Create a full state table including 2s, 4s, 7s, and 8s. Also create a full scale
ID: 1718454 • Letter: C
Question
Create a full state table including 2s, 4s, 7s, and 8s. Also create a full scale T-s diagram including all the states.
Explanation / Answer
The most prominent physical feature of a modern steam power plant (other than its smokestack) is the steam generator, or boiler, as seen in Figure 2.1. There the combustion, in air, of a fossil fuel such as oil, natural gas, or coal produces hot combustion gases that transfer heat to water passing through tubes in the steam generator. The heat transfer to the incoming water (feedwater) first increases its temperature until it becomes a saturated liquid, then evaporates it to form saturated vapor, and usually then further raises its temperature to create superheated steam. Steam power plants such as that shown in Figure 2.1, operate on sophisticated variants of the Rankine cycle. These are considered later. First, letís examine the simple Rankine cycle from which the cycles of large steam power plants are derived. In the simple Rankine cycle, steam flows to a turbine, where part of its energy is converted to mechanical energy that is transmitted by rotating shaft to drive an electrical generator. The reduced-energy steam flowing out of the turbine condenses to liuid water in the condenser. A feedwater pump returns the condensed liquid (condensate) to the steam generator. The heat rejected from the steam entering the condenser is transferred to a separate cooling water loop that in turn delivers the rejected energy to a neighboring lake or river or to the atmosphere.
Rankine-Cycle Analysis
In analyses of heat engine cycles it is usually assumed that the components of the
engine are joined by conduits that allow transport of the working fluid from the exit of
one component to the entrance of the next, with no intervening state change. It will be
seen later that this simplification can be removed when necessary.
It is also assumed that all flows of mass and energy are steady, so that the steady
state conservation equations are applicable. This is appropriate to most situations
because power plants usually operate at steady conditions for significant lengths of
time. Thus, transients at startup and shutdown are special cases that will not be
considered here.
Consider again the Rankine cycle shown in Figure 2.2. Control of the flow can be
exercised by a throttle valve placed at the entrance to the turbine (state 1). Partial valve
closure would reduce both the steam flow to the turbine and the resulting power
output. We usually refer to the temperature and pressure at the entrance to the turbine
as throttle conditions. In the ideal Rankine cycle shown, steam expands adiabatically
and reversibly, or isentropically, through the turbine to a lower temperature and
pressure at the condenser entrance. Applying the steady-flow form of the First Law of
Thermodynamics [Equation (1.10)] for an isentropic turbine we obtain:
q = 0 = h2 ñ h1 + wt
[Btu/lbm | kJ/kg]
where we neglect the usually small kinetic and potential energy differences between the
inlet and outlet. This equation shows that the turbine work per unit mass passing
through the turbine is simply the difference between the entrance enthalpy and the
lower exit enthalpy:
wt
= h1 ñ h2 [Btu/lbm | kJ/kg] (2.1)
The power delivered by the turbine to an external load, such as an electrical generator,
is given by the following:
Turbine Power = mswt = ms(h1 ñ h2) [Btu/hr | kW]
where ms [lbm /hr | kg/s] is the mass flow of steam though the power plant.
Applying the steady-flow First Law of Thermodynamics to the steam generator,
we see that shaft work is zero and thus that the steam generator heat transfer is
qa = h1 ñ h4 [Btu/lbm | kJ/kg] (2.2)
The condenser usually is a large shell-and-tube heat exchanger positioned below or
adjacent to the turbine in order to directly receive the large flow rate of low-pressure
turbine exit steam and convert it to liquid water. External cooling water is pumped
through thousands of tubes in the condenser to transport the heat of condensation of
the steam away from the plant. On leaving the condenser, the condensed liquid (called
condensate) is at a low temperature and pressure compared with throttle conditions.
Continued removal of low-specific-volume liquid formed by condensation of the highspecific-volume
steam may be thought of as creating and maintaining the low pressure
in the condenser. The phase change in turn depends on the transfer of heat released to
the external cooling water. Thus the rejection of heat to the surroundings by the
cooling water is essential to maintaining the low pressure in the condenser. Applying
the steady-flow First Law of Thermodynamics to the condensing steam yields:
qc = h3 ñ h2 [Btu/lbm | kJ/kg] (2.3)
The condenser heat transfer qc is negative because h2 > h3. Thus, consistent with sign
convention, qc represents an outflow of heat from the condensing steam. This heat is
absorbed by the cooling water passing through the condenser tubes. The condensercooling-water
temperature rise and mass-flow rate mc are related to the rejected heat
by:
ms|qc| = mc cwater(Tout - Tin) [Btu/hr | kW]
where cwater is the heat capacity of the cooling water [Btu/lbm-R | kJ/kg-K]. The
condenser cooling water may be drawn from a river or a lake at the temperature Tin and
returned downstream at Tout, or it may be circulated through cooling towers where heat
is rejected from the cooling water to the atmosphere.
We can express the condenser heat transfer in terms of an overall heat transfer
coefficient, U, the mean cooling water temperature, Tm = (Tout + Tin)/2, and the
condensing temperature T3:
ms|qc| = UA(T3 - Tm) [Btu/hr | kJ/s]
It is seen for given heat rejection rate, the condenser size represented by the tube
surface area A depends inversely on (a) the temperature difference between the
condensing steam and the cooling water, and (b) the overall heat-transfer coefficient.
For a fixed average temperature difference between the two fluids on opposite
sides of the condenser tube walls, the temperature of the available cooling water
controls the condensing temperature and hence the pressure of the condensing steam.
The states at the inlets and exits of the components, following the notation of Figure 2.2, are listed in the following table. The enthalpy and entropy of state 1 may be obtained directly from tables or charts for superheated steam (such as those in Appendices B and C) at the throttle conditions. A Mollier chart is usually more convenient than tables in dealing with turbine inlet and exit conditions. For an ideal isentropic turbine, the entropy is the same at state 2 as at state 1. Thus state 2 may be obtained from the throttle entropy (s2 = s1 = 1.5603 Btu/lbm-R) and the condenser pressure (1 psia). In general, this state may be in either the superheatedsteam region or the mixed-steam-and-liquid region of the Mollier and T-s diagrams. In the present case it is well into the mixed region, with a temperature of 101.74° F and an enthalpy of 871 Btu/lbm.
The enthalpy, h3 = 69.73 Btu/lbm, and other properties at the pump inlet are obtained
from saturated-liquid tables, at the condenser pressure. The steady-flow First Law of
Thermodynamics, in the form of Equation (2.4), indicates that neglecting isentropic
pump work is equivalent to neglecting the pump enthalpy rise. Thus in this case
Equation (2.4) implies that h3 and h4 shown in Figure (2.2) are almost equal. Thus we
take h4 = h3 as a convenient approximation.
State Temperature
(°F)
Pressure
(psia)
Entropy
(Btu/lbm-°R)
Enthalpy
(Btu/lbm)
1 1000.0 2000 1.5603 1474.1
2 101.74 1 1.5603 871.0
3 101.74 1 0.1326 69.73
4 101.74 2000 0.1326 69.73
The turbine work is
h1 ñ h2 = 1474.1 ñ 871 = 603.1 Btu/lbm.
The heat added in the steam generator is
h1 ñ h4 = 1474.1 ñ 69.73 = 1404.37 Btu/lbm.
The thermal efficiency is the net work per heat added = 603.1/1404.37 = 0.4294
(42.94%). This corresponds to a heat rate of 3413/0.4294 = 7946 Btu/kW-hr. As
expected, the efficiency is significantly below the value of the Carnot efficiency
of 1 ñ (460 + 101.74)/(460 + 1000) = 0.6152 (61.52%), based on a source
temperature of T1 and a sink temperature of T3.
The quality of the steam at the turbine exit is
(s2 ñ sl
)/(sv ñ sl
) = (1.5603 ñ 0.1326)/(1.9781 ñ 0.1326) = 0.7736
Here v and l indicate saturated vapor and liquid states, respectively, at pressure p2.
Note that the quality could also have been obtained from the Mollier chart for steam as
1 - M, where M is the steam moisture fraction at entropy s2 and pressure p2.
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