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Oil is cooled in a an heat exchanger from a temperature 105 C to a temperature of 50 C Page 1 of 3 ENSY 5000 Fundamentals of Energy System Integration 2022 Fall Homework Set 10 ...

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Oil is cooled in a an heat exchanger from a temperature 105 C to a temperature of 50 C
Page 1 of 3


ENSY 5000 Fundamentals of Energy System Integration
2022 Fall Homework Set 10


Answer Questions 1 to 4 according to given information below.

A thermal storage system contains 2000 kg liquid water + antifreeze solution at high pressure in a closed rigid
tank. It uses internal energy change of the liquid due to its temperature change as the energy storage
mechanism. Assume pressure of storage liquid is constant and it stays in liquid phase in the tank regardless of
its temperature and it behaves like an incompressible substance with constant density of 1000 kg/m3 and
constant specific heat of 4 kJ/kg∙K. Assume temperature of the liquid in the tank is spatially uniform at any time
(but the temperature varies over time due to usage or energy storage). This thermal storage system provides
heating to a room which is kept at constant 25oC (298K). This is achieved via a closed loop heating system
which circulates another liquid water+antifreeze solution between storage tank and the room using two isolated
heat exchangers. Those heat exchangers have a compact design having sufficiently large surface area such that
temperature of the liquid leaving the heat exchanger in the room (state 1) is always equal to the room
temperature (T1=TR), and temperature of liquid leaving the exchanger in the storage tank is always equal to the
storage temperature (T2=TS). The pump has an advanced control mechanism to adjust the mass flow rate of the
liquid circulating within the heat exchangers to provide constant heating rate to the room (�̇�??? = 5 kW)
egardless of storage tank temperature, ????. Outside atmospheric air is at 7oC (280K) temperature and 1 bar
pressure which can be used as dead (reference) state.





Q1: During charging phase, the pump is not operating, and
therefore there is no heat taken from the storage tank (i.e., QS=0).
Assume, initially, liquid in the storage tank is at 37oC (310 K)
temperature. Suppose energy at a rate �̇�????? is available from the
excess energy supply and it is transfe
ed to the tank until storage
liquid temperature reaches to 227oC (500K) using an electrical
heater placed in the storage tank using excess electrical power
coming from a power source. Suppose thermal resistance between
the liquid inside of the tank and air outside of the tank is ????.
CHARGING PHASE

Therefore, the rate of heat through the tank wall is �̇�?∞ = (???? − ??∞)/????.
A) Note that �̇�????? and ???? are not given. Do not try to find a numerical answer yet. Hence, we are looking for
a parametric solution that includes t, ????, �̇�????? as independent variables. Determine variation of the
temperature of the liquid in the tank as a function of time by applying energy balance equation in rate
form.
Page 2 of 3

B) Suppose thermal resistance, ???? = 0.1 K/W and �̇�????? = 2.5 kW.
a. Plot variation of tank temperature with respect to time until final temperature (500 K) is reached.
. Use your solution in part A to determine how long it takes to reach final temperature.
c. Determine total energy input between initial and final states, i.e., ?????? = ∫ �̇�?????????
d. Determine the storage efficiency (EUF, or 1st law efficiency). Consider energy stored in the tank
as product of this process, and energy supplied by the excess supply as the required input
e. If charging continues indefinitely after storage tank passes 500 K temperature, what would be
steady state temperature of the liquid in the storage tank. Solve this using 2 approaches
i. Equation you obtained for Ts(t) in part A.
ii. Start from the rate form of energy balance equation, apply simplifications for steady-state
operation, and solve for steady-state temperature.
iii. Explain any difference between these two solutions in parts i) and ii).
C) Suppose thermal resistance, ???? = 0.1 K/W and �̇�????? = 50 kW.
a. Plot variation of tank temperature with respect to time until final temperature (500 K) is reached.
. Use your equation in part A to determine how long it takes to reach final temperature.
c. Determine total energy input between initial and final states, i.e., ?????? = ∫ �̇�?????????
d. Determine the storage efficiency (EUF, or 1st law efficiency). Consider energy stored in the tank
as product of this process, and energy supplied by the excess supply as the required input
D) Compare results of part C and D. Explain the differences.

Q2: Assume liquid in the storage tank is initially at 37oC (310 K) temperature and tank temperature reaches
227oC (500K) , at the end of process like in Q1. You can think this process as charging phase of this energy
storage system as in Q1. Storage tank is in a transient operation. However, we are only interested what happens
etween initial and final states. Thus, use integral form of balance equations when needed instead of rate forms
(this is the difference between Q1 and Q2). Ignore the thermal resistance given in Q1 and assume that heat from
storage tank to su
ounding (outside atmospheric air) is 5% of the total energy coming into the storage tank
from electrical power source.
A) (6pt) Determine the change in internal energy, change in entropy, and change in exergy of storage liquid
for this process from initial state to final state.
B) (4pt) Make a sketch of your system indicating system states involved and relevant energy flows for the
solution of following parts.
C) (5pt) Using energy balance equation, find the total energy (coming from excess power supply) used to
charge this storage system, and total heat going from storage tank to su
ounding (outside air), in MJ.
D) (5pt) Use entropy balance equation to find entropy production for the entire charging process in kJ/K.
E) (2pt) Find the exergy destruction for the entire charging process, in MJ.
F) (5pt) Find the energy utilization factor or 1st law efficiency of this charging system. Treat stored energy
in the tank as desired product and electrical energy used to charge as required input. Note that this is also
charging efficiency of this storage system.
G) (5pt) Find the exergetic (2nd law) efficiency of this type of charging process. Treat stored exergy in the
tank as product and exergy of energy input as required input. Note that this is also exergetic efficiency
of charging for this storage system.




Page 3 of 3

Q3: Liquid in the storage system is heated from initial at 37oC (310 K) temperature to final 227oC (500K) as in
Q2. Instead of charging thermal storage system using excess electricity as in Q2, consider charging it using an
excess heat source at 600 K. Assume heat from storage tank to su
ounding (outside atmospheric air) is 5% of
the total energy coming into the storage tank from the heat source.
A) (4pt) Make a sketch of your system indicating system states involved and relevant energy flows for the
solution of following parts.
B) (5pt) Using energy balance equation, find the total energy (coming from excess heat supply) used to
charge this storage system, and total heat going from storage tank to su
ounding (outside air), in MJ.
C) (5pt) Use entropy balance equation to find entropy production for the entire charging process in kJ/K.
D) (2pt) Find the exergy destruction for the entire charging process, in MJ.
E) (5pt) Find the energy utilization factor or 1st law efficiency of this charging system. Treat stored energy
in the tank as desired product and heat used to charge as required input. Note that this is also charging
efficiency of this storage system.
H) (5pt) Find the exergetic (2nd law) efficiency of this type of charging process. Note that this is also
exergetic efficiency of charging for this storage system.


Q4: Now, we will focus on discharging phase which is controlled by
the heat delivery system as shown on the right. Assume there is no
excess energy coming into the storage tank, but the pump is on, and
oom is heated using the energy in the storage tank (see figure on the
ight). Assume state 1 and state 2 pressures are the same, properties
of the circulating liquid are the same as liquid in the storage tank.
Controller of the pump adjust the pump power to deliver constant �̇�??? = 5 kW heating to the room regardless of
the temperature of the storage tank. For simplicity, assume that power required by the pump is proportional to
the mass flow rate of the liquid circulating between the heat exchangers such that �̇�??? = ??�̇�?, where constant of
proportionality is ?? = 2 kJ/kg.

To answer following parts, consider a time snapshot (t), when storage temperature is TS but unknown (i.e.,
Using rate form of balance equations will make solution easier). Assume within a small time-interval including
the time of snapshot (t) this energy delivery system operates under steady-state conditions.
A) Find an expression for required mass flow rate (�̇�?) of the circulating liquid controlled by the pump. In
other words, you will find �̇�? as function of ????. Make a proper system sketch suitable for your solution
indicating all energy flows and relevant states. Make a plot of �̇�? vs ???? for the temperature range from
????=310 K to ???? = 500 K.
B) Find an expression for the rate of heat taken from the storage (�̇�???) as function of storage tank
temperature ????. Make a plot of �̇�??? vs ????. Make a proper system sketch suitable for your solution
indicating all energy flows and relevant states.
C) Find an expression for energy utilization factor (1st law efficiency, EUF or ??) as a function of TS.
Consider rate of heat delivered to the room as desired output (product), rate of energy taken from the
storage tank and power input to the pump as required inputs. Make a plot of this efficiency vs ????. Make
a proper system sketch suitable for your solution indicating all energy flows and relevant states.
D) Find an expression for exergetic (2st law) efficiency as a function of TS. Make a plot of this efficiency vs
????.
E) Is there any optimum value of ???? that maximize 1st and/or 2nd law efficiencies?
Answered 3 days After Dec 07, 2022

Solution

Banasree answered on Dec 11 2022
35 Votes
1.A)
Ein = QH -QL
QL = Magnitude of the heat removed.
QH = Magnitude of the heat rejected
We can rewrite it in terms of thermal resistance Rt = T1-T2/Heat flow(P)
Ohm’s law
t = P*Rt
B)
a.
.)
500 K final temp = 0.02K/W = Rt
c.)
Ein = = Ein – Eout = 2500 – 500 = 2000 = 2K
d.
∆E = Q-W
2 = 5 – W
W = 3
e.i)
Q∞ = (Ts -T∞)/Rt
Ts = (Rt x Q∞) + T∞
ii)
T = T1 – (T1-T2)L * x
iii)
at a steady state with dissimilar temperatures on either side and no internal heat generation. Therefore, the temperature distribution and the heat flux through it.
C)
a.
)
500 K final temp = 0.001K/W = Rt
c)
Ein = = Ein – Eout = 50000 – 500 = 49500 = 49.5K
d)
∆E = Q-W
49.5 = 5 – W
W = 44.5
D) Both the graph had showed a different plot of the tank temperature and the final temperature. If you carefully observe those plots, it is evident that in the first case optimal is not yet meet however in second case optimal was meet.
2.
A)
Let consider the system under steady state
Then,
Qin +Win + = Qout + Wout +
Where,
Θ = h+ke+pe
B)
C)
Energy balance
Ein – Eout = dEsystem/dt = 0(Steady)
Qin + Win + =Qout +Wout +
C) constant specific heat of 4 kJ/kg∙K
q = 4(500-310) = 760kj.k = 0.76 MJ
D)
T1 = 310 K = 318 KJ/Kg
T2 = 500 K = 513.71 KJ/Kg
E)
Ex dest = T∞ * Sgen
= 83.44 x 298 = 24.86MJ
F)
Energy efficiency = 1 – 310/500 = 0.38
G)
Ex1 = q(1 – 310/500) = 4 * 0.38 = 152Kj
Ex efficiency = 152/24860 =...
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