5 m62/s^2$

u = 477 m/s