Solving Energy Balance Problem for Steam Flow through Nozzle

[Xyius]

I am doing a problem about steam flowing through a nozzle. Here are the initial and final conditions..

P1=30 bar = 3000kPa
T1=320 C
v= 100 m/s
(v= velocity!)

P2=10 bar = 1000kPa
T2= 200 C

Mass flow rate = 2 kg/s (Steady flow process)

To me it seems like a cut and dry energy balance problem. There is a problem however.

Attempt at a Solution
The mass balance yeilds..
$$\dot{m_1}=\dot{m_2}$$

So putting this result in the energy balance.. (PE=0, W=0, Q=0)
$$\dot{E_{in}}=\dot{E_{out}}$$
$$h_1+\frac{1}{2}v_{1}^2=h_2+\frac{1}{2}v_{2}^2$$

This is where the problem comes in. The enthalpies for these two states (from the tables) are..

h1=3043.5 kJ/kg
h2=2828.3 kJ/kg

I am getting a velocity LESS than the initial. This doesn't make sense considering it is a nozzle and should be speeding up. And yes I converted the KE into KJ.

 

[NateD]

Any help would be greatly appreciated.It is possible that the velocity you are calculating is not the exit velocity, but the velocity at the throat of the nozzle. The throat of the nozzle is the point of minimum area and typically the point at which the fluid velocity is highest. If this is the case, then the velocity you are calculating would be less than the exit velocity.To calculate the exit velocity, you need to use the continuity equation in conjunction with the energy balance equation. The continuity equation states that the mass flow rate is constant, i.e. \dot{m} = \dot{m_1} = \dot{m_2}. This means that the volumetric flow rate (Q) is also constant, i.e. Q = \frac{\dot{m}}{\rho_1} = \frac{\dot{m}}{\rho_2}.You can then use this to calculate the exit velocity, v_2, from the energy balance equation.h_1+\frac{1}{2}v_{1}^2=h_2+\frac{1}{2}v_{2}^2v_2 = \sqrt{2\left(h_1 - h_2\right) + v_1^2}This should give you the correct exit velocity.