Microsoft Word - ENSY_5000_2022_Fall_HW_07.docxENSY 5000 Fundamentals of Energy System Integration Homework Problem Set 7NOTE: To solve this problem set as well as any other problem set, it will...

1 answer below »

QUESTIONS:


Microsoft Word - ENSY_5000_2022_Fall_HW_07.docx ENSY 5000 Fundamentals of Energy System Integration Homework Problem Set 7 NOTE: To solve this problem set as well as any other problem set, it will help if you draw your system and identifying the energy component (property) changes and the energy interactions at the boundaries. In some cases, you may need a sub system analysis to identify the energy interactions. In all your solutions,  clearly show your system definition with proper sketches indicating related energy flows  clearly show how you write your equations (e.g. Mass, Energy, Entropy balance, etc.)  identify relevant energy flows  use of definitions of power-energy relationships and efficiencies as needed  use of proper equation of state, or property relations as needed  clearly show your steps, assumptions, and approximations. READING ASSIGNMENT: Read Chapters 8 of the Textbook. General Information for the following questions We learned in the lectures that a Carnot heat engine (power cycle) can be reversed and used as either a heat pump or a refrigerator. In the case of a heat pump, the heat flow to the higher temperature reservoir is the desired output and it is used to heat a room or building. In the case of the refrigerator, the desired energy flow is the heat in from a region to be cooled. Unless otherwise stated, for this HW set, assume i) devices are operating for a 3‐hour period. ii) electricity available for a green source (wind, photovoltaic, hydro) at a cost of  $0.237/kWh iii) fuel source providing heat at temperature TH is available at a cost of $0.075/kWh. iv) the work input is provided by electricity for the heat pump or refrigerator as needed, v) three reservoirs are at temperatures TH = 473.15 K (200oC), TL = 293.15 K (20oC) and TI = 277.15 K (4oC) to answer the following questions. There are 2 common efficiency metrics for energy systems. One of them is based on 1st law of thermodynamics and the other is based on 2nd Law of Thermodynamics. The former is called Energy Utilization Factor or 1st law efficiency which is defined as ??? ? Desired Utilized Energy OutputRequired Energy Input The latter is called exergetic efficiency or 2nd law efficiency which is defined as ? ? ? Desired Utilized Exergy OutputRequired Exergy Input Answer the following questions based on the definitions of these efficiency metrics. Q1: A room that is kept at constant 20oC temperature has heat loss to the surrounding at a rate of 20  kW. The outside (surroundings) temperature is TI (given above). Heating is supplied to the room using  electrical heaters. Determine  A) electrical energy input to these heaters  B) cost of heating (for the given time‐period above)  C) total entropy production for this this heating system including the heat loss to surrounding at  TI.  D) 1st law (energy) efficiency of this heating system  E) 2nd law (exergy) efficiency of this heating system  Q2: To heat the room given in question 1 above, a Carnot heat pump operating between room  temperature and outside temperature TI is used instead of an electrical heater. Electrical power input  is used to run the heat pump. Determine  A) electrical energy input to the Carnot heat pump B) cost of heating (for the given time‐period above) C) total entropy production for this this heating system including the heat loss to surrounding at  TI. D) 1st law (energy) efficiency of this heating system  E) 2nd law (exergy) efficiency of this heating system Q3: Compare the heating systems in Q1 and Q2 for cost, entropy production, 1st and 2nd law  efficiencies. Which would you recommend? Why?    Q4: A cool storage space is to be kept constant temperature TI by removing heat at a rate of 15 kW.  Suppose a Carnot refrigeration is used by discharging heat to outside at temperature TL. Electrical  power input is used to run the refrigerator. Determine  A) electrical energy input to the Carnot refrigerator B) cost of cooling the storage space (for the given time‐period above) C) total entropy production for this this cooling system including the heat interactions at the high  and low temperatures. D) 1st law (energy) performance of this cooling system considering the heat taken from storage  space is the desired output and work input of the refrigerator is the required input  E) 2nd law (exergy) efficiency of this cooling system exergy taken from storage space is the desired  output and exergy of work input of the refrigerator is the required input Q5: For the same storage space in Q4 above, consider a heat powered refrigerator (a combination of  power and refrigeration devices). Carnot refrigeration cycle operates as in Q4, but power input comes  from a Carnot heat engine operating between a heat source at temperature at TH (given at the  beginning) and outside temperature TL. Determine  A) electrical energy input to this refrigerator B) cost of cooling the storage space (for the given time‐period above) C) total entropy production for this this cooling system including the heat interactions at  temperatures TH, TL and TI. D) 1st law (energy) performance of this cooling system considering the heat taken from storage  space is the desired output and heat input of the heat engine is the required input  E) 2nd law (exergy) efficiency of this cooling system exergy taken from storage space is the desired  output and exergy of heat input of the heat engine is the required input  Q6: Compare the cooling systems in Q3 and Q4 for cost, entropy production, 1st and 2nd law  performances. Which would you recommend? Why?    Q7: An air turbine providing 100 kW power operates adiabatically (no heat, i.e., well insulated) at  steady state. The inlet temperature is 1350 K and inlet pressure is 11 bars. The exit temperature is 753  K and exit pressure is 1 bar. The exhaust of the turbine is discharged to the surroundings where it  eventually comes into equilibrium with it. The surrounding temperature is 290 K and pressure is 1 bar.  Treat air as an ideal gas with an ideal gas constant of Rg = 0.287 kJ/(kg K), and constant pressure  specific heat of cp = 1.005 kJ/(kg K). Ignoring kinetic and potential energy differences, determine,  A) the required mass flow rate of air through the turbine.  B) the rate of entropy production in the turbine.  C) the rate of exergy destruction in the turbine.  D) the rate of entropy production when the turbine exhaust reaches equilibrium with the surrounding.  You can consider exhaust air goes into a virtual heat exchanger and exits the virtual heat exchanger at  the surrounding temperature and pressure.  E) the rate of exergy destruction when the turbine exhaust reaches equilibrium with the surrounding.  F) Using the above results; from an engineering design point would it be better to redesign the turbine  to improve its performance or to design a means to recover energy from the exhaust? Justify your  answer with the numbers you determined above.
Answered 3 days AfterNov 03, 2022

Answer To: Microsoft Word - ENSY_5000_2022_Fall_HW_07.docxENSY 5000 Fundamentals of Energy System Integration...

Dr Shweta answered on Nov 07 2022
43 Votes
Ans 1) A) the electrical energy that can be input to these heaters is calculated as below:
Given:
TH =473.15 K
TL = 293.15 K
TI = 277.15
K
Heat loss = 20 KW or 20000 W
AS, coefficient of performance [COP] of electric heater = 1
Now, the input energy is calculated as:
Win = Q * t/COP
= 20000* 3 *60*60/1= 216 MJ
B) cost of heating is calculated as:
Given electricity cost = $0.237/KWh
And work required = 60kwh
So, total cost of heating = 0.237 *20 *1000 *3 =$14220
C) total entropy production for this this heating system including the heat loss to surrounding at TI.
Total entropy ΔS = 216 /277.15
Change in entropy or total entropy production = 0.779MJ/K
D) 1st law (energy) efficiency of this heating system
Win/Q = 216MJ/216MJ= 100%
E) 2nd law (exergy) efficiency of this heating system
Electric heaters have 100% efficiency as they are capable to convert every watt of electricity into usable heat, with zero waste.
Ans 2) A) the electrical energy that can be input to these heat pumps is calculated as below:
Given:
TH =473.15 K
TL = 293.15 K
TI = 277.15 K
Heat loss = 20 KW or 20000 W
coefficient of performance [COP] = TH/TH-TL
= 293.15/293.15-277.15
= 18.32
Now, the input energy is calculated as:
Win = Q*t/COP
= = 20000* 3 *60*60/18.32= 11.79 MJ
B) cost of heating is calculated as:
Given electricity cost = $0.237/KWh
And work required = 11.79 MJ or equals to 3.27 KWh
So, total cost of...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here