Exercise 1 1. C 2. B 3. F 4. E Exercise 2
1. A portion of the corrections spreadsheet calculation is shown below. Equations shown on the plots of correction curves are used to calculate the correction ratios. Corrected load = measured load divided by the ratios for ambient temperature, ambient pressure and relative humidity.
Elapsed Time
GTG Output
Inlet Temp
Rel.
Humidity Amb Press
Amb Temp
Amp Press
Rel.
Humidity Slope
Rel.
Humidity Intercept
Relative Humidity
Corrected load
Loss
(day) (MW) (F) (%) (Psia) (Ratio) (Ratio) Ratio (MW) (MW)
0.0000 177.6 46.3 93.2 14.83 1.0103 1.0060 0.0001 0.9999 1.0000 174.7 4.4 0.0417 177.1 46.3 94.3 14.83 1.0103 1.0060 0.0001 0.9999 1.0000 174.2 4.8 0.0833 176.3 48.4 94.1 14.83 1.0050 1.0058 0.0001 0.9999 1.0000 174.4 4.6 0.1250 175.5 50.5 90.4 14.82 0.9997 1.0056 0.0001 0.9999 1.0000 174.6 4.5 0.1667 174.5 52.6 82.1 14.82 0.9941 1.0054 0.0001 0.9999 1.0000 174.6 4.4 0.2083 173.1 54.7 76.3 14.81 0.9883 1.0051 0.0002 0.9999 1.0000 174.3 4.7 0.2500 172.2 56.8 77.6 14.81 0.9824 1.0049 0.0002 0.9998 1.0000 174.4 4.6 0.2917 171.7 56.8 73.7 14.80 0.9824 1.0047 0.0002 0.9998 1.0000 174.0 5.0
2. Rate of change of load ~0.1 MW/day. See Figure 3.10.
3. Lost Profit Opportunity (LPO) for a 48‐hour water‐wash outage:
LPO SS MDC WWD FC* * $259 000,
where:
SS: spark spread or gross profit margin = market value of power – operating heat rate (MMBtu/MWh) * gas price ($/MMBtu);
MDC: maximum dependable capability = 284 MW;
WWD: water wash duration = 48 hours;
FC: fixed costs of water wash = $20 000.
4. Integrate the approximate loss curve and add the LPO and sum for one year:
Loss 1 2/ m t WWD/24 b t WWD/24 SS CF MDC SS WWD FC* * * 3655/t where:
t = days;
m = slope of the fouling loss curve (MW/day);
b = intercept of the fouling loss curve (MW).
5. The optimum interval between water washes:
t* WWD/24 2 2 b WWD/* 24 MDC WWD/CF/* 24 FC/CF/SS/24 /m 0 5. t* (optimum wash cycle) = 114 days
6. Savings at optimum rather than at 5 MW:
(a) Losses at the calculated optimum: $1.63 million per year.
(b) Losses at 5 MW loss – 51 days: $2.20 million per year.
(c) Savings = $0.58 million per year.
y = 0.1025x – 0.3843 R2= 0.7512
–2.0 0.0 2.0 4.0 6.0 8.0 10.0 12.0
0 20 40 60 80
Lost output (MW)
Elapsed time (days) Compressor fouling:
Lost output before and after water Post-outage
Pre-outage
Linear (post-outage)
Figure 3.10 Answer: section 3.2.2, number 2.
7. Data scatter could be caused by:
(a) The effects of daily online water washing.
(b) Variable weather conditions that were not culled from the data – sudden showers, for example, that would quickly lower ambient temperature and raise relative humidity.
(c) The effect of pulling liquid through the filter during rain showers.
(d) Correction factors not used in the analysis.
(e) Reducing the effects of data scatter requires more data and a longer sampling time prior to determining the optimum period between washes.
Exercise 3
1. A section of the spreadsheet calculation used to calculate the net present value of the filter installation and savings is shown in Figure 3.11. As shown, the optimum water‐wash period for the HEPA filter has been recalculated to almost a year. Including the outage duration the net present value of the filter upgrade is over $600 000.
2. Since the net present value is positive at the required rate of return a recommendation to replace the filter would be advised.
HEPA differential cost Filter guarantee period Escalation
Tax rate Discount rate
Filter replacement option Year
Natural gas escalation (%/yr) Power price escalation (%/yr) Natural gas price ($/MWh) Power price ($/MWh) Production price ($/MWh) LPC ($MWh)
Assumed rate of change for HEPA (MW/day) Optimum water wash frequency (days) Annual losses current filter ($’000) Projected annual losses HEPA ($’000) Filter differential cost ($’000) Pre-tax savings ($000)
Straight line depreciation of new filter (’000) Taxes ($000)
Annual savings after tax ($’000)
Filter capital cost ($’000) Outage extension cost ($’000) Total filter cost (’000)
NPV@ 11% ($’000)
100,000 2.50%
28.00%
11.00%
0.75%
0.75%
$4.50
$46.22
$28.67
$17.55 0.0103
361.3
$1,627
$521
$ 100
1,007
$
$250
$212
$757
$5,000
$1,716
$6,716
$173.06
$865
$242
$250 1,115
$525
$1,639 $1,651
$528
$250 $250
$244 $216
$873 $773
1,123
$ $ 1,023
108
$532
$1,663
$
$17.68 $17.81
$28.88 $29.10
$46.56 $46.91 $47.26
$29.32
$17.95
$4.53 $4.57 $4.60
0.75%
0.75%
0.75%
0.75%
0.75%
0.75%
$
1 2 3 4
$ 3 years
Figure 3.11 Net present value spreadsheet excerpt.
References
Kurz, R., Brun, K., and Mokhatab, S. (2011) Gas turbine compressor blade fouling mechanisms. Pipeline and Gas Journal 238(9), http://www.pgjonline.com/gas‐turbine‐compressor‐blade‐fouling‐mechanisms?page=show (accessed January 20, 2016).
Stubbs, C. (2009) Implementing Performance Based Maintenance, Saving Energy and Improving Uptime. Genentech, http://www.cypressenvirosystems.com/files/pdf/Genentech_Article_Final_Rev.pdf (accessed January 20, 2016).
Wilcox, M., Baldwin, R., Garcia‐Hernandez, A., and Brun, K. (2010) Guideline for Gas Turbine Inlet Air Filtration Systems, Gas Turbine Research Council, Southwest Research Institute, http://www.gmrc.org/documents/GUIDE LINEFORGASTURBINEINLETAIRFILTRATIONSYSTEMS.pdf (accessed January 20, 2016).
Case Studies in Mechanical Engineering: Decision Making, Thermodynamics, Fluid Mechanics and Heat Transfer, First Edition. Stuart Sabol.
© 2016 John Wiley & Sons, Ltd. Published 2016 by John Wiley & Sons, Ltd.
Companion website: www.wiley.com/go/sabol/mechanical
Flow Instrument Degradation, Use and Placement
The owner of a nuclear power plant has a responsibility to operate the facility in strict accordance with permits granted by local, state, and federal authorities including provisions set forth in the Final Safety Analysis Report (FSAR). One such provision is to operate the nuclear reactor at no more than 100% of its licensed thermal output power (MWt). One method of determining the reactor core power level is the continuous measurement of the thermal input to the steam power cycle – steam calorimetry.
The full‐time operation of a nuclear power plant at 100% reactor, or core, power is often the desire of its owner due to the low marginal cost of production. High‐accuracy measurements of core power ensure that the owner operates as close to the full power level as possible, and reduces the size of emergency systems designed to cool the reactor under the most extreme conditions. Steam‐side calorimetry is very often one of the means of measuring the reactor’s core power and may be compared with others, depending on its established accuracy. The primary contributors to the uncertainty in steam side calorimetry are the measurements of feedwater flow and temperature.
A regulated utility owns and operates a nuclear power plant consisting of two pressurize water‐reactor (PWR) nuclear units, each producing 780 MW of electric power (MWe). Fifteen years after initial operations, the units are operating reliably for long periods at full load. The exception to the good performance is that the electric output is about 2% below what is expected. Records show that the output fell slowly over the past 15 years but it seems to have leveled off at a loss of about 2%. After a major refueling outage of one of the two units, which included an overhaul of the steam turbine generator, and inspection of the steam generating unit (SGU,) the output loss of 2% was confirmed by a performance test. Circumstantial evi- dence of a systematic loss in output was provided by the rate of fuel depletion during the previous fuel cycle.
Case 4
The plant is an important source of revenue for the utility and its continued superior performance is important for investors who purchased bonds for its construction. The state’s public utility commission (PUC) is keenly aware of the facility and its performance. The PUC demands that the utility operate it in a safe, reliable, and efficient manner for the overall benefit of the electric rate payers. Therefore, even the slight loss of output is of paramount importance, which led the senior vice‐president of operations to ask the engineering department to investigate the problem and provide a solution.