8.7 The Brayton Cycle with Regeneration In the Brayton cycle analysis given in Example 8-12 the exhaust temperature from the gas turbine is 992 ºF = 533 ºC. This represents an enormous amount of high temperature energy rejection to the atmosphere. While all heat power cycles must have some heat rejection, if the rejection temperature is “high enough,” we may be able to utilize some of that energy to produce additional useful work. Note that in this example the temperature at the compressor exit/combustor inlet is 745 ºF = 396 ºC. This suggests that if we insert a heat exchanger (called a regenerator) into this cycle with the turbine exhaust flow on one side and the compressor exhaust flow on the other (as shown on Figure S8-1a), then we could recover some of the energy from the hot turbine exhaust. (See Figure S8-1b.) The recovered heat, Q saved = m ( h4 − hy ) = m ( hx − h2 ) , would be used to increase the temperature (preheat) of the air flowing into the combustor. Because the required inlet temperature to the turbine would not change with the addition of the regenerator, less energy would have to be added in the combustor, Q in = m ( h3 − hx ) . Hence, cycle efficiency would increase.

a)

1

b) Figure S8-1 a) Brayton cycle schematic b) T-s diagram with regeneration Referring to the T-s diagram in Figure S8-1b, the highest temperature (Tx) that the air from the regenerator could ever attain would be T4, the turbine exhaust gas temperature; that is, for all situations Tx ≤ T4. For temperature Tx to exceed T4, heat transfer would have to occur from a lower temperature to a higher temperature, which would be a violation of the second law of thermodynamics. An ideal regenerator would result in the preheat temperature equaling the turbine exhaust temperature (Tx = T4). However, in practice, we would never have an ideal regenerator because the surface area required would be infinite. We need a method to quantify the performance of a regenerator and to determine its effect on the Brayton cycle. To do this, we define a heat exchanger effectiveness, ε, as:

ε=

Q Actual Heat Transfer Rate = act Maximum Possible Heat Transfer Rate Q max

Eq. S8-1

The magnitude of effectiveness can range from 0 (no heat transfer) to 1 (ideal or maximum possible heat transfer). The actual heat transfer rate in the regenerator shown in Figure S8-1 is calculated with the energy equation: p (Tx − T2 ) or Q act = m ( hx − h2 ) = mc p (T4 − Ty ) Q act = m ( h4 − hy ) = mc

Eq. S8-2

Note that the use of cP ∆T assumes an ideal gas with constant specific heat. To evaluate the maximum possible heat transfer rate (without violating the second law of thermodynamics), we need to examine how the temperatures of the two single-phase fluids behave. The maximum possible heat transfer rate would occur when one of the fluids undergoes the maximum possible temperature rise or fall. That is, the highest possible outlet temperature for the cold fluid would equal TH,in and the lowest possible outlet temperature for the hot fluid would equal TC,in. Hence, the maximum possible temperature change of either fluid would be (TH,in - TC,in). Which fluid could undergo this maximum temperature change? We define the heat capacity rate p . In general, this product is different for the hot (CH) and cold (CC) fluids flowing through a as C = mc heat exchanger; that is, CH does not have to equal CC. For example, if CH > CC, then we designate the

2

larger heat capacity rate Cmax = CH and the smaller heat capacity rate Cmin = CC. Because conservation of energy must be satisfied

Q act = CH ∆TH = CC ∆TC Consequently, if CH > CC , then the cold fluid would undergo a larger actual temperature change ( ∆TC ) than the hot fluid ( ∆TH ); that is, ∆TC > ∆TH . As can be seen, the fluid with the minimum heat capacity rate will undergo a greater temperature change than the fluid with the maximum heat capacity rate. Thus, only the fluid with the minimum heat capacity rate, Cmin, could experience the maximum possible temperature change TH,in - TC,in, which then leads to the maximum possible heat transfer rate Q max = m ( h4 − h2 ) = Cmin (TH ,in − TC ,in ) = Cmin (T4 − T2 )

Eq. S8-3

The same result is obtained if we assumed CC > CH. For the regenerator, the mass flow rate is the same on both sides. Therefore, substituting Eq. S8-2 and Eq. S8-3 into Eq. S8-1we obtain Q

h −h

h −h

ε = act = x 2 = 4 y Qmax h4 − h2 h4 − h2

Eq. S8-4

When the cold-air-standard analysis assumptions are invoked, then Eq. S8-4 reduces to

ε=

T4 − Ty hx − h2 c p (Tx − T2 ) Tx − T2 = = or, likewise, ε = h4 − h2 c p (T4 − T2 ) T4 − T2 T4 − T2

Eq. S8-5

Beginning with the basic definition of cycle thermal efficiency and following a cold-air-standard analysis similar to that used to develop the cycle efficiency for an ideal Brayton cycle (isentropic compressor and turbine) without regeneration (Eq. 8-33), the thermal efficiency of an ideal Brayton cycle (ηT = ηC = 100%) with an ideal regenerator (ε = 1) can be derived as: ⎛ T1 ⎞ ( k −1) k ⎟ rp ⎝ T3 ⎠

ηcycle = 1 − ⎜

Eq. S8-6

Note that not only is the cycle thermal efficiency dependent on the pressure ratio, but the maximum-tominimum temperature ratio in the cycle also affects the efficiency. For any Brayton cycle (isentropic or non-isentropic compressor or turbine), the addition of a regenerator does not change the inlet or outlet temperatures of the compressor or turbine. Thus, ηcycle = 1 − Q out Q in = 1 − ( hy − h1 ) ( h3 − hx ) . Using Eq. S8-5, we can obtain expressions for hy and hx in terms of ε, h2, and h4. Substitute these back into the expression for cycle efficiency and rearrange, and the effect of any regenerator ( ε ≤ 1 ) on the cycle thermal efficiency can be shown to be:

ηcycle

⎡ h4 − h1 ⎤ ⎢h − h ⎥ −ε = 1− ⎣ 4 2 ⎦ ⎡ h3 − h2 ⎤ ⎢h − h ⎥ −ε ⎣ 4 2⎦

Eq. S8-7

3

If we assume a cold-air-standard analysis, Eq. S8-7can be rewritten in terms of temperatures; in that case, when ε = 1, Eq. S8-7 and Eq. S8-6 give the same results, and Figure S8-2 illustrates the effects of pressure ratio and temperature ratio on the ideal Brayton cycle efficiency. Note that regeneration has a greater effect when the unregenerated cycle efficiency is low (i.e., at lower pressure and maximum-to-minimum temperature ratios).

Figure S8-2 Effect of regeneration on ideal Brayton cycle efficiency using a cold-air-standard anaylysis (k = 1.4)

The increased cycle thermal efficiency associated with the addition of a regenerator does not come without cost. Trade-offs must be examined to determine the economics of the modification. For a fixed net power output, the increased thermal efficiency would result in a lower mass flowrate, smaller pieces of equipment, and lower fuel costs. Capital cost per piece of equipment and fuel cost would decrease. The regenerator would add capital cost, as well as weight, maintenance costs, and pressure drop (which we have ignored throughout our analyses). Regenerators with higher effectivenesses tend to be more expensive. Hence, life cycle cost (capital cost plus operating cost) must be examined to determine what net financial gain might be realized with a regenerator and what effectiveness level can be afforded. Two other modifications can be made to the Brayton cycle to improve cycle performance. (See Figure S8-3a.) Instead of a one-stage expansion through the turbine, multiple stages with reheat (similar to that used in the Rankine cycle) can be used. This arrangement increases the work produced per unit mass by the cycle. (See Figure S8-3b.) However, the cycle thermal efficiency may not increase because of the additional heat input required. Because the exit temperature from the turbine is higher when reheat is used compared to no reheat, the potential gain through the use of regeneration is increased.

4

a)

b) Figure S8-3 Brayton cycle with reheat and intercooling. a) system schematic and b) T-s diagram A second modification is to use multistage compression with intercooling. (Again, see Figure S8-3a.) Intercooling reduces the work per unit mass needed to compress a gas between two pressures and reduces the compressor outlet temperature. Hence, while the net work per unit mass from a cycle will be increased, efficiency may not be improved because the lower compressor outlet temperature requires that more heat be added in the combustor/heat addition process.

Example S8-1 Brayton cycle with regeneration In a Brayton cycle, air at 100 kPa and 27 °C enters the compressor and leaves at 710 kPa. The air is heated to 825 °C in the combustor. The turbine and compressor are isentropic. Determine: a) the cycle thermal efficiency b) the cycle thermal efficiency if a regenerator with an effectiveness of 85% is added to the system.

5

Approach: The solution approach to part (a) of this problem is similar to that used in previous examples. The use of the regenerator in part (b) requires more thought. The addition of the regenerator does not affect the performance of the compressor or turbine, so those analyses remain the same. The only change when we add the regenerator is to the inlet temperature to the combustor, which changes the heat input required to raise the air temperature to the turbine inlet condition. The inlet temperature to the combustor (outlet temperature from the regenerator) is obtained with conservation of energy applied to the regenerator and use of the regenerator effectiveness.

Solution:

Assumptions:

The cycle thermal efficiency is W W − W ηcycle = net = T C Q Q in

in

Referring to the figure above, the turbine and compressor powers are evaluated assuming steady, negligible potential and kinetic energy effects, and adiabatic [A1], [A2], and [A3] applied to control volumes drawn around each device, so that from conservation of mass and energy, we obtain: W = WT = m ( h3 − h4 ) and W = −WC = m ( h1 − h2 ) Because we have not designated this problem as a cold-air-standard analysis, we must take into account the temperature dependence of the specific heats. Sufficient information is given to evaluate all four enthalpies as was done in Method 1, Example 8-11. Mass flow rate is not needed: it cancels out of the efficiency. For part (a), there is no regeneration, so applying conservation of mass and energy to the control volume around the combustor and assuming steady, negligible potential and kinetic energy effects, and no work [A1], [A2], and [A4], we obtain: Q = Q in = m ( h3 − h2 )

A1. The complete system and individual components are steady. A2. Potential and kinetic energy effects are negligible. A3. The compressor and turbine are adiabatic. A4. No work occurs in the combustor.

When we add the regenerator, the calculations for neither the compressor nor the turbine are affected. The only change from the above analysis is to the input heat transfer rate Qin . Application of conservation of energy to the combustor with the same assumptions as before: Q = Q in = m ( h3 − hx ) because the regenerator has raised the air temperature entering the regenerator from T2 to Tx. We evaluate hx using the definition of heat exchanger effectiveness: Q h −h or hx = h2 + ε ( h4 − h2 ) ε = act = x 2 Qmax h4 − h2 Using Table A-9, we evaluate the two inlet enthalpies:

6

State 1: T1 = 27 ºC = 300 K → h1 = 300.19 kJ/kg Pr1 = 1.386 State 3: T3 = 825 ºC = 1098 K → h3 ≈ 1160 kJ/kg Pr 3 = 167 Assuming an isentropic process [A5] and using the definition of relative pressure ⎛ P4 ⎞ ⎛P ⎞ ⎛ 710 ⎞ ⎛ 100 ⎞ Pr 2 = Pr1 ⎜ 2 ⎟ = 1.386 ⎜ ⎟ = 23.5 ⎟ = 9.84 and Pr 4 = Pr 3 ⎜ ⎟ = 167 ⎜ ⎝ 710 ⎠ ⎝ 100 ⎠ ⎝ P1 ⎠ ⎝ P3 ⎠

A5. The compressor and turbine are isentropic.

From Table A-9 by interpolation a Pr2, h2s = 526 kJ/kg and T2s = 522.3 K; likewise, h4s = 673.6 kJ/kg and T4s = 663 K = 390 ºC. Note that the subscript s is used to indicate this is for an isentropic process. Therefore, without a regenerator, and using these values of enthalpy we obtain: WC kJ kJ = h2 − h1 = ( 526-300 ) =226 kg kg m WT kJ kJ = h3 − h4 = (1160-673.6 ) =484.4 m kg kg The input heat transfer rate is: Q in kJ kJ = h3 − h2 = (1160-526 ) =634 kg kg m Finally, the cycle thermal efficiency is kJ ( 486.4 − 226 ) Wnet kg ηcycle = = = 0.410 kJ Qin 634 kg For Part (b), we now add a regenerator with ε = 0.85. From the above equations: kJ hx = h2 + ε ( h4 − h2 ) = 526 + 0.85 ( 673.6 − 526 ) = 651.5 kg As noted above, only the heat input is affected, so that: Q in kJ kJ = h3 − hx = (1160-651.5 ) =508.5 kg kg m and, finally, ( 486.4-226 ) kJ kg W ηcycle = net = =0.512 508.5 kJ kg Q in

Comments: The regenerator raises the cycle thermal efficiency by a dramatic amount. The temperature of the exhaust gas leaving the regenerator can be found with kJ hy = h4 − ε ( h4 − h2 ) = 673.6 − 0.85 ( 673.6 − 526 ) = 548.1 . kg By interpolation, Ty = 543.6 K = 270.6 ºC which is substantially cooler than that with no regenerator. Note that if we did a cold-air-standard analysis with no regenerator and isentropic compressor and turbine, Eq. 833 would give us ηcycle = 0.429. If we did a cold-air-standard analysis, with a perfect regenerator (ε = 100%) and isentropic compressor and turbine, Eq. S8-6 would give us ηcycle = 0.522. The solution we obtain falls between these two ideal values, as would be expected.

7

8.8 Combined Cycles and Cogeneration The exhaust temperature from a Brayton cycle can be high. (See Example 8-12 where the turbine outlet temperature is 992 ºF = 533 ºC.) This represents a substantial quantity of energy. In Section 8.7 we showed how some of that energy can be recovered through the use of a regenerator. There is a second approach to using this energy as well. Note that the highest temperatures in a typical Rankine cycle are not much higher (See Example 8-8.) than the Brayton cycle exhaust temperature; indeed, the maximum water temperature in the Rankine cycle boiler often does not exceed about 600 ºC. Thus, if we use the exhaust gases from a Brayton cycle as the heat input to a Rankine cycle, we can capture some of the energy that otherwise would be lost. There is a second advantage. The Carnot cycle thermal efficiency increases when the temperature of the heat addition increases or when the temperature of the heat rejection decreases. The Brayton cycle has a much higher high temperature heat addition than the Rankine cycle, and the Rankine cycle has a much lower low temperature heat rejection than the Brayton cycle. Hence, the combined cycle (See Figure S8-4.) has the very attractive attributes of the individual cycles, and the combined cycle efficiency is greater than either of the individual cycles.

Figure S8-4 Simple combined cycle (Brayton-Rankine) power plant For the combined cycle shown in Figure S8-4, the cycle thermal efficiency is defined in the same manner as for the individual cycles:

ηcycle =

Wnet , Brayton + Wnet , Rankine Q

Eq. S8-8

in

The net power produced by each cycle is the difference between the gross power output from the turbines minus the compressor and pump power inputs. In the combined cycle shown, only one heat input is used, the heat input to the Brayton cycle, but there are two heat rejection processes, one at the condenser in the Rankine cycle and the other for the Brayton cycle downstream of the heat exchanger that couples together the two cycles. If insufficient energy can be recovered from the Brayton cycle exhaust (perhaps due to low temperature or flow rate), then a second heat input could be added to the Rankine cycle after the hot gas/steam heat exchanger and before the steam turbine, and Eq. S8-8 would have to be modified as appropriate.

8

Example S8-2 Combined cycle power plant A combined Brayton-Rankine cycle power plant has a net power output of 25 MW. Air enters the compressor ( ηC = 0.87) of the gas turbine cycle at 100 kPa, 300 K, and is compressed to 1500 kPa. The conditions at the inlet to the turbine ( ηT = 0.90) in the gas cycle are 1500 kPa, 1400 K; the exit pressure is 100 kPa. The air then passes through the interconnecting heat exchanger, and is finally discharged at 500 K. Steam enters the turbine ( ηT = 0.92) of the vapor power cycle at 10 MPa, 450 ºC, expands to the condenser pressure of 10 kPa, and exits the condenser as a saturated liquid. The pump isentropic efficiency is 84%. Determine: a) the thermal efficiency of the combined cycle b) the mass flow rates of the air and steam (in kg/s).

Approach: The analysis of this combined cycle power plant is similar to that used for both the Rankine and Brayton cycles. The combined cycle efficiency is ηcycle = Wnet Q in . We need to evaluate the net power output from both cycles, which is done as before. The main difference is in the evaluation of the heat input, which occurs only in the Brayton cycle. However, neither the air nor the steam flow rate is given. We use the given net power output and energy balances around all the turbines and pumps. Doing that, we still have two unknowns (the flow rates) and one equation. The second equation must incorporate both flows. The one device where both flows interact is the heat exchanger connecting the two cycles. Application of conservation of mass and energy to that heat exchanger provides the second equation we need.

9

Solution:

Assumptions:

The combined cycle thermal efficiency is: W + W W ηcycle = net = net , Brayton net , Rankine Q Q in

in

The net power is given. The heat addition rate is calculated by applying conservation of mass and energy to the air passing through the input heat exchanger in the Brayton cycle and assuming steady, negligible potential and kinetic energy effects, and no work [A1], [A2], and [A3] to obtain: m 3 = m 2 = m air and Q = Q in = m air ( h3 − h2 ) The two enthalpies can be evaluated with the given information. The air mass flow rate is determined in two steps. First, the ratio of the air and steam mass flow rates can be evaluated by applying conservation of mass and energy to the interconnecting heat exchanger, because it is the only component in the combined cycle that contains both flows. Assuming steady, negligible potential and kinetic energy effects, no work, and adiabatic [A1], [A2], [A3], and [A4], we obtain: m 5 = m 4 = m air and m C = m B = m steam m steam h4 − h5 = m air hC − hB The enthalpies can be evaluated with the given information. The second step in evaluating the air mass flow rate is to combine the above mass flow rate ratio with the given net power output and energy balances around the turbines, pump, and compressor. We begin with: W = W + W

m air ( h4 − h5 ) + m steam ( hB − hC ) = 0 or

net

net , Brayton

A1. The complete system and individual components are steady. A2. Potential and kinetic energy effects are negligible. A3. No work occurs in any heat exchanger. A4. The pump, turbines, and compressor are adiabatic, as is the interconnecting heat exchanger.

net , Rankine

Applying conservation of mass and energy to the turbine and compressor of the Brayton cycle assuming steady, negligible potential and kinetic energy effects, and adiabatic [A1], [A2], and [A4], we obtain: Wnet , Brayton = WT − WC = m air ⎡⎣( h3 − h4 ) − ( h2 − h1 ) ⎤⎦ We do the same with the Rankine cycle: Wnet , Rankine = WT − W P = m steam ⎡⎣( hC − hD ) − ( hB − hA ) ⎤⎦ All the enthalpies in these two equations can be evaluated with the given information. Now, combining the last four equations: Wnet = m air ⎣⎡( h3 − h4 ) − ( h2 − h1 ) ⎦⎤ + m steam ⎣⎡( hC − hD ) − ( hB − hA ) ⎤⎦

⎧⎪ ⎫⎪ ⎛ m ⎞ = m air ⎨ ⎡⎣( h3 − h4 ) − ( h2 − h1 ) ⎤⎦ + ⎜ steam ⎟ ⎡⎣( hC − hD ) − ( hB − hA ) ⎤⎦ ⎬ ⎝ m air ⎠ ⎩⎪ ⎭⎪ The only unknown in this equation is the air mass flow rate. Using Table A-9, we evaluate the two inlet enthalpies in the Brayton cycle: State 1: T1 = 300 K → h1 = 300.19 kJ/kg Pr1 = 1.386 State 3: T3 = 1400 K → h3 = 1515.42 kJ/kg Pr 3 = 450.5 Using the definition of relative pressure and initially assuming the turbines and compressor are isentropic [A5]: ⎛ P4 ⎞ ⎛P ⎞ ⎛ 1500 ⎞ ⎛ 100 ⎞ Pr 2 = Pr1 ⎜ 2 ⎟ = 1.386 ⎜ ⎟ = 20.79 and Pr 4 = Pr 3 ⎜ ⎟ = 450.5 ⎜ ⎟ = 30.03 ⎝ 100 ⎠ ⎝ 1500 ⎠ ⎝ P1 ⎠ ⎝ P3 ⎠ From Table A-9 at this value of Pr2 by interpolation, h2s = 650.5 kJ/kg; likewise, h4s = 721.7 kJ/kg. Note that the subscript s indicates this is for an isentropic process. We now use isentropic efficiency to evaluate the actual exit enthalpies:

A5. Initially assume the pump, turbines, and compressor are isentropic to evaluate ideal performance.

10

State 2: h2 = h1 +

h2 s − h1

ηC

= 300.19 +

650.5 − 300.19 kJ = 702.9 0.87 kg

State 4: h4 = h3 − ηT ( h3 − h4 s ) = 1515.42 − 0.90 (1515.42 − 721.7 ) = 801.1

kJ . By kg

interpolation, we obtain T4 = 780 K. State 5: 500 K → h5 = 503.02 kJ/kg The steam properties are determined from Tables A-11 and A-12. State A: PA = 10 kPa, saturated liquid → hA = h fA ( PA ) = 191.83 kJ/kg, v A = v fA ( PA ) = 0.001010 m3/kg

State B: PB = 10 MPa → First, we evaluate the pump as if it is ideal [A5], sB = s A so that the exit enthalpy can be evaluated [A6]: hBs ≈ hA + v A ( PB − PA ) ⎛ 1kN m 2 ⎞ ⎛ 1kJ ⎞ kJ ⎛ m3 ⎞ + ⎜ 0.001010 ⎟ (10000-10 ) kPa ⎜ ⎟⎜ ⎟ kg ⎝ kg ⎠ ⎝ 1kPa ⎠ ⎝ 1kNm ⎠ kJ hBs = 191.83 + 10.1 = 201.9 kg Second, we use isentropic efficiency h − hA 201.9 − 191.83 kJ hB = hA + Bs = 191.83 + = 203.9 ηP 0.84 kg

A6. The subcooled liquid approximation is valid.

= 191.83

State C: PC = 10 MPa, TC = 450 ºC superheated vapor → hC = 3240.9 kJ/kg, sC = 6.4190 kJ/kg·K State D: PC = 10 kPa →First, we evaluate the turbine as if it were ideal (isentropic), sD = sC so that hDs = hDs ( PD , sD = sC ) Note that xDs =

hDs − h fD hgD − h fD

=

sDs − s fD sgD − s fD

=

6.4190 − 0.6493 = 0.769 . Solving for hDs 8.1502 − 0.6493

hDs = h fD + xDs h fgD = 1407.56 + 0.769 (1317.1) = 2420.6

kJ kg

Second, we use isentropic efficiency hD = hC − ηT ( hC − hDs ) = 3240.9 − 0.92 ( 3240.9 − 2420.6 ) = 2486.3

m air

kJ kg

The ratio of steam to air mass flow rates is: m steam h4 − h5 801.1 − 503.02 = = = 0.098 m air hC − hB 3240.9 − 203.9 Rearranging the equation for net power output, we obtain: Wnet = ⎛ m ⎞ ⎡⎣( h3 − h4 ) − ( h2 − h1 ) ⎤⎦ + ⎜ steam ⎟ ⎡⎣( hC − hD ) − ( hB − hA ) ⎤⎦ m ⎝ air ⎠ =

25,000 kW kJ {⎡⎣(1515.42-801.1) - ( 702.9-300.19 )⎤⎦ + ( 0.98) ⎡⎣(3240.9-2486.3) - ( 203.9-191.83)⎤⎦} kg

=65.0 kg s Therefore, the steam mass flow rate is: ⎛ m ⎞ m steam = m air ⎜ steam ⎟ = ( 65.0 kg s )( 0.098 ) =6.37 kg s m ⎝ air ⎠

11

The input heat transfer rate is: kg ⎞ kJ ⎛ 1 kW ⎞ ⎛ Q in = m air ( h3 − h2 ) = ⎜ 65.0 ⎟ (1515.53-702.9 ) ⎜ ⎟ =52,850 kW s ⎠ kg ⎝ 1kJ s ⎠ ⎝ Finally, the combined cycle thermal efficiency is: W 25,000 kW ηcycle = net = = 0.473 Qin 52,850 kW

Comments: We can calculate the efficiencies of the individual cycles. For the Brayton cycle, Wnet , Brayton = 20,270 kW and Qin , Brayton = 52,850 kW, so that ηcycle , Brayton = 0.383. For the Rankine cycle, Wnet , Rankine = 4,730 kW and Q in , Rankine = m water ( hC − hD ) = 19,350 kW, so that ηcycle , Rankine = 0.244. As can be seen, the synergistic effect of coupling the two individual cycles to take advantage of their favorable characteristics resulted in the combined cycle efficiency greater than that of either individual cycle. Some industrial processes require both steam and electricity. For example, in an oil refinery electric motors are used to drive pumps and compressors; steam is used in heat exchangers to raise the oil temperature. In cities, electricity is needed to power the lights, computers, and countless other devices; steam is needed, for example, for heating buildings or for providing heat to run an absorption refrigeration plant (a refrigeration cycle that uses chemical reactions rather than vapor compression). In these uses, often the required steam temperature is low. The electricity and steam can be produced in separate plants. However, burning natural gas with its high combustion temperature to produce low-temperature steam is an inefficient use of energy. A process called cogeneration (Figure S8-5) is used in which electricity and steam are produced in the same plant, and the cost to do this is often much less than if two separate plants (one strictly for electricity and one strictly for steam) were used. For industrial use, cogeneration is called combined heat and power (CHP) while for community use it is called district heating.

Figure S8-5 Schematic of a simple cogeneration system A measure of overall plant efficiency or energy utilization can be developed with the basic definition, Eq. S8-8, used above in terms of energy we want to use compared to the energy we purchase. In a cogeneration plant, the energy we use now includes the net work output and the process heat delivered. The energy we purchase is still the heat input to the system. Hence, plant efficiency is defined as:

η plant =

Wnet + Q delivered Q = 1 − out Qin Q in

Eq. S8-9

12