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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...

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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 = XXXXXXXXXXK (200oC), TL = XXXXXXXXXXK (20oC) and TI
= XXXXXXXXXXK (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 su
ounding at a rate of 20 
kW. The outside (su
oundings) 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 su
ounding 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 su
ounding 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 
eginning) 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 tu
ine 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 tu
ine is discharged to the su
oundings where it 
eventually comes into equili
ium with it. The su
ounding 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 tu
ine. 
B) the rate of entropy production in the tu
ine. 
C) the rate of exergy destruction in the tu
ine. 
D) the rate of entropy production when the tu
ine exhaust reaches equili
ium with the su
ounding. 
You can consider exhaust air goes into a virtual heat exchanger and exits the virtual heat exchanger at 
the su
ounding temperature and pressure. 
E) the rate of exergy destruction when the tu
ine exhaust reaches equili
ium with the su
ounding. 
F) Using the above results; from an engineering design point would it be better to redesign the tu
ine 
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 After Nov 03, 2022

Solution

Dr Shweta answered on Nov 07 2022
57 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 su
ounding 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...
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