Chapter 7 Vapor and Gas Power Systems

In this chapter, we will study the basic components of common industrial power and refrigeration systems. These systems are essentially thermodyanmic cycles in which a working fluid is alternatively vaporized and condensed as it flows through a set of four processes and returns to its initial state. We will use the first and second laws of thermodynamics to analyze the performance of ower and refrigeration cycles.

7.1 Rankine Cycle

We can use a Rankine cycle to convert a fossil-fuel, nuclear, or solar power source into net electrical power. Figure 7.1-1a shows the components of a Rankine cycle and Figure 7.1-1b identifies the states of the Rankine cycle on a Ts diagram.

Wt Turbine 1

2 Fuel QC air QH Rankine cycle Boiler 4 Condenser

Wp Pump Figure 7.1-1a The ideal Rankine cycle

Figure 7.1-1b The path of an ideal Rankine cycle1

1 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 395 7-1

In the ideal Rankine cycle, the working fluid undergoes four reversible processes:

Process 1-2: Isentropic expansion of the working fluid from the turbine from saturated vapor at state 1 or superheated vapor at state 1’ to the condenser pressure.

Process 2-3: Heat transfer from the working fluid as it flows at constant pressure through the condenser with saturated liquid at state 3.

Process 3-4: Isentropic compression in the pump to state 4 in the compressed liquid region.

Process 4-1: Heat transfer to the working fluid as it flows at constant pressure through the boiler to complete the cycle.

We will consider the Rankine cycle with water as the working fluid.

Process 1-2: Assuming that bulk kinetic and potential energy and heat transfer are negligible, the power produced by the turbine is given by

" Wt = m" (h1 − h2) (7.1-1)

In this equation m" is the mass flow rate of the working fluid. If the steam enters as a superheated vapor, it does not condense significantly in the turbine. If the steam is saturated as it enters the turbine, a significant of liquid is formed which causes erosion and wear of the turbine blades.

Process 2-3: The steam enters the condenser and exits in state 3 as saturated liquid water. The change of phase occurs at constant pressure with the heat removed from the flowing steam given by

" Qc = m" (h3 − h2) (7.1-2)

Process 3-4: A pump raises the pressure of the liquid. High-pressure water exits the pump in state 4 where the work delivered to the liquid is given by

" Wp = m" (h4 − h3) ≈ m" v3(p4 − p3) (7.1-3)

In this equation the pump work is integrated from

" Wp 4 = — vdp ≈ v3(p4 − p3) m" 3

The liquid volume is assumed to be constant. Since the specific volume of the liquid is significantly less than that of the vapor, the work required by the pump is much less than that produced by the turbine. Typically, a small fraction of the power produced by the turbine is used to compress the liquid, and the remaining power is the net power obtained by the cycle.

7-2 Process 4-1: The high-pressure liquid is brought back to a saturated or superheated vapor state in the boiler where the rate of heat transfer to the working fluid is given by

" QH = m" (h1 − h4) (7.1-4)

The vapor exits the boiler in state 1 and the cycle is repeated.

Example 7.1-12. ------Steam enters the turbine in a power plant at 600oC and 10 MPa and is condensed at a pressure of 0.1 MPa. Assume the plant can be treated as an ideal Rankine cycle. Determine the power produced per kg of steam and the efficiency of the cycle. Determine the efficiency of a Carnot cycle operated between these two temperatures.

Solution ------Table E7.4-1 States of the ideal Rankine cycle Specific Internal Specific Specific Temp Pressure Volume Energy Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg kJ/kg/K 1 600 10 0.03837 3242 3625 6.903 Dense Fluid (T>TC) 2 99.62 0.1 1.566 2349 2505 6.903 0.9246 Liquid Vapor Mixture 3 99.62 0.1 0.001043 417.3 417.4 1.303 0 Saturated Liquid 4 100.3 10 0.001039 417.4 427.7 1.303 Compressed Liquid

1 600oC, 10 MPa

4 0.1 MPa 3 2

Figure E7.1-1 The processes on Ts diagram.

Table E7.1-1 lists the states of steam in the ideal Rankine cycle with the bold values are the two properties used to defined the states. The work produced by the turbine per kg of steam is

Wt = (h1 − h2) = 3625 − 2505 = 1120 kJ/kg

2 Koretsky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 140 7-3 The work received by the pump is

Wp = (h4 − h3) = 427.7 − 417.4 = 10.3 kJ/kg

The net work produced by the plant is

Wcycle = Wt − Wp = 1120 − 10.3 = 1109.7 kJ/kg

The rate of heat transfer to the working fluid is given by

QH = (h1 − h4) = 3625 − 427.7 = 3197.3 kJ/kg

The efficiency of the cycle is

W 1109.7 η = cycle = = 0.347 QH 3197.3

The efficiency of a Carnot cycle operated between these two temperatures is

TL 99.6 + 273.15 ηCarnot = 1 − = 1 − = 0.573 TH 600 + 273.15 ------

Figure 7.1-2 Ts diagram showing irreversibilities in pump and turbine3.

The isentropic turbine and pump efficiencies are given by:

h1 − h2 h4s − h3 ηt = , and ηp = h1 − h2s h4 − h3

The actual work obtained from the turbine is less than the isentropic work and the actual work required for the pump is larger than the isentropic work.

3 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 402 7-4

Example 7.1-24. ------Steam is the working fluid in a modified Rankine cycle where the turbine and pump each have an isentropic efficiency of 85%. Saturated vapor enters the turbine at 8.0 MPa and saturated liquid exits the condenser at a pressure of 0.008 MPa. The net power output of the cycle is 100 MW. Determine for the modified cycle (a) the thermal efficiency, (b) the mass " flow rate of the steam, in kg/h, (c) the rate of heat transfer, QH , into the working fluid as it " passes through the boiler, in MW, (d) the rate of heat transfer, Qc , from the condensing steam as it passes through the condenser, in MW, (e) the mass flow rate of the condenser cooling water, in kg/h, if cooling water enters the condenser at 15°C and exits at 35°C.

Solution ------

Figure E7.1-2a The ideal Rankine cycle

Figure E7.1-2b The modified Rankine cycle5

Figure E7.1-2a shows the four states in the ideal Rankine cycle with the steam properties listed in Table E7.1-2. Figure E7.1-2b shows the modified Rankine cycle on the Ts diagram with the dash line represent the irreversibilities in the turbine and the pump. These irreversibilities cause an increase in entropy across the turbine and the pump.

4 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 396 5 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 403 7-5 Table E7.1-2 Steam properties at various states in the Rankine cycle. Specific Internal Specific Specific State Temp Pressure Volume Energy Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg kJ/kg/K 1 295.1 8 0.02352 2570 2758 5.743 1 Saturated Vapor 2s 41.51 0.008 12.21 1697 1795 5.743 0.6745 Liquid Vapor Mixture 3 41.51 0.008 0.001008 173.8 173.9 0.5925 0 Saturated Liquid 4s 41.76 8 0.001005 173.9 181.9 0.5925 Compressed Liquid

The isentropic work produced by the turbine per kg of steam is

" W / m" = (h1 − h2s) = 2758 − 1795 = 963 kJ/kg ( t )s

The specific enthalpy at the turbine exit, state 2, can be determined using the turbine efficiency

" h1 − h2 Wt / m" ηt = = " h1 − h2s W / m" ( t )s

h2 = h1 − ηt(h1 − h2s) = 2758 − 0.85(2758 − 1795) = 1939.5 kJ/kg

The specific enthalpy at the pump exit, state 4, can be determined using the turbine efficiency

h4s − h3 ηp = Ω h4 = h3 + (h4s − h3)/ηt h4 − h3

h4 = 173.9 + (181.9 − 173.9)/0.85 = 183.3 kJ/kg

The net power produced by the cycle is

" " " Wcycle = Wt − Wp = m" [(h1 − h2) − (h4 − h3)]

The rate of heat transfer to the working fluid is given by

" QH = m" (h1 − h4)

(a) The thermal efficiency is

" Wcycle (h1 − h2 ) − (h4 − h3 ) η = " = QH h1 − h4

(2758 −1939.5) − (183.3−173.9) η = = 0.314 2758 −183.3

7-6 (b) Determine the mass flow rate of the steam

Solving with numerical values, we obtain the net power produced by the cycle per kg/s of steam

" Wcycle / m" = (h1 − h2) − (h4 − h3) = (2758 − 1939.5) − (183.3 − 173.9) = 809.1 kJ/kg

" Since Wcycle = 100 MW, the mass flow rate of the steam is

(100 MW)(1000 kW/MW) m" = W" /809.1 kJ/kg = = 123.59 kg/s cycle 809.1 kJ/kg

m" = (123.59 kg/s)(3600 s/h) = 4.449×105 kg/h

" (c) Determine the rate of heat transfer to the steam, QH

" 5 QH = m" (h1 − h4) = (123.59 kg/s) (2758 − 183.3) kJ/kg = 3.182×10 kW

" 5 3 QH = (3.182×10 kW)( 1 MW/10 kW) = 318.2 MW

" (d) Determine the rate of heat transfer to the condensing water, Qc

" 5 Qc = m" (h2 − h3) = (123.59 kg/s) (1939.5 − 173.9) kJ/kg = 2.182×10 kW

" 5 3 Qc = (2.182×10 kW)( 1 MW/10 kW) = 218.2 MW

(e) Determine the mass flow rate of the condenser cooling water, in kg/h, if cooling water enters the condenser at 15°C and exits at 35°C.

Specific Internal Specific Specific Temp Pressure Volume Energy Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg kJ/kg/K 1 15 0.001705 0.001001 62.98 62.98 0.2245 0 Saturated Liquid 2 35 0.005628 0.001006 146.7 146.7 0.5052 0 Saturated Liquid

The mass flow rate of the condensing cooling water is given by

" 5 Qc 2.182×10 kJ/s 3 m" cw = = = 2.61×10 kg/s hcw,out − hcw,in (146.7 - 62.98) kJ/kg

3 6 m" cw = (2.61×10 kg/s)(3600 s/h) =9.39×10 kg/h

7-7 7.2 Refrigeration Cycle

The most common refrigeration cycle is the vapor compression cycle shown in Figure 7.2-1. In step 4 → 1, heat is removed at the temperature TL from the system being refrigerated by the evaporation of a liquid under the pressure PL. In step 1 → 2, saturated vapor at PL is compress isentropically to PH where it becomes superheated vapor. In step 2 → 3, heat QH is transferred to the surrounding by condensation at TH. In step 3 → 4, the cycle is closed by throttling the liquid to the lower pressure PL.

PH Q H 2 3 2 Condenser

Liquid PL Throttle 3 valve Compressor Evaporator Isenthalp 4 1 Vapor 1 W 4 QL a b c s Figure 7.2-1 A vapor-compression refrigeration cycle and its Ts diagram.

The heat transfer between the system and the surroundings can be obtained from the Ts diagram. Since Q = — Tds , the heat effect is the area under the curve representing the path. In

Figure 7.2-1 the heat QH transferred from the refrigerator to the high temperature environment is the area 2-3-a-c, which is negative. The heat QL removed from the low temperature system is the area 4-1-c-b and is positive. For the cyclic process

∆U = QL + W − QH = 0 Ω W = QH − QL (7.2-1)

The efficiency of the refrigeration cycle, called the coefficient of performance COP, is given by

Q COP = L (7.2-2) W

The work required for the refrigeration cycle can be obtained from the Ts diagram

W = QH − QL = (Area 2-3-a-c) − (Area 4-1-c-b)

W = QH − QL = (Area 1-2-3-4) + (Area 3-4-b-c)

The refrigeration cycle shown in Figure 7.2-1 is a semi-reversible cycle since all steps except throttling are reversible.

7-8 Example 7.2-15------A vapor compression refrigeration process using NH3 as the working fluid is to operate between 20 and 80oF. Determine the coefficient of performance for the semi-reversible operation.

Solution ------T

Liquid PL 3 80oF h4 h1 h2 Evaporator Isenthalp

4 1 o Vapor 20 F 1 QL W 4 a b c s

We will first locate the four states of the refrigeration cycle:

o State 3: Saturated liquid at 80 F, h3 = h4 = 131.7 Btu/lb. o State 1: Saturated vapor at 20 F, h1 = 616.8 Btu/lb. o State 4: Liquid and vapor mixture at h4 = 131.7 Btu/lb and 20 C o State 2: Superheated vapor at s2 = s1 = 1.295 Btu/lb⋅ R, PH = P2 = 153.1 psia

Making energy balance around the evaporator yields

0 = h4 − h1 + QL Ω QL = h1 − h4 = 616.8 − 131.7 = 485.1 Btu/lb

Making energy balance around the compressor yields

0 = h1 − h2 + W Ω W = h2 − h1 = 686.6 − 616.8 = 69.8 Btu/lb

The coefficient of performance is then

Q 485.1 COP = L = = 6.95 W 69.8

5 Kyle, B.G., Chemical and Process Thermodynamics, Prentice Hall, 1999, pg. 656 7-9 Example 7.2-2------A vapor-compression refrigeration process using ammonia as the working fluid is to operate between 0oC and 30oC. In step 4 → 1 heat is supplied to the fluid at 0oC under the pressure p1. The saturated vapor at P1 is then compressed isentropically to p2, where it becomes superheated vapor, state 2. Removal of heat from this vapor leads to cooling at constant pressure followed by condensation at 30oC, step 2 → 3. Throttling the saturated liquid at state 3 to the lower pressure at state 4 closes the cycle. The four states are given (not in any particular order) as follow T(oC) p(MPa) h(kJ/kg) s(kJ/kg⋅K) 0 .4293 1442.32 5.3313 0 1.2315 30 1.1668 322.52 1.2005 69.8 1581.49 Determine (a) The heat transfer (kJ/kg) in step 4 → 1, (b) the heat transfer (kJ/kg) in step 2 → 3, and (c) the work supplied (kJ/kg) by the compressor. Solution ------T

PH Q H 2 3 2 Condenser

Liquid PL Throttle 3 valve Compressor Evaporator Isenthalp 4 1 Vapor 1 W 4 QL a b c s

Table E7.2-2 The four states in proper order State T(oC) p(MPa) h(kJ/kg) s(kJ/kg⋅K) 1 0 0.4293 1442.32 5.3313 2 69.8 1.1668 1581.49 5.3313 3 30 1.1668 322.52 1.2005 4 0 0.4293 322.52 1.2315

(a) The heat transfer (kJ/kg) in step 4 → 1

QL = 1442.32 − 322.52 = 1119.8 kJ/kg

(b) The heat transfer (kJ/kg) in step 2 → 3

QH = 1581.49 − 322.52 = 1259.0 kJ/kg

Wc = 1581.49 − 1442.32 = 139.17 kJ/kg

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