Respuesta :
Answer:
a) ΔEC=-23.4kW
b)W=12106.2kW
c)[tex]A=0.01297m^2[/tex]
Explanation:
A)
The kinetic energy is defined as:
[tex]\frac{m*vel^2}{2}[/tex] (vel is the velocity, to differentiate with v, specific volume).
The kinetic energy change will be: Δ ([tex]\frac{mvel^2}{2}[/tex])=[tex]\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}[/tex]
Δ ([tex]\frac{mvel^2}{2}[/tex])=[tex]\frac{m}{2}*(vel_2^2-vel_1^2)[/tex]
Where 1 and 2 subscripts mean initial and final state respectively.
Δ([tex]\frac{mvel^2}{2}[/tex])=[tex]\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW[/tex]
This amount is negative because the steam is losing that energy.
B)
Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).
[tex]H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)[/tex]
We already know the last quantity: [tex]\frac{m}{2} *(vel_1^2-vel_2^2)[/tex]=[tex]-[/tex]Δ ([tex]\frac{mvel^2}{2}[/tex])=23400W
For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, [tex]h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}[/tex]
The exit state is a liquid-vapor mixture, so its enthalpy is:
[tex]h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}[/tex]
Finally, the work can be obtained:
[tex]W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW[/tex]
C) For the area, consider the equation of mass flow:
[tex]m=p*vel*A[/tex] where [tex]p[/tex] is the density, and A the area. The density is the inverse of the specific volume, so [tex]m=\frac{vel*A}{v}[/tex]
The specific volume of the inlet steam can be read also from the steam tables, and its value is: [tex]0.08643\frac{m^3}{kg}[/tex], so:
[tex]A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2[/tex]
Calculating the work done by the turbine at the given mass flow rate
gives the power output of the turbine.
Responses:
(a) -23.4 kJ
(b) The power output of the turbine is approximately 12.12 MW
(c) The turbine inlet area is 0.01296615 m²
Which methods evaluate the values of the steam turbine?
(a) Kinetic energy, K.E. = [tex]\mathbf{\frac{1}{2} \cdot m \cdot v^2}[/tex]
Where;
m = The mass
v = Velocity
The change in kinetic energy is given by the difference in the kinetic
energy of the input and output as follows;
ΔK.E. = K.E.₂ - K.E.₁
[tex]K.E._1 = \dfrac{1}{2} \times 12 \times 80^2 = \mathbf{38400}[/tex]
K.E.₁ = 38400 J = 38.4 kJ
[tex]K.E._2 = \dfrac{1}{2} \times 12 \times 50^2 = \mathbf{15000}[/tex]
K.E.₂ = 15000 J = 15 kJ
Which gives;
ΔK.E. = 15000 - 38400 = -23400
- The change in kinetic energy of the steam is -23,400 J = -23.4 kJ
(b) From the system energy balance, we have;
The energy input into the turbine = The energy output from the turbine
Which gives;
The heat + Kinetic energy at input = The heat + Kinetic energy at output + Work output of the turbine
Let h₁ represent the specific enthalpy at input and let h₂ represent the
specific enthalpy of the steam at the exit.
Mass × Specific enthalpy = Heat in the system
Which gives;
m·h₁ + K.E.₁ = m·h₂ + K.E.₂ + W
From steam tables, at 4 MPa, 500°C, h₁ = 3445.84 kJ·kg⁻¹
At the exit, we have;
Pressure = 30 kPa
Quality, x = 92% (water and steam)
[tex]h_2 = \mathbf{ h_f +x \cdot h_{fg}}[/tex]
At 30 kPa, [tex]h_f[/tex] = 289.299 kJ·kg⁻¹
[tex]h_{fg}[/tex] = 2624.55 kJ·kg⁻¹
Which gives;
h₂ = 289.299 + 0.92 × 2335.32 = 2437.7934
The heat content at exit, h₂ = 2437.7934 kJ·kg⁻¹
Therefore;
12 × 3445.84 + 38,400 = 12 × 2437.7934 + 15000 + W
W = (12 × 3445.84 + 38.4) - (12 × 2437.7934 + 15) = 12119.9592
Given that the mass flow rate in kg/s is used to for the above calculation
for the work done, W, we have;
- The power output of the turbine = 12119.9592 kW ≈ 12.12 MW
(c) The mass flow rate, [tex]\mathbf{\dot m}[/tex], is given by the following equation;
[tex]\dot m[/tex] = ρ·v·A
Where;
ρ = The density of the fluid
v = The velocity
A = The cross sectional area
Where;
[tex]\rho = \mathbf{ \dfrac{1}{Specific \ volume}} = \dfrac{1}{\nu }[/tex]
The specific volume at 4 MPa, 500°C is
[tex]\nu[/tex] = 0.0864410 m³·kg⁻¹
Therefore;
[tex]\dot m = \mathbf{\dfrac{v \cdot A}{\nu}}[/tex]
[tex]A = \dfrac{ \dot m \cdot \nu}{v}[/tex]
[tex]A = \mathbf{\dfrac{ 12 \times 0.0864410}{80}} = 0.01296615[/tex]
- The turbine inlet area, A = 0.01296615 m²
Learn more about steam turbines here:
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