We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.
For 1bar=1000kPa:
[tex]T_{sat}=179.88\°c[/tex]
[tex]H_{fg} = 2014.6kJ/kg[/tex]
[tex]c_p=4.18 kJkg^{-1}{K^{-1}[/tex]
[tex]\nu_g = 0.19436m^3/kg[/tex]
Replacing,
[tex]\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})[/tex]
[tex]\Delta h = 2014.6+4.18(179.88-24)[/tex]
[tex]\Delta h=2666.17kJ/kg[/tex]
With the specific volume we know can calculate the mass flow, that is
[tex]\dot{m}=\frac{\frac{15000}{3600}}{0.19436}[/tex]
[tex]\dot{m} = 21.4378kg/s[/tex]
Then the heat required in input is,
[tex]Q=\dot{m}\Delta h[/tex]
[tex]Q=21.4378*2666.17[/tex]
[tex]Q=57157.036kW[/tex]
With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,
[tex]V= \frac{\dotV}{A}[/tex]
[tex]V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}[/tex]
[tex]V=235.79m/s[/tex]
Finally we can apply the steady flow energy equation, that is
[tex]\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2[/tex]
Re-arrange for Q,
[tex]Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})[/tex]
[tex]Q=\dot{m}(\Delta h-\frac{V^2}{2000})[/tex]
[tex]Q= (21.4378)(2666.17-\frac{235.79^2}{2000})[/tex]
[tex]Q= 56560.88kW[/tex]
We can note that consider the Kinetic Energy will decrease the heat input.
The enthalpy change and heat input are equal to 2666.18 kJ/kg and 56,542.41 kW respectively.
Given the following data:
From the steam table, the following parameters are associated with saturated water at a pressure of 1000 kPa (10 bar):
[tex]V_g =0.19436 \;m^3/kg[/tex]
[tex]H_{fg}=2014.6\;kJ/kg\\\\T_{sat}=452.88\;K\\\\C_p=4.18\;kJ/kgK\\\\\rho = 5.144 \;kg/m^3[/tex]
Mathematically, enthalpy change is calculated by using this formula:
[tex]\Delta H=H_{fg}+C_p(T_{sat}-T_{in})\\\\\Delta H=2014.6+4.18(452.88-297)\\\\\Delta H=2014.6+651.58\\\\\Delta H=2666.18\;kJ/kg[/tex]
First of all, we would determine the mass flow rate by using this formula:
[tex]M=Av = \rho V\\\\M=5.144 \times \frac{15000}{3600} \\\\M=5.144 \times 4.167[/tex]
M = 21.43 kg/s.
Next, we would determine velocity of water by using this formula:
[tex]v=\frac{V}{A} =\frac{V}{\frac{\pi r^2}{4} } \\\\v=\frac{4.167}{\frac{3.142 \times 0.15^2}{4} } \\\\v=\frac{4.167}{0.0177}[/tex]
v = 235.42 m/s.
Mathematically, the heat input for an adiabatic process is given by this formula:
[tex]Q=Mh_2-M(h_1+\frac{v^2}{2000} )\\\\Q=M(h_2-h_1-\frac{v^2}{2000})\\\\Q=M(\Delta H-\frac{v^2}{2000})\\\\Q=21.43(2666.18-\frac{235.42^2}{2000})\\\\Q=21.43(2666.18-27.71)\\\\Q=21.43 \times 2638.47[/tex]
Q = 56,542.41 kW.
Read more on adiabatic process here: https://brainly.com/question/3962272
