Steam Plant System Analysis: Applying Thermodynamic Energy Equations

This report provides a detailed analysis of a steam plant system, focusing on the application of thermodynamic principles and energy equations. It begins by defining a closed system and contrasting it with the components of a steam power plant, such as turbines, pumps, and heat exchangers. The report then applies the Steady Flow Energy Equation to analyze the mass flow rate and energy changes within the system, considering factors such as heat supplied, work done, internal energy, and kinetic/potential energy. A simple Rankine cycle steam plant system is examined, and the First Law of Thermodynamics is applied to control volumes within the pump, boiler, turbine, and condenser. The report calculates thermodynamic efficiency based on enthalpy values and concludes with references to relevant research on steam power plant system reliability. This document is available on Desklib, a platform offering a wide range of study tools and resources for students.

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Steam Plant System
Name
Institutional Affiliation

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1. A system is considered to be closed if no mass crosses the boundaries. However, from the
continuity equation, heat and work can cross the boundaries. Compared to a closed system, a
steam power plant consists of components such as turbines, pumps, and heat exchangers
(Mangara & Finkelstein, 2014, p. 88). In the turbine at high-pressure, there is expansion of the
saturated steam received from the steam generator. Work is delivered in the shaft representing an
isentropic process. The expanded and moist steam leaves from the HP turbine and the moisture
separator heater dries and superheat the moist steam. The low-pressure turbine expands the
superheated steam from the moisture separator reheater to produce a shaft work output resulting
into a constant entropy process (Mangara & Finkelstein, 2014, p. 98). The steam is leaves from
the turbine and goes to the condenser. While the conditions of the vacuum remain constant, heat
is rejected and transferred to the cooling water. The condensate and feedwater pump compresses
the feedwater as a liquid while the feedwater is preheated by the feedwater heaters. After which,
addition of heat to the refrigerant in the steam generator takes place while pressure remains
constant. From the above cycle of movement of the working fluid, one would observe that the
mass of the working fluid does not cross the boundary while heat and work cross the boundaries.
Therefore, this is a typical example of a closed system.
2. For a simple Rankine Cycle steam plant system
The rate of mass flow at the inlet = the mass flow at the outlet
Given that ρ1 represent the density of the fluid at inlet and ρ2 is the density at outlet, while V1
represents the velocity of the fluid at inlet through cross-section area A1 and V2 represents the
velocity of the fluid at the outlet through a cross section A2.
A1V1ρ1 = A2V2ρ2

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Change in E is always a constant and = change in U – Change in W
Change in U = Sum of Q – Sum of W, where U is the internal energy, Q is the heat supplied
while W is the work done on the system.
Q- Ẇ = E (in) – E (out)
Q- Ẇ = mgz + 1/2mv2 + mu
But E (crossing) + E (entering) = E (exiting)
(Q- Ẇ) + m{1/2v12 + u1 + gz1}= m{1/2v22 + u2 + gz2}
Since mass flow rate (m)= ρAV and Ẇ = P1A1V1 – P2A2V2 + WS
Q- P2A2V2 + P1A1V1 - WS + ρ1A1V1{1/2V12 + U1 + gz1} = ρ2A2V2{1/2V22 + U2 + gz2}
But since ρ is constant for steady flow,
Q - WS = P2A2V2 – P1A1V1 + m{(V22- V12)/2 + U2- U1 + g(z2-z1)}
Q - WS = m{(V22- V12)/2 + U2 – U1 + g + P2/ρ - P1/ρ}
But since m= ρAV, then P/ρ= m and h = U + P/ρ
Therefore,
Q - WS = m{(h2 – h1) + (V22 – V12)/2 + g(z2 – z1)}

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3. A simple Rankine steam plant system in figure1 consists of a pump, boiler, turbine, and
condenser based on the application of the First law of thermodynamics to the control volume.
Since the specific entropy is constant in an isentropic expansion (s3=s4 for the 4Mpa). The vapor
quality is known with an x=0.9
State 2 enthalpy is picked directly from the tables as shown.
bar=4MPa
T2=550
P3=0.8bar=0.08bar
From the steam table,
4MPa lies between 3.973 and 4.688 with h-values lying 2801.5 and 2796.9kJ/kg. From
extrapolation, h2l =2799.5kJ/kg
Again from steam table at a pressure of 0.08 bars, h3v=2665kJ/kg and h3l=395kJ/kg
h3wet=h3v*x+(1-x) h3lv=2665*0.9+(1-0.9) *395=2436kJ/kg
Work done by the steam
Wt=∆ h=2665−2436=229 kJ / kg
Since state 3 is fixed by the pressure and p2=4MPa and for isentropic compression, s3=s4 for
0.08Mpa from which hsubcooled =401kJ/kg from the steam tables

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Work done by the pump Wp=399-395=4kJ/kg
Net heat added=h2v-hsub=2799.5-401=2398.5 =Qin
Thermodynamic efficiency
Ꞃ=Wt −℘
Qin
Ꞃ=229−4
2398.5 =9.42 %

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Reference
Mangara, B. T., & Finkelstein, M. (2014). On the improvement of steam power plant system
reliability. IECON 2014 - 40th Annual Conference of the IEEE Industrial Electronics
Society. doi:10.1109/iecon.2014.7049323