Question

Saturated water with a quality of 0.55 and a temperature of 120 oC enters an uninsulated diffuser at a velocity of 180 m/s. The mass flow rate is 1.5 kg/s. If the water is to exit as saturated water vapor at a temperature of 120 oC, with negligible velocity, determine the necessary heat transfer rate.

Answer 1

Answer:

$Q_{in}=146.22kW$

Explanation:

Hello,

In this case, the energy balance for the given uninsulated diffuser is:

$m_{in}h_{in}+m_{in}\frac{v_{in}^2}{2}+Q_{in} =m_{out}h_{out}+m_{out}\frac{v_{out}^2}{2}$

Thus, we shall remember that the inlet mass equals the outlet mass and the outlet velocity is negligible, for that reason we obtain:

$mh_{in}+m\frac{v_{in}^2}{2}+Q_{in} =mh_{out}$

Hence, solving for the heat, we need the enthalpy at the inlet, which at 120°C for a liquid-vapor mixture results:

$h_{in}=503.81\frac{kJ}{kg} +0.55*2202.1\frac{kJ}{kg}=1714.965\frac{kJ}{kg}$

Moreover, at 120 °C the outlet enthlapy as saturated steam is:

$2706.0\frac{kJ}{kg}$

(Values extracted from Cengel, Thermodynamics 7th edition). In such a way, the required heat inlet is:

$Q_{in}=m(h_{out}-h_{in}-\frac{v_{in}^2}{2})=0.15\frac{kg}{s}(2706.0\frac{kJ}{kg}-1714.965\frac{kJ}{kg}-\frac{(180m/s)^2}{2*1000})$

$Q_{in}=146.22kW$

Notice that the term having the velocity should be divided by 1000 to obtain  it in kJ/kg.

Best regards.