1. Steam Plant System Name Institutional Affiliation. 2

1
Steam Plant System
Name
Institutional Affiliation

2
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)}