Homework 4 solution

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© 2023 Kenneth Bryden 1 ME 433 Homework 4 - Biomass 1. As-received poplar wood chips have a higher heating value of 14.70 MJ/kg, a moisture content of 18.0%, and an ash content of 7.0%. The ultimate composition of the wood chips is 52.00% C, 6.00% H, and 42.00% O on a dry ash-free basis. a. What is the moisture content of the wood chips on a dry ash-free basis (% wt)? 0.18 kg ! ! " kg "#$%&’( × 1 kg "#$%&’( (1 − 0.18 − 0.07)kg (") = 24.00% = 24.0 ± 0.1% ⇐ b. What is the higher heating value of the wood chips on a dry ash-free basis (MJ/kg daf )? 14.7 MJ kg "#$%&’( × kg "#$%&’( (1 − 0.18 − 0.07)kg (") = 19.60 MJ/kg (") = 19.6 ± 0.1 MJ/kg (") ⇐ c. How many metric tons of CO 2 are produced from combustion of 1.00 metric ton of as- received wood chips (Mg CO2 /Mg as-recd )? (1 − 0.18 − 0.07)Mg (") Mg "#$%&’( × 0.52 Mg * Mg (") × 3.667 Mg *+ ! Mg * = 1.430 Mg *+ ! Mg "#$%&’( = 1.43 ± 0.01 Mg *+ ! /Mg "#$%&’( d. If the poplar chips are dried and then used in a power system that has an overall efficiency of 28.0% based on higher heating value, what are the carbon emissions per MWh of power produced (Mg CO2 /MWh e )? 1.430 Mg *+ ! Mg "#$%&’( × kg "#$%&’( 14.7 MJ ,- × Mg "#$%&’( 1000 kg "#$%&’( × 3600 MJ ,- MWh ,- × MWh ,- 0.28 MWh & = 1.251 Mg *+ ! MWh & = 1.25 ± 0.01 Mg *+ ! /MWh & ⇐ 2. Giant miscanthus is a perennial grass which grows to heights of 10-12 feet and yields 14.0 tons of dry biomass per acre. Assume the grass has an as-received bulk density of 4.00 lbm/ft 3 and an as-received moisture and ash content of 16.0% and 5.0%, respectively. The dry ash-free composition of giant miscanthus is 47.0% C, 9.0% H, and 44.0% O. The biomass has a higher heating value of 6000 Btu/lbm, as-received. A local farm wishes to grow giant miscanthus and use it as fuel in a 500-kW biomass power system. The proposed system will have a capacity factor of 10.0% and an overall efficiency of 19.0%. a. How much CO 2 is released by burning an acre of giant miscanthus (Mg CO2 /ac)? 14 t (%. ac × (1 − 0.16 − 0.05) t (") (1 − 0.16) t (%. × 0.47 t * t (") × 3.667 t *+ ! t * × Mg 1.102 t = 20.59 Mg *+ ! ac = 20.6 ± 0.1 Mg *+ ! /ac ⇐

© 2023 Kenneth Bryden 2 b. What is the minimum number of acres needed to grow the needed giant miscanthus to fuel the power system each year (ac/y)? 500 kW %",&( 1 × 0.10 MW (&/01&%&( MW %",&( × 8760 h y = 438,000 kWh/y 438,000 kWh & y × lbm "#$%&’( 6000 Btu ,- × t 2000 lbm × 3412 Btu ,- kWh ,- × kWh ,- 0.19 kWh & = 655.5 t "#$%&’( y 655.5 t "#$%&’( y × ac 14 t (%. × (1 − 0.16) t (%. t "#$%&’( = 39.33 ac/y = 39.3 ± 0.1 ac/y ⇐ c. How many tons of ash will the system generate each year (t ash /y)? 655.5 t "#$%&’( y × 0.05 t "#- t "#$%&’( = 32.78 t "#- /y = 32.8 ± 0.1 t "#- /y ⇐ d. The giant miscanthus will be dried before storage, what volume of storage is needed to store a one-year supply of fuel (ft 3 )? 655.5 t "#$%&’( y × 2000 lbm "#$%&’( t "#$%&’( × ft 2 4 lbm "#$%&’( = 327,800 ft 2 = 328,000 ± 1000 ft 2 ⇐ 3. Corn stover fuels a 25.0 MW Rankine cycle power plant dedicated to electricity generation. The plant has a capacity factor 91.0% and an overall efficiency of 33.0% based on higher heating value. The corn stover yields 5.00 t/ac on an as-received basis. 60.0% of the corn stover is harvested. The remaining stover is left on the land for conservation purposes. The as-received stover costs $50.00 per ton and contains 23.0% moisture and 8.0% ash. The stover is first dried to 3.0% as-received moisture content before being used to fuel the power plant. ultimate analysis (dry, ash-free) C 46.5% H 5.6% O 47.9% HHV (dry, ash-free) 19.4 MJ/kg a. What is the as-received higher heating value of the corn stover (MJ/kg as-recd )? 19.4 MJ kg (") × (1 − 0.23 − 0.08) kg (") kg "#$%&’( = 13.39 MJ/kg "#$%&’( = 13.4 ± 0.1 MJ/kg "#$%&’( ⇐

© 2023 Kenneth Bryden 3 b. What is the area of land required to harvest the annual corn stover needed to fuel the plant (ac/y)? To solve this problem, we first need to determine how much as-received corn stover is needed annually to fuel the plant (mg as-recd /y) and then determine the land required to provide the corn stover. In determining the how much corn stover is needed annually to fuel the plant, the question that naturally arises when we say the plant has “an overall efficiency of 33% based on HHV, what is the basis for this HHV (i.e., is it dry, ash-free; dry; or as-received). The answer is that the basis doesn’t matter. It only matters that the efficiency is not based on LHV. One way to understand this is to consider that the “basis” is the weight basis as such does not impact the energy part of the analysis. To further explore this, consider the following 25 MW 1 × 8760 hr y × 0.91 MWh (&/ MWh %",&( Energy delivered annual by the plant × MWh ,- 0.33 MWh & × 3600 MJ MWh × kg "#$%&’( 13.39 MJ × Mg 10 2 kg amount of fuel required to deliver 1 MWh = 162,400 Mg "#$%&’( /y In this equation we have used 13.39 MJ/kg as-recd to obtain the needed answer. But let’s dissect the second part of the analysis a little further. MWh ,- 0.33 MWh & × 3600 MJ MWh × kg "#$%&’( 13.39 MJ × Mg 10 2 kg = 0.08872 Mg "#$%&’( MWh & if we use the dry, ash-free HHV but need the answer as Mg as-recd /y this is MWh ,- 0.33 MWh & × 3600 MJ MWh × kg (") 19.4 MJ × kg "#$%&’( (1 − 0.23 − 0.08)kg (") × Mg 10 2 kg = 0.08875 Mg "#$%&’( MWh & The same analysis could be done for using a dry basis. The takeaway is that efficiency is a thermal issue not a mass issue and as a result efficiency is only impacted by whether the HHV or the LHV of the fuel is considered, not the mass basis of the fuel. With the amount of as-received corn stover is needed annually to fuel the plant fuel we can now determine the area of land required to harvest the annual corn stover needed to fuel the plant (ac/y)? 162,400 Mg "#$%&’( y × 1.102 t "#$%&’( Mg "#$%&’( × ac 0.60(5) t "#$%&’( = 59,650 ac y = 59,600 ± 100 ac/y ⇐

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© 2023 Kenneth Bryden 4 4. Located in Kettle Falls, Washington, the Kettle Falls Generating Station began operation in 1983. The 47.0 MW power plant generates electricity via a Rankine steam cycle fueled by direct combustion of wood waste from an adjacent mill. Assume the wood waste has 30.0% moisture content and 5.0% ash, as-received and that based on life cycle analysis the wood waste is 50.0% carbon neutral (e.g., when land use changes, harvesting, and carbon uptake by the wood are considered, the net CO 2 emissions are 50.0% of the direct or gross CO 2 emissions from combustion of the wood waste). The wood-fired power generation plant has an efficiency of 26.0% and a capacity factor of 91.0%. During periods of high demand, the plant supplements its electricity generation with a 7.0 MW natural gas turbine with a heat rate of 8000 Btu/kWh. Assume the plant supplies electrical power to 43,500 homes, each using 9,000 kWh of electricity annually. The natural gas is 80.0% methane (CH 4 ), 10.0% ethane (C 2 H 6 ), 5.0% carbon dioxide (CO 2 ), and 5.0% nitrogen (N 2 ) by volume. The wood waste and natural gas have higher heating values of 12.8 MJ/kg (as-received) and 48.3 MJ/kg, respectively. Neglect transmission losses. The ultimate composition of the dry, ash-free wood waste is given in the table below. Ultimate Composition (wt% dry ash-free) C 50.0% H 6.0% O 44.0% a. How much electricity does the wood-fired portion of the Kettle Falls Generating Station generate each year (MWh/y)? 47 MW 1 × 8760 hr y × 0.91 MWh (&/ MWh %",&( = 374,700 MWh/y = 375,000 ± 1000 MWh/y ⇐ b. How much natural gas is needed to fuel the gas turbine each year (MBtu/y)? 43,500 homes 1 × 9000 kWh home ∙ y × MWh 10 2 kWh = 391,500 MWh/y (391,500 − 374,700) MWh y × 10 2 kWh MWh × 8000 Btu ,-&%K"/ kWh & × MBtu 10 L Btu = 134,400 MBtu y = 134,000 ± 1000 MBtu/y ⇐ c. How many tons of as-received wood waste are required to fuel the plant each year (T as- rec-d /y)? 374,000 MWh & y × 3600 MJ MWh × MJ ,-&%K"/ 0.26 MJ & × kg "#$%&’( 12.8 MJ × 2.204 lbm kg × t 2000 lbm = 446,700 t "#$%&’( /y = 447,000 ± 1000 t "#$%&’( /y

© 2023 Kenneth Bryden 5 d. What is the mass fraction of carbon in the natural gas (% carbon)? The molar composition of a kgmole of natural gas is 0.80 CH M + 0.10 C N H L + 0.05 CO N + 0.05 N N from this the molecular weight of natural gas is 0.80 kmol *!M kmol O", × 16.04 kg *!M kgmol *!M + 0.10 kg *N!L kgmol O", × 30.07 kg *N!L kmol *N!L + 0.05 kg *+N kgmol O", × 44.01 kg *+N kmol *+N + 0.05 kg PN kgmol O", × 28.01 kg PN kmol PN = 19.44 kg/kgmol O", and the mass of carbon in a kgmol of natural gas is 0.80 kmol *!M kmol O", × 12.01 kg * kgmol *!M + 0.10 kg *N!L kgmol O", × 24.02 kg * kmol *N!L + 0.05 kg *+N kgmol O", × 12.01 kg * kmol *+N = 12.61 kg * /kgmol O", from this the mass fraction of carbon, y c , in the natural gas is ࠵? Q = 12.61 kg * /kgmol O", 19.44 kg O", /kgmol O", = 64.87% = 64.9 ± 0.1% ⇐ e. How many metric tons of CO 2 are emitted from the wood-fired portion of the power plant annually (Mg CO2 /y)? 446,700 t "#$%&’( y × Mg "#$%&’( 1.102 t "#$%&’( × (1 − 0.30 − 0.05) Mg (") Mg "#$%&’( × 0.50 Mg * Mg (") × 3.667 Mg *+ 2 | #$%&’( Mg * × 0.5Mg *+ 2 |&*$++$,- Mg *+ 2 |#$%&’( = 241,500 Mg *+ 2 y = 242,000 ± 1000 Mg *+ 2 /y ⇐ f. How many metric tons of CO 2 are emitted from the power plant annually (Mg CO2 /y)? Include the emissions resulting from the wood waste and the emissions from the natural gas combustion. The CO 2 emissions from the gas turbine are 134,400 MBtu y × 1055 MJ MBtu × kg *! . 48.3 MJ × 0.6487 kg * kg O", × 3.667 kg *+ 2 kg * × Mg 10 2 kg = 6983 Mg *+ 2 /y Combining the emissions from the wood-fired Rankine cycle plant and the gas turbine 241,500 Mg *+ 2 y + 6983 Mg *+ 2 y = 248,500 Mg *+ 2 /y = 248,000 ± 1000 Mg *+ 2 /y ⇐

© 2023 Kenneth Bryden 6 5. A gasification/power power production energy complex produces 850,000 MWh of electricity and 60,000 Mg biochar (dry, ash-free basis) annually. The resultant biochar is composed of 85.0% carbon and 6.0% ash on a dry basis. Assume that 80.0% of the biochar remains in the soil on a long-term basis. How much CO 2 is sequestered per MWh of electricity produced (Mg CO2 /MWh)? 60 × 10 2 Mg ’-"%((")) y × Mg ’-"%((%.) (1 − 0.06) Mg ’-"%((")) = 63.83 × 10 2 Mg ’-"%((%.) y 63.83 × 10 2 Mg ’-"%((%.) y × 0.85 Mg * Mg ’-"%((%.) × 0.8 Mg *(’"U,) Mg * × 3.667 Mg *+ 2 Mg * × y 850 × 10 2 MWh (&/ = 0.1873 Mg *+ 2 MWh R ’"U, = 0.187 ± 0.001 Mg *+ 2 MWh R ’"U,

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