Answer:
a) [tex]\dot m = 16.168\,\frac{kg}{s}[/tex], b) [tex]v_{out} = 680.590\,\frac{m}{s}[/tex], c) [tex]\dot W_{out} = 18276.307\,kW[/tex]
Explanation:
A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:
Mass Balance
[tex]\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0[/tex]
Energy Balance
[tex]-q_{loss} - w_{out} + h_{in} - h_{out} = 0[/tex]
Specific volumes and enthalpies are obtained from property tables for steam:
Inlet (Superheated Steam)
[tex]\nu_{in} = 0.055665\,\frac{m^{3}}{kg}[/tex]
[tex]h_{in} = 3650.6\,\frac{kJ}{kg}[/tex]
Outlet (Liquid-Vapor Mix)
[tex]\nu_{out} = 5.89328\,\frac{m^{3}}{kg}[/tex]
[tex]h_{out} = 2500.2\,\frac{kJ}{kg}[/tex]
a) The mass flow rate of the steam is:
[tex]\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}[/tex]
[tex]\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }[/tex]
[tex]\dot m = 16.168\,\frac{kg}{s}[/tex]
b) The exit velocity of steam is:
[tex]\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}[/tex]
[tex]v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}[/tex]
[tex]v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}[/tex]
[tex]v_{out} = 680.590\,\frac{m}{s}[/tex]
c) The power output of the steam turbine is:
[tex]\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})[/tex]
[tex]\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)[/tex]
[tex]\dot W_{out} = 18276.307\,kW[/tex]
Answer:
The mass flow rate = 16.17kg/sec
The exit velocity = 680.84 m/sec
[tex]W_o_u_t=14562.37kw[/tex]
Explanation:
Attached is a copy of the diagram to this question
Let's write out the given data
T= 600⁰C
[tex]v_1=0.055665m^3/kg\\h_1=3650.60kJ/kg\\at p = 25kpa\\v_f=0.00102m^3/kg\\v_g=6.2034m^3/kg\\hfg=2345.5kJ/kg\\hf=271.96kJ/kg[/tex]
The enthalpy at exit is
[tex]h_2=271.96+\frac{95}{100} *2345.50=2500.185 kJ/kg\\[/tex]
The specific volume is
[tex]v_2=v_f+x_2vfg\\\\v_2=0.001020+\frac{95}{100}(6.2034-0.00102)=5.893m^3/kg[/tex]
a) To calculate the mass flow rate
[tex]m=\frac{V_1m}{v_1}=\frac{60*0.015}{0.055665}\\m=16.17kg/s[/tex]
The mass flow rate is 16.17kg/s
b) Exit velocity
[tex]v_2=\frac{mV_2}{A_2}=\frac{16.17*5.893}{0.14}=680.64m/sec[/tex]
The exit velocity is 680.64m/s
c) The power output
[tex]W_o_u_t=-mQ_o_u_t+m(h_1-h_2+\frac{v_1^2-v_2^2}{2})\\ W_o_u_t=-(16.17*20)+16.17(3650.60-2500.185+\frac{60^2-680.64^2}{2}*\frac{1}{1000} \\W_o_u_t=14562.37kw[/tex]
The power output of the machine is 14562.37kw
