CEE4210-HW6-F2022-SOLUTION

docx

School

Ohio State University *

*We aren’t endorsed by this school

Course

4210

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

docx

Pages

Uploaded by AdmiralTree11989

CEE 4210/6210 Renewable Energy Systems – Fall 2022 Assignment 6 SOLUTION Q1. Heat pumps (20 points) A heat pump is to heat a building during a winter heating season from December 1 to March 15 where the heat requirement averages 21 kW for the entire period (i.e., on a 24-hour basis over the entire number of days elapsed). Assume that the building will be heated with an ideal ground-source heat pump that starts with refrigerant 134a in condition of 100% saturated vapor at pressure of P = 0.36 MPa and temperature of T = 5.8°C. The remaining values you need to analyze the heat pump cycle are in the table below. (Hint: You won’t need to use all the values in the table. It is a test of your understanding of the heat pump cycle to see that you can select the values that you do use. The refrigerant is compressed, then passes through a heat exchanger to return to saturated liquid state, then expanded through a throttling device that maintains constant enthalpy, then evaporated in another heat exchanger to return to the beginning of the cycle as a saturated vapor. Electricity for this residential application costs $0.18/kWh. Questions: (a) Sketch the values in the table onto a T-s diagram to visualize how the cycle works. It does not need to be to scale, a hand sketch is fine; (b) calculate the coefficient of performance of this heat pump; (c) the flow rate in kg/second of the R-134a refrigerant required to maintain 21 kW of heat output; (d) the cost of electricity to provide the required heat for the heating season, considering only the energy consumption of the compressor and ignoring electricity used for pumping, friction and thermal losses, etc. Table of values for R-134a: Point Temp Pressure Enthalpy Entropy DegC MPa kJ/kg kJ/kgDegK 1 5.8 0.36 253.81 0.9284 2 57.2 1.4 282.00 0.9284 3 52.4 1.4 276.12 0.9105 4 52.4 1.4 127.22 0.4532 5 5.8 0.36 121.23 0.4532 6 5.8 0.36 127.22 0.4748 1

Solution: Part a. Sketch will look something like this, but with values labeled: Part b. COP is based on compressor work and heat output: Compressor: W = 282 − 253.81 = 28.19 kJ / kg Heat output: Q = 282 − 127.22 = 154.78 kJ / kg COP value: COP = Q out W = 154.78 28.19 = 5.49 Part c. The heating rate of 21 kW is equivalent to 21 kJ/sec. Therefore, the flow rate is q = 21 kJ / s 154.78 kJ / kg = 0.136 kg / s Part d. The heating season is the entire months from December through February plus 15 days in March, so the number of days is 31 + 31 + 28 + 15 = 105. This amount is equivalent to 2520 hours. Power required by the compressor: P = Q COP = 21 5.49 = 3.83 kW . Therefore the total electricity consumed and cost are: E = Pt = 3.83 kW × 2520 h = 9639 kWh Cost = 9639 kWh×$ 0.18 = $ 1735 2

Q2 Effluent Thermal Energy Recovery (ETER) (20 points) The Ithaca Area Wastewater Treatment Facility, or IAWWTF, is a wastewater treatment plant (WWTP) that typically treats 6.5 million gallons of wastewater per day, or, in metric units, ~24.6 million liters per day. Suppose they decide to build up an Effluent Thermal Energy Recovery (ETER) system that diverts 7.6 million liters/day into a heat exchanger to extract heat for both the thermal needs of the plant and to provide to sales to surrounding customers for both building heating and process heat. The working fluid in the heat pump loop has a specific heat of 0.961 kJ/kg°C. The specific heat value of water is 4.186 kJ/kg°C, and you can assume a mass of water of 1 kg per 1 liter. The effluent flow from the WWTP passes through the heat exchanger on the cold side of the ETER system, entering at 10°C and leaving at 7°C. The working fluid enters the heat exchanger on the hot side at 57°C and exits at 41°C. Based on the energy input to the compressor and the heat delivered, and ignoring all other energy inputs and losses, the COP of the system is COP = 3.2. The plant pays $0.105/kWh for electricity. A) What is the rate at which the ETER system delivers heat from the heat exchanger on the hot side, in kW? B) What is the rate at which the refrigerant must flow in kg/s through the heat exchanger on the hot side to deliver the required heat, based on the amount of temperature change? C) Suppose the system runs at this rate continuously for a 24-hour period. What is the cost of the electricity for those 24 hours to operate the compressor? D) Suppose the WWTP could instead purchase natural gas for $11 per GJ and burn it in an 80%-efficient system to provide the same heat as the heat pump. Use a calculation of the 24-hour cost of natural gas to determine which option is cheaper, heat pump or natural gas. Hint: Solving this problem does not require the use of a temperature-entropy diagram. Solution: The solution will actually be provided in the opposite order from which the problem was first posed, first the heat output, then the refrigerant flow rate, then the cost. Part a. Heat removed per day is: 7.6E6 kg× ( 10 − 7 ) × 4.186 = 95.4 GJ / day . Therefore, the compressor work is: W = Q COP − 1 = 95.4 3.2 − 1 = 43.4 GJ / day The heat output is then Q = 95.4 + 4 3.4 = 138.8 GJ / day It can be confirmed that the COP is correct: COP = 138.8 43.4 = 3.2 . There are 86400 seconds per day, so dividing heat output by seconds gives the rate in kW: P = 138.8 GJ 86400 s = 1606 kW Part b. We can use the temperature change and specific heat to convert power to fluid flow: q = 1606 kJ / s 0.961 ( 57 − 41 ) = 104.5 kg / s Part c. The compressor power requirement is: P / COP = 1606 kW / 3.2 = 501.8 kW . Therefore the total electricity consumption per day and cost per day are: 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

501.8 kW × 24 = 12045 kWh / day 12045 ×$ 0.105 = $ 1265 / day Part d. 1606 kW ( 24 h ) = 38544 kWh = 138.8 GJ Including losses, 138.8 GJ ÷ 0.8 = 173.4 GJ . Multiplying by fuel cost of $11/GJ gives $1908/day, so the ETER system is cheaper. Q3 Thermodynamics of a microturbine (20 points) A microturbine has a cycle that starts with air at ambient pressure and 295 K with a relative pressure of 1.3068, compresses the air by a ratio of 3.5 to 1, then heats it to a temperature of 580 K with a relative pressure of 14.38. The enthalpy values at the compressor entrance and at the highest temperature and pressure in the cycle are 295.2 and 586.0 kJ/kg, respectively. The turbine has an output of 65 kW e and the actual efficiency value is 75.8% of the ideal value based on thermodynamic analysis. The energy for the microturbine is provided from biogas generated in an anaerobic biodigester. Over a 24-hour day, the microturbine operates with a capacity factor of 94.5%. A) What is the ideal efficiency value of the microturbine? B) What is the quantity of biogas required to operate the turbine in one 24-hour day, measured in therms? (1 therm = 10 5 Btu). Tables for interpolating the enthalpy values at the compressor exit and turbine exit are given in the table below, and note that this table is available for both the compression of the gas in the compressor and expansion of the gas in the turbine. Temp (K) h (kJ/kg) P-r 400 400.98 3.806 410 411.12 4.153 420 421.26 4.522 430 431.43 4.915 Solution: Part a. Based on the 3.5 : 1 ratio, the relative pressure at State 2 is 1.3068 × 3.5 = 4.57 . Interpolating between the given values gives: 4.57 − 4.522 4.915 − 4.522 = 0.13 h 2 = 421.26 + 0.13 ( 431.43 − 421.26 ) = 422.6 kJ / kg Process for calculating h4 is similar. Based on the 3.5 : 1 ratio, the relative pressure at State 2 is 14.38 ÷ 3.5 = 4.109 . Interpolating between the given values gives: 4.109 − 3.806 4.153 − 3.806 = 0.87 h 4 = 400.98 + 0.87 ( 411.12 − 400.98 ) = 409.8 kJ / kg 4

We can now create a complete table of enthalpy values for all four states, as follows: State h (kJ/kg) 1 295.2 2 422.6 3 586.0 4 409.8 Turbine work, compressor work, and heat added are the following: w t = 586.0 − 409.8 = 176.2 kJ / kg w c = 422.6 − 295.2 = 127.4 kJ / kg Q ¿ = 586.0 − 422.6 = 163.4 kJ / kg Finally the thermal efficiency ignoring losses is as follows: η th = w t − w c Q ¿ = 176.2 − 127.4 163.4 = 0.299 = 29.9% Answer: Ideal efficiency is 29.9% Part b. To calculate biogas required, first calculate actual efficiency: 0.299 × 0.7 45 = 0.226 = 22.6% With a 94.5% capacity factor, the output in 24 hours will be: 65 kW × 0.94.5 × 24 h = 1474.2 kWh The input energy required in units of kWh or therms is the following: 14 74.2 0.226 = 6523 kWh 6523 × 3.6 = 23.5 GJ Answer to Part b: 23.5 GJ. Q4 Economics of investment in biogas electricity generation at IAWWTF (20 points) The Ithaca wastewater plant (IAWWTF) pays $0.105/kWh for electricity that it does not generate itself, being a large commercial user that therefore receives a favorable rate. It installs four 65-kW microturbines of the type analyzed in Q3 at the cost of $3100 per kW. The investment must be repaid over 12 years with a discount rate of 7%. Assume that the turbines operate at the same capacity factor as in Q3 for 365 days per year. All operating cost including maintenance and any other costs are incurred at a rate of $0.02 per kWh produced. How much money could the plant expect to save per year by generating its own electricity with this system instead of purchasing the electricity from the grid? Solution: 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

The cost of the system is 4 × 65 kW ×$ 3100 = $ 806 K Annualizing gives ( $ 806 K , 6% , 15 y ) = $ 83 K The total production based on Q3 is 1.474 MWh / d× 365 × 4 = 2152 MWh / yr The total cost per year including operating cost is then: $ 83000 + 2152 MWh / yr×$ 20 / MWh = $ 126 K To procure the same power from the grid would cost: 2152 ×$ 0.105 × 1000 = $ 226 K Savings are therefore $ 226 K − $ 126 K = $ 100 K / year . Answer: ~$100,000 saved per year. Q5. Energy from municipal solid waste (MSW) (20 points) Consider a hypothetical WTE plant to be built in the greater Buffalo, NY, metro area. This area has a population of about 1.2 million people, according to the U.S. census. Suppose the plant has a nameplate capacity of 65 MW electric , and over the period of a year, 100% output is not possible, so assume an 88% capacity factor to account for downtime, etc. The thermal efficiency of the plant is based on the enthalpy values in the table below, where water is compressed from states (1) to (2), heated and converted to saturated vapor at (3), and then expanded in a turbine to (4). (Hint: pressure values are provided for reference only; you do not need to use them in this problem.) You can assume the plant achieves the thermal efficiency derived from the values in the table, i.e., the process is isentropic and you can ignore losses. The generator efficiency of the plant is 98% and the efficiency of the boiler that conducts the waste is 70%, lower than typical values for a coal-fired plant due to challenges with combusting waste. Municipal solid waste (MSW) can be treated as having an average energy content of 12.6 GJ/tonne (1 metric tonne = 1,000 kg), and the average person in the U.S. generates ~2 kg of MSW per day, including both what they throw out directly and waste directed into the waste stream on their behalf (e.g., from supermarkets, etc.). Suppose the plant is able to obtain as much MSW as it needs from the population of the Buffalo region to meet all its combustion needs. What percent of the Buffalo region’s MSW is consumed by the plant? Table of values: State: Enthalpy (kj/kg) Pressure (kPa) 1. Condensed water 340.54 50 2. Compressed water 346.67 6000 3. Saturated vapor 2784.60 6000 4. Expanded water-vapor mix 2041.62 50 7

Solution: The total electricity produced is 65 MW × 0.88 × 8760 = 501.1 GWh . When converted to Joules, this amount is equivalent to ~1.80 PJ or Petajoules. Based on the table of values, the pump work is 346.67 – 340.54 = 6.13 kJ/kg, the boiler input is 2784.6 – 346.67 = 2437.9 kJ/kg, and the turbine output is 2784.6 – 2041.62 = ~743.0. Therefore the turbine efficiency is: η = wt −℘ q = 743.0 − 6.13 2437.0 = 30.2% The overall efficiency is then 0.7 x 0.302 x 0.98 = ~20.7%. Therefore, the energy input from MSW into the plant per year is Q = 1.80 PJ 0.207 = 8.71 PJ / yr Converting to tons of MSW per year gives (8.71 PJ)/(12.6 GJ/tonne) = ~6.916 x 10 5 tonnes/year. The average resident generates 2 kg per day, equivalent to 0.730 tonnes/year. Therefore, the population generates in one year ( 0.73 tonne ) ( 1.2 M ) = 876,000 tonne and the fraction used is: 691600 876000 = 79% 8