Question

Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.

Answer 1

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

$T_{sat}=179.88\°c$

$H_{fg} = 2014.6kJ/kg$

$c_p=4.18 kJkg^{-1}{K^{-1}$

$\nu_g = 0.19436m^3/kg$

Replacing,

$\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})$

$\Delta h = 2014.6+4.18(179.88-24)$

$\Delta h=2666.17kJ/kg$

With the specific volume we know can calculate the mass flow, that is

$\dot{m}=\frac{\frac{15000}{3600}}{0.19436}$

$\dot{m} = 21.4378kg/s$

Then the heat required in input is,

$Q=\dot{m}\Delta h$

$Q=21.4378*2666.17$

$Q=57157.036kW$

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

$V= \frac{\dotV}{A}$

$V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}$

$V=235.79m/s$

Finally we can apply the steady flow energy equation, that is

$\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2$

Re-arrange for Q,

$Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})$

$Q=\dot{m}(\Delta h-\frac{V^2}{2000})$

$Q= (21.4378)(2666.17-\frac{235.79^2}{2000})$

$Q= 56560.88kW$

We can note that consider the Kinetic Energy will decrease the heat input.