Oil is cooled in a an heat exchanger from a temperature 105 C to a temperature of 50 CPage 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) regardless 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 transferred 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. b. 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. b. 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 between 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 surrounding (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 surrounding (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 surrounding (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 surrounding (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 room is heated using the energy in the storage tank (see figure on the right). 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 AfterDec 07, 2022

Answer To: Oil is cooled in a an heat exchanger from a temperature 105 C to a temperature of 50 CPage 1 of 3...

Banasree answered on Dec 11 2022
47 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.
b.)
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.
b)
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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