Respuesta :
Answer:
a. [tex]306.78^oC[/tex]
b. [tex]Q_{heater}=1666.8kW[/tex]
c. See explanation.
d. Its result is not significant in comparison with the involved enthalpies and the produced work (see explanation).
Explanation:
Hello,
a. In this case, as superheated steam enters to the turbine, the energy balance for the given mass rate is:
[tex]h_{in}m=W_{out}+h_{out}m[/tex]
At 40 bar and 500 °C, the corresponding enthalpy is 3446.0kJ/kg, therefore, the enthalpy at the outlet is:
[tex]h_{out}=\frac{h_{in}m-W_{out}}{m} =\frac{3446.0kJ/kg*250\frac{kg}{min}*\frac{1min}{60s}-1500\frac{kJ}{s} }{250\frac{kg}{min}*\frac{1min}{60s}} \\h_{out}=3086kJ/kg[/tex]
Now, at 5 bar and 3086 kJ/kg, the temperature corresponds to the temperature of a supersaturated vapor as well, thus interpolating from Cengel's table A-6, the outlet temperature is 306.78 °C.
b. Afterwards, the heat required at the heater involves just sensible heat from 306.78 °C to 500 °C whose difference implies that polynomial heat capacity must be used, therefore, the heat is computed via:
[tex]Q_{heater}=m\int\limits^{500^oC}_{306.78^oC} {A+BT+CT^2+DT^3} \, dT[/tex]
The values for water given in Cengel are in kJ/kmol, therefore, a conversion is done by using its molar mass, thus:
[tex]Q_{heater}=250\frac{kg}{min}*\frac{1min}{60s} [32.24(773.15K-579.93K)+\frac{0.1923x10^{-2}}{2} (773.15K^2-579.93K^2)+\frac{1.055x10^{-5}}{3} (773.15K^3-579.93K^3)+\frac{-3.595x10^{-9}}{4} (773.15K^4-579.93K^4)]\\Q_{heater}=30002.8\frac{kW*kg}{kmol} *\frac{1kmol}{18kg} \\Q_{heater}=1666.8kW[/tex]
c. Then, an overall energy balance is:
[tex]h_{in}m+Q{in}=h_{out}m+W_{out}[/tex]
In this case, [tex]h_{in}[/tex] equal the inlet enthalpy in the a. section, [tex]Q{in}[/tex] was computed in the previous section and [tex]W_{out}[/tex] is given in the statement, just [tex]h_{out}[/tex] is computed as follows, considering that at 5 bar and 500 °C the water is steam:
[tex]h_{out}=3484.5kJ/kg[/tex]
Therefore, the energy balance is satisfied as shown below:
[tex]3446.0\frac{kJ}{kg} *250\frac{kg}{min}*\frac{1min}{60s} +1666.8kW=3484.5\frac{kJ}{kg} *250\frac{kg}{min}*\frac{1min}{60s} +1500kW\\16025.1kW=16018.75kW[/tex]
It is seen that both of the sides are roughly the same, therefore, the energy balance is satisfied.
d. Finally, since kinetic energy is defined as:
[tex]KE=\frac{1}{2}mV^2[/tex]
Whereas [tex]V[/tex] is the fluid velocity, it could be computed by considering the specific volume extracted from the same tables at the entrance and the exit of the turbine which are:
[tex]v_{in}=0.08644m^3/kg\\v_{out}=0.529056m^3/kg[/tex]
Now, the velocities are:
[tex]V_{in}=\frac{0.08644\frac{m^3}{kg} *250\frac{kg}{min}*\frac{1min}{60s}}{\pi(0.5m^2/4)}=1.834m/s\\V_{out}=\frac{0.529056\frac{m^3}{kg} *250\frac{kg}{min}*\frac{1min}{60s}}{\pi(0.5m^2/4)}=11.23m/s[/tex]
Therefore, the change in kinetic energy is:
[tex]\Delta KE=\frac{1}{2} m(V_{out}^2-V_{in}^2)=(\frac{1}{2})250\frac{kg}{min}* \frac{1min}{60s}[(11.22\frac{m}{s} )^2-(1.834\frac{m}{s} )^2]\\\Delta KE=255.26kW[/tex]
Summing it up, such result is not significant in comparison with the involved enthalpies and the produced work, in such a way, it is reasonable to neglect it along the calculations.
Best regards.
