write the steady flow energy equation for a single entry steam and single exit stream
Answers
Answered by
4
The Steady Flow Energy Equation (SFEE) is used for open systems to determine the total energy flows.
It is assumed that the mass flow through the system is constant.
It is also assumed that the total energy input to the system is equal to the total energy output.
The energies that are included are;
internal, flow, kinetic, potential, heat and work.
The equation is shown below where suffix 1 is the entrance and suffix 2 the exit from the system.
where:
u=internal energy (J)
P=pressure (N/m2)
v=volume (m3)
C=velocity (m/s)
g=acceleration due to gravity (m/s2)
Z=height above a datum (m)
Q=heat flow (J)
W=work (J)
The term P .vrepresents the displacement or flow energy.
The termC2 / 2represents the kinetic energy.
The term g. Zrepresents the potential energy.
In thermodynamics the changes in potential energy are usually small except for example a water reservoir supplying water to a low level turbine.
In the following examples we can omit the potential energy term thus simplifying the equation to;
u1+P1v1++Q=u2+P2v2++W
Also the term( u+P.v ) is also known as specific enthalpy (h), so the equation is now written;
h1++Q=h2++W
Example 1
In a steady flow open system a fluid flows at a rate of 4 kg/s.
It enters the system at a pressure of 6 bar, a velocity of 220 m/s, internal energy
2200 kJ/kg and specific volume 0.42 m3/kg.
It leaves the system at a pressure of 1.5 bar, a velocity of 145 m/s, internal energy
1650 kJ/kg and specific volume 1.5 m3/kg.
During its passage through the system, the fluid has a loss by heat transfer of 40kJ/kg to the surroundings.
Determine the power of the system, stating whether it is from or to the system.
Neglect any change in potential energy.
u1+P1v1++Q=u2+P2v2++W
Power from the system (kW)= Work output (kJ/kg) x mass flow rate of fluid (kg/s)
Work output (kJ/kg) can be found by rearranging the SFEE.
We can work in kilojoules (kJ). This means that thekinetic energy section of the SFEE will be divided by 103 to put it in kJ.
W=( u1-u2)+( P1v1-P2v2 ) +()+Q
2 x 103
W=( 2200-1650 )+( 600x0.42-150x1.5 )+()– 40
W= 550+27+13.69-40
W=550.69kJ/kg
This is positive so it is energy output from the system.
Power from the system (kW)=Wxm
=550.69x4 kg/s
=2202.76 kW
It is assumed that the mass flow through the system is constant.
It is also assumed that the total energy input to the system is equal to the total energy output.
The energies that are included are;
internal, flow, kinetic, potential, heat and work.
The equation is shown below where suffix 1 is the entrance and suffix 2 the exit from the system.
where:
u=internal energy (J)
P=pressure (N/m2)
v=volume (m3)
C=velocity (m/s)
g=acceleration due to gravity (m/s2)
Z=height above a datum (m)
Q=heat flow (J)
W=work (J)
The term P .vrepresents the displacement or flow energy.
The termC2 / 2represents the kinetic energy.
The term g. Zrepresents the potential energy.
In thermodynamics the changes in potential energy are usually small except for example a water reservoir supplying water to a low level turbine.
In the following examples we can omit the potential energy term thus simplifying the equation to;
u1+P1v1++Q=u2+P2v2++W
Also the term( u+P.v ) is also known as specific enthalpy (h), so the equation is now written;
h1++Q=h2++W
Example 1
In a steady flow open system a fluid flows at a rate of 4 kg/s.
It enters the system at a pressure of 6 bar, a velocity of 220 m/s, internal energy
2200 kJ/kg and specific volume 0.42 m3/kg.
It leaves the system at a pressure of 1.5 bar, a velocity of 145 m/s, internal energy
1650 kJ/kg and specific volume 1.5 m3/kg.
During its passage through the system, the fluid has a loss by heat transfer of 40kJ/kg to the surroundings.
Determine the power of the system, stating whether it is from or to the system.
Neglect any change in potential energy.
u1+P1v1++Q=u2+P2v2++W
Power from the system (kW)= Work output (kJ/kg) x mass flow rate of fluid (kg/s)
Work output (kJ/kg) can be found by rearranging the SFEE.
We can work in kilojoules (kJ). This means that thekinetic energy section of the SFEE will be divided by 103 to put it in kJ.
W=( u1-u2)+( P1v1-P2v2 ) +()+Q
2 x 103
W=( 2200-1650 )+( 600x0.42-150x1.5 )+()– 40
W= 550+27+13.69-40
W=550.69kJ/kg
This is positive so it is energy output from the system.
Power from the system (kW)=Wxm
=550.69x4 kg/s
=2202.76 kW
Attachments:
Similar questions