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Group Activity 1: Off-design of a Heat Exchanger Mechanical Engineering - Energy Systems Authors: •Group xx: xxxxx xxxxx (xxxxxxxx), xxxxx xxxxx (xxxxxxxx) •Group xx: xxxxx xxxxx (xxxxxxxx), Fabio Santoro (xxxxxxxx) Lecturer:Prof. Stefano Consonni Teaching Assistants:Riccardo Cremona, Hamidreza Heydari and Nima Razmjoo Academic year:2025-2026Contents 1 Introduction and Data2 2 Design Analysis4 3 Off-design – Scenario A6 4 Off-design – Scenario B11 5 Further considerations between all scenarios15 GA Assessment16 References16 1 Energy Systems- Report GA1 - Groups xx and xx 1.Introduction and Data The scope of this activity is the design of a counter-flow shell & tube heat exchanger (HE) used to heat up a stream of water from environmental temperature to saturated vapor conditions (Figure1). The heat source is hot geothermal brine (water with dissolved salts).Figure 1:Example scheme of a counter-flow shell & tube heat exchanger with one shell pass and one tube pass In the following table, we can find all data necessary to design the device. The brine is defined as hot side (H) and the water as cold side (C): Table 1:General dataNameU.M.Hot Side (H)Cold Side (C) ˙mkg/s 25 5 pbar 50 3 T in◦ C 260 25c pkJ/(kg·K) 4.25 4.2h inkJ/kg - 105.11*The water is heated until it reaches complete vapor saturation. Since it enters the heat exchanger at sub-cooled liquid conditions, the exchanger is divided into two sections: the first section, labeled P RE, where the water is heated to the saturated liquid state, and the second section, labeledE V A, where phase change occurs as the water evaporates from saturated liquid to saturated vapor. These two sections are characterized by different heat exchange coefficients with the respective values of UP RE= 1.5kW/(m2 K) andUE V A= 3kW/(m2 K). 2 Energy Systems- Report GA1 - Groups xx and xx Table2reports selected values from a saturated water table, while Figure2shows the effective- ness–NTU (ε–NTU) relationship for a counter-flow heat exchanger necessary for the off-design phase. Table 2:Properties of Saturated Water (Liquid–Vapor): Pressure Table. Crop from [2]pTh sat,Lh sat,V[bar][ ◦ C][kJ/kg][kJ/kg] 5 151.8 640.2 2748.1 4 143.6 604.7 2738.1 3 133.5 561.5 2724.9 2 120.2 504.7 2706.2 1 99.6 417.4 2674.9 Figure 2:ε−N T Udiagram for counter-flow heat exchanger [1] On the results diagrams we plotted also the saturated water-vapor curves taken from the tables [2]. In order to overlap the same diagram, it was necessary to transform the generic enthalpy valuehsatwith thermal power with the following relation:hsat−→˙m C·(h sat−h in,C). NOTE:For the estimation of the water-vapor saturated properties we used the thermodynamics add- inThermotablefor Excel provided by a research team [3]. With this tool we are able to compute any value for water-vapor: temperature, pressure, enthalpy, and entropy. To maintain coherence with the results, we updated the initial value of the enthalpy of the cold stream in Table1with the value provided by the add-in. As an example, the Excel formula is: hin,C=@h_pT_H2O(p;T)=@h_pT_H2O(300;25)≈105.11kJ/kg where the pressurep= 3 bar = 300 kPamust be expressed in kilopascals, andT= 25◦ C is the water temperature. 3 Energy Systems- Report GA1 - Groups xx and xx 2.Design Analysis In the design analysis we are required to determine every unknown temperatures, draw theT−˙ Q diagram and the heat exchange surface area. First, let’s consider a genericT−˙ Qdiagram. All the framed quantities are unknownFigure 3:GenericT− ˙ Qdiagram From Table2, we can extract the saturation values for a pressurep= 3bar:Tsat=T out,C=@Tsat_p_H2O(p=3bar)≈133.53◦ C hsat,L=@h_px_H2O(p=3bar;x=0)≈561.46kJ/kg hsat,V=@h_px_H2O(p=3bar;x=1)≈2724.89kJ/kg Now we can compute the heat transfer powers˙ QE V Aand˙ QP RE: ˙ QE V A= ˙m C·(h sat,V−h sat,L)≈10817.18kW ˙ QP RE= ˙m C·(h sat,L−h in,C)≈2281.71kW ˙ QT OT=˙ QE V A+˙ QP RE≈13098.89kW Most of the heat power is used for the evaporation phase,≈82.58%of the total power. With the powers we can determine the remaining unknown temperatures with energy balance: ˙ QE V A= ˙m H·c p,H·(T in,H−T mid,H)⇒T mid,H=T in,H−˙ QE V A˙m H·c p,H≈158.19 ◦ C ˙ QP RE= ˙m H·c p,H·(T mid,H−T out,H)⇒T out,H=T mid,H−˙ QP RE˙m H·c p,H≈136.72 ◦ C The remaining part is to compute the heat transfer area with the formula:˙ Q=U·A·LM T D·F For the counter-flow single pass configuration, the value of the correction factor isF= 1. 4T Q T in,H T mid,H T out,H T out,C =T sat T in,C Q EVA Q PRE Energy Systems- Report GA1 - Groups xx and xx With all the temperatures, we evaluate theLM T Ds and the areas: LM T DE V A=(T in,H−T sat)−(T mid,H−T sat)ln  Tin,H−T satT mid,H−T sat ≈62.28KA E V A=˙ QE V AU E V A·LM T D E V A≈57.89m 2 LM T DP RE=(T mid,H−T sat)−(T out,H−T in,C)ln  Tmid,H−T satT out,H−T in,C ≈57.63KA P RE=˙ QP REU P RE·LM T D P RE≈26.40m 2 The total area isAT OT≈84.288m2 with the evaporation part that covers approximately68.7%Figure 4:Design Analysis: Result T-Q Diagram This is the finalT−˙ Qdiagram for the design of the heat exchanger. In red we have represented the brine which is cooling and in blue we heating water. We plotted also the saturated curve from the water-vapor table. We can see that the evaporation process goes from the left side of the saturated curve to the right one (complete evaporation). The pre-heating stage from 25 to 133.53◦ C curve lays on the saturated liquid water curve. The values of the heat capacity of the two fluids are: CH= ˙m H·c p,H= 106.25kW/K andC C= ˙m C·c p,C= 21kW/K. In the pre-heating stage we can see that the slope of the water is higher than the brine in fact, the heat capacity of the water is much lower so, the representation is coherent, instead, in the evaporation phase the specific heat and heat capacity of the water goes to infinite (horizontal line). 5 Energy Systems- Report GA1 - Groups xx and xx 3.Off-design – Scenario A The scenario A of the off-design consider the presence of fouling in the heat exchanger after years of operation, this causes value drop for the heat transfer coefficientUof50%and35%respectively for pre-heating and evaporation sections: Uof f A P RE=Udes P RE·(1−0.5) = 0.75kW/(m2 K)Uof f A E V A=Udes E V A·(1−0.35) = 1.95kW/(m2 K) We can take the scheme for the design case as a reference (see Figure3); however, in this scenario, it is necessary to evaluate the new operating pressure of the cold stream under the constraint that the water must reach the saturated vapor condition. To perform the off-design analysis at the new conditions, a numerical optimization implicit algorithm was developed and implemented in Excel. Due to its complexity, it is not possible to provide a graphical diagram; instead, a simplified pseudo-code representation is proposed below:Algorithm 1Off-Design scenario A numerical optimization algorithm Step 1A: Data definition from design analysis and new constants Step 2A: Definition of numerical optimization variables:Tof f A satandTof f A mid,H Step 3A: Computation of thermodynamic properties at saturation pof f A sat←Tof f A sat hof f A sat,L←Tof f A sat hof f A sat,V←Tof f A sat Step 4A: Computation of heat transfer powers˙ Qof f A E V A←hof f A sat,L,hof f A sat,V ˙ Qof f A P RE←hof f A sat,L Step 5A: Computation of hot stream temperatures Tof f A H,out← ˙ Qof f A P RE,Tof f A mid,H Tof f A in,H,calc← ˙ Qof f A E V A,Tof f A mid,H Step 6A: Computation of LMTD for each section of the heat exchanger LM T Dof f A E V A←Tof f A in,H,calc,Tof f A mid,H,Tof f A sat LM T Dof f A P RE←Tof f A H,out,Tof f A mid,H,Tof f A sat Step 7A: Computation of the heat transfer surface areas: Aof f A E V A← ˙ Qof f A E V A,LM T Dof f A E V A Aof f A P RE← ˙ Qof f A P RE,LM T Dof f A P RE Step 8A: Computation of the number of thermal units: N T Uof f A E V A←Aof f A E V A N T Uof f A P RE←Aof f A P RE Step 9A: Computation of the effectiveness withε−N T Umethod: εof f A E V A,N T U←N T Uof f A E V A εof f A P RE ,N T U←N T Uof f A P RE Step 10A: Computation of the effectiveness with infinite transfer area definition: εof f A E V A,temp←Tof f A in,H,calc,Tof f A mid,H,Tof f A sat εof f A P RE ,temp←Tof f A mid,H,Tof f A sat Step 11A: Constraints and Ob jective function evaluationThe numerical problem is formulated as a constrained optimization using the reduced gradient method. The ob jective function to be minimized and the inequality constraints are described in detail later, when each step of the algorithm is discussed. For each step, the final results of the corresponding variables are presented along with relevant comments. During optimization, the solver changes the value of the variables and repeat every steps from 2A to 11A till it reaches convergence. 6 Energy Systems- Report GA1 - Groups xx and xx Step 1A: Data definition from design analysis and new constants For simplicity, we state that all quantities not labeled with"offA"refer to the design analysis discussed in the previous section. The new values of the heat transfer coefficients,U, for the pre-heating and evaporation sections of the heat exchanger have already been defined at the beginning of this section. Step 2A: Definition of numerical optimization variables This numerical optimization problem requires two variables to be fully defined. We selected two temperatures: the saturation temperature of the cold stream,Tof f A sat, and the temperature reached by the hot stream to supply the heat necessary for the phase change from saturated liquid to saturated vapor,Tof f A mid,H. Both variables are continuous and positive. The final values of these temperatures at convergence are: Tof f A sat≈98.903◦ CTof f A mid,H≈153.677◦ C(variables A) We can clearly observe that the saturation temperature is below the value at atmospheric conditions, this implies that the cold stream side of the heat exchanger operates at a pressure inferior to 1 atm. Step 3A: Computation of the thermodynamic properties at saturation With the value of the saturation temperature we can evaluate all the thermodynamic properties of the water at saturation conditions: pof f A sat=@psat_T_H2O(Tof f A sat)*0.01≈0.975bar= 0.9624atm hof f A sat,L=@h_Tx_H2O(Tof f A sat;0)≈414.47kJ/kg hof f A sat,V=@h_Tx_H2O(Tof f A sat;1)≈2673.84kJ/kg(3.1A) (3.2A) (3.3A) the pressure is below the atmosphere. Step 4A: Computation of the heat transfer powers˙ Qof f A E V A= ˙m C· hof f A sat,V−hof f A sat,L ≈11296.83kW ˙ Qof f A P RE= ˙m C· hof f A sat,L−h in,C ≈1546.78kW ˙ Qof f A T OT= ˙ Qof f A P RE+ ˙ Qof f A E V A≈12843.62kW(4.1A) (4.2A) (4.3A) In this off-design scenario the overall heat power decreased respect to the design case by approximately 1.95%due to the presence of fouling in the heat exchanger, but the impact of the evaporation over the whole process increased to87.96%. Step 5A: Computation of the hot stream temperatures Tof f A in,H,calc=Tof f A mid,H+˙ Qof f A E V AC H≈260.00 ◦ C Tof f A out,H=Tof f A mid,H−˙ Qof f A P REC H≈139.12 ◦ C(5.1A) (5.2A) The output temperature of the hot stream is higher than the design case and this is coherent with a reduction of the overall heat power transferred. If we start from the same temperature and we exchange less heat, keeping the heat capacity the same (the red line slope doesn’t change), the temperature difference decreases. 7 Energy Systems- Report GA1 - Groups xx and xx Step 6A: Computation of the LMTD of each section LM T Dof f A E V A=(T of f A in,H,calc−Tof f A sat)−(Tof f A mid,H−Tof f A sat)ln Tof f A in,H,calc−Tof f A satT of f A mid,H−Tof f A sat! ≈98.558K LM T Dof f A P RE=(T of f A mid,H−Tof f A sat)−(Tof f A out,H−T in,C)ln Tof f A mid,H−Tof f A satT of f A out,H−T in,C! ≈80.849K(6.1A) (6.2A) these values are higher compared to the ones in the design analysis because the hot and cold streams lines are more distanced. Step 7A: Computation of the heat transfer surface areas Aof f A E V A=˙ Qof f A E V AU of f A E V A·LM T Dof f A E V A≈58.780m 2 Aof f A P RE=˙ Qof f A P REU of f A P RE·LM T Dof f A P RE≈25.509m 2 Aof f A T OT=Aof f A P RE+Aof f A E V A≈84.289m2(7.1A) (7.2A) (7.3A) In this off-design scenario the portion of area related to the evaporation section is increased to 69.74%, an higher value of the design case. Step 8A: Computation of the number of thermal units The values of the heat capacities Cof f A min,P RE=C C= 21kW/KCof f A min,E V A=C H= 106.25kW/K Cof f A max,P RE=C H= 106.25kW/KCof f A max,E V A=∞kW/K Cof f A min,P REC of f A max,P RE≈0.198 C of f A min,E V AC of f A max,E V A= 0 the values of the NTUs:N T UE V A=U of f A E V A·Aof f A E V AC of f A min,E V A≈1.079 N T UP RE=U of f A P RE·Aof f A P REC of f A min,P RE≈0.911(8.1A) (8.2A) Step 9A: Computation of the effectiveness with NTU definitionεof f A E V A,N T U= 1−e−N T U of f A E V A≈0.6600 εof f A P RE ,N T U=1−e −h N T Uof f A P RE (1−C minC maxi1− C minC max·e−h N T Uof f A P RE (1−C minC maxi ≈0.5731(9.1A) (9.2A) 8 Energy Systems- Report GA1 - Groups xx and xx Step 10A: Computation of the effectiveness with inf. transfer area definition Figure 5:Infinite Area effectiveness diagrams (left for pre-heating and right for evaporation) Referring to the diagrams in the picture, we can apply theε−N T Umethod to evaluate the effectiveness with an alternative approach: εof f A E V A,temp=∆T CH∆T max=T of f A in,H,calc−Tof f A mid,HT of f A in,H,calc−Tof f A sat≈0.6600 εof f A P RE ,temp=∆T CC∆T max= T of f A sat−T in,CT of f A mid,H−T in,C≈0.5743(10.1A) (10.2A) The values of the effectiveness both for the pre-heating and evaporation are coherent with the ones which are extracted approximately from the diagram2. It is the same also for the step 9A expressions. Step 11A: Ob jective function and constraints evaluation The ob jective function of this numerical problem consists into the minimization of the sum of absolute error of the value of the effectiveness computed in step 9A and 10A. Here the expression and final value:min(X ∆errε) =|εof f A E V A,N T U−εof f A E V A,temp|+|εof f A P RE ,N T U−εof f A P RE ,temp| ≈0.00% + 0.12%≈0.12% the overall difference is less than 1% we can say that is due to numerical approximations. We can also double-check the values of the variables with explicit formulations by combining the expressions (10.1A) and (10.2A) but using the value ofεof the expressions (9.1A) and (9.2A): Tof f A sat,calc=ε of f A P RE ,N T U 1−εof f A E V A,N T U ·Tof f A in,H,calc+ 1−εof f A P RE ,N T U ·Tin,C1−ε of f A P RE ,N T U·εof f A E V A,N T U≈98.647 ◦ C Tof f ,A mid,H,calc=ε of f A E V A,N T U 1−εof f A P RE ,N T U ·Tin,C+ 1−εof f A E V A,N T U ·Tof f A in,H,calc1−ε of f A P RE ,N T U·εof f A E V A,N T U≈153.508 ◦ C These temperatures differ to the variables of 0.255◦ C and 0.169◦ C respectively for the saturation and mid hot stream temperature. 9T Q T mid,H T out,H T in,C Q PRE T sat Q PRE (A=∞) Q T in,H T out,C =T sat Q EVA T mid,H Q EVA (A=∞) Energy Systems- Report GA1 - Groups xx and xx The numerical problem is constrained by two inequality constraints, which are necessary to impose the boundary conditions of the system. From a theoretical point of view, equality constraints should be used; however, since the problem is solved numerically, they would be too restrictive. Therefore, inequality constraints are preferred in numerical optimization problems. The first constraint is referred to the input temperature of the hot stream from the geothermal source. We need to impose that the absolute difference of the value computed in (5.1) is very close to the value given by the data: |Tof f A in,H,calc−T in,H| ≤1E−4(constraint temperature A) The second constraint is needed to impose that the total surface area of the heat exchanger doesn’t vary from the design to the off-design scenario and this is true because we didn’t change the device: |Aof f A T OT−A T OT| ≤1E−3(constraint area A) Here there is the finalT−˙ Qdiagram in which we compare the results of the off-design scenario A to the design analysis in the previous section. The continuous lines represent the design case while, the dashed ones represent the off-design scenario. We see clearly that the saturation pressure and temperature decreased from the design case in fact, the horizontal phase change in the water saturation diagram is below but it still connects the saturated liquid and vapor. The total power is lower in fact the horizontal distance between the points on the right is minimal but, the distance in the saturated liquid is higher. This is coherent with the increased impact of the evaporation power over the total.Figure 6:Off-Design scenario A: Result T-Q Diagram 10 Energy Systems- Report GA1 - Groups xx and xx 4.Off-design – Scenario B Now in the scenario B, we consider that the temperature of the geothermal source has decreased from 260 to 210◦ C due to depletion. In this case we maintain all the conditions from the design analysis but, with a reduction of the inlet hot stream temperature toTof f B in,H= 210◦ C and with an unknown mass flow rate of water. As already mentioned for scenario A, it is also necessary in this case to develop a numerical optimization algorithm to solve the problem. During optimization, the solver changes the value of the variables and repeat every steps from 2B to 11B till it reaches convergence. The corresponding pseudo-code is reported below:Algorithm 2Off-Design scenario B numerical optimization algorithm Step 1B: Data definition from design analysis and new constants Step 2B: Definition of numerical optimization variable:Tof f B mid,H Step 3B: Computation of cold stream mass flow and heat capacity ˙mof f B C←Tof f B mid,H Cof f B C←˙mof f B C Step 4B: Computation of heat transfer powers ˙ Qof f B E V A←˙mof f B C ˙ Qof f B P RE←˙mof f B C Step 5B: Computation of hot stream outlet temperature Tof f B H,out← ˙ Qof f B P RE,Tof f B mid,H Step 6B: Computation of LMTD for each section of the heat exchanger LM T Dof f B E V A←Tof f B mid,H LM T Dof f B P RE←Tof f B mid,H,Tof f B H,out Step 7B: Computation of the heat transfer surface areas: Aof f B E V A← ˙ Qof f B E V A,LM T Dof f B E V A Aof f B P RE← ˙ Qof f B P RE,LM T Dof f B P RE Step 8B: Computation of the number of thermal units: N T Uof f B E V A←Aof f B E V A N T Uof f B P RE←Aof f B P RE,Cof f B C Step 9B: Computation of the effectiveness withε−N T Umethod: εof f B E V A,N T U←N T Uof f B E V A εof f B P RE ,N T U←N T Uof f B P RE Step 10B: Computation of the effectiveness with infinite transfer area definition: εof f B E V A,temp←Tof f B mid,H εof f B P RE ,temp←Tof f B mid,H Step 11B: Constraints and Ob jective function evaluationStep 1B: Data definition from design analysis and new constants For simplicity, we state that all quantities not labeled with"offB"refer to the design analysis discussed in the previous section. The new value of the surface inlet temperature of the hot stream has already been defined at the beginning of this section. Step 2B: Definition of numerical optimization variable This numerical optimization problem requires one variable to be fully defined. We selected the tem- perature reached by the hot stream to supply the heat necessary for the phase change from saturated liquid to saturated vapor,Tof f B mid,H. This temperature is continuous and positive. 11 Energy Systems- Report GA1 - Groups xx and xx The final value of this variable at convergence is: Tof f A mid,H≈146.42◦ C(variable B) Step 3B: Computation of the cold stream mass flow and heat capacity With the value of the mid temperature of the hot stream we can write the energy balance on the evaporation side to obtain the expression of the mass flow and, subsequetly, the heat capacity: ˙mof f B C(h sat,V−h sat,L) = ˙m Hc p,H Tof f B in,H−Tof f B mid,H ↓ ˙mof f B C= ˙m Hc p,HT of f B in,H−Tof f B mid,Hh sat,V−h sat,L≈3.123kg/s Cof f B C= ˙mof f B C·c p,C≈13.12kW/K(3.1B) (3.2B) Step 4B: Computation of the heat transfer powers˙ Qof f B E V A= ˙mof f B C·(h sat,V−h sat,L)≈6755.79kW ˙ Qof f B P RE= ˙mof f B C·(h sat,L−h in,C)≈1425.03kW ˙ Qof f B T OT= ˙ Qof f B P RE+ ˙ Qof f B E V A≈8180.82kW(4.1B) (4.2B) (4.3B) In this off-design scenario the overall heat power decreased respect to the design case by approximately 37.55%due to the mass flow reduction, the impact of the evaporation over the whole process is82.58%. This value coincides with the design case. It is also important to note that the relative decrease in the total heat transfer power matches the relative reduction in the water mass flow rate: 1−˙m of f B C˙m C= 1−˙ Qof f B T OT˙ QT OT≈37.55% Step 5B: Computation of the hot stream outlet temperatureTof f B out,H=Tof f B mid,H−˙ Qof f B P REC H≈133.00 ◦ C(5.1B) The output temperature of the hot stream is a bit lower than the design case. Step 6B: Computation of the LMTD of each section LM T Dof f B E V A=(T of f V in,H−T sat)−(Tof f B mid,H−T sat)ln Tof f B in,H−T satT of f B mid,H−T sat! ≈35.712K LM T Dof f B P RE=(T of f B mid,H−T sat)−(Tof f B out,H−T in,C)ln Tof f B mid,H−T satT of f B out,H−T in,C! ≈44.745K(6.1B) (6.2B) Contrary to the design analysis and the off-design scenario A, in this case the LMTD value of the pre-heating section is higher than that of the evaporation section. The significant reduction in the evaporation LMTD compared to the design case is due to the fact that the initial temperature of the hot stream decreases while the saturation temperature remains constant, resulting in the two lines being closer together. 12 Energy Systems- Report GA1 - Groups xx and xx Step 7B: Computation of the heat transfer surface areas Aof f B E V A=˙ Qof f B E V AU of f B E V A·LM T Dof f B E V A≈63.058m 2 Aof f B P RE=˙ Qof f B P REU P RE·LM T Dof f B P RE≈21.232m 2 Aof f B T OT=Aof f B P RE+Aof f B E V A≈84.289m2(7.1B) (7.2B) (7.3B) In this off-design scenario the portion of area related to the evaporation section is increased to 74.81%, an higher value of the off-design scenario A. Step 8B: Computation of the number of thermal units The values of the heat capacities Cof f B min,P RE=Cof f B C= 13.115kW/KCof f B min,E V A=C H= 106.25kW/K Cof f B max,P RE=C H= 106.25kW/KCof f B max,E V A=∞kW/K Cof f B min,P REC of f B max,P RE≈0.123 C of f B min,E V AC of f B max,E V A= 0 the values of the NTUs:N T UE V A=U E V A·Aof f B E V AC of f B min,E V A≈1.780 N T UP RE=U P RE·Aof f B P REC of f B min,P RE≈2.428(8.1B) (8.2B) Step 9B: Computation of the effectiveness with NTU definitionεof f B E V A,N T U= 1−e−N T U of f B E V A≈0.8314 εof f B P RE ,N T U=1−e −h N T Uof f B P RE (1−C minC maxi1− C minC max·e−h N T Uof f B P RE (1−C minC maxi ≈0.8941(9.1B) (9.2B) Step 10B: Computation of the effectiveness with inf. transfer area definition We can refer to the same diagrams of Figure5of scenario A, we can apply the same expressions: εof f B E V A,temp=∆T CH∆T max=T of f B in,H−Tof f B mid,HT of f B in,H−T sat≈0.8314 εof f B P RE ,temp=∆T CC∆T max= T sat−T in,CT of f B mid,H−T in,C≈0.8938(10.1B) (10.2B) The values of the effectiveness both for the pre-heating and evaporation are coherent with the ones which are extracted approximately from the diagram2. It is the same also for the step 9B expressions. Step 11B: Ob jective function and constraint evaluation The ob jective function of this numerical problem consists into the minimization of the sum of absolute error of the value of the effectiveness computed in step 9B and 10B. Here the expression and final value:min(X ∆errε) =|εof f B E V A,N T U−εof f B E V A,temp|+|εof f B P RE ,N T U−εof f B P RE ,temp| ≈0.00% + 0.03%≈0.03% 13 Energy Systems- Report GA1 - Groups xx and xx the overall difference is less than 0.1% we can say that is due to numerical approximations. We can also double-check the values of the variables with explicit formulations by combining the expressions (10.1B) and (10.2B) but using the value ofεof the expressions (9.1B) and (9.2B): Tof f ,B mid,H,calc=ε of f B E V A,N T U 1−εof f B P RE ,N T U ·Tin,C+ 1−εof f B E V A,N T U ·Tof f B in,H1−ε of f B P RE ,N T U·εof f B E V A,N T U≈146.531 ◦ C This temperature differ to the variable of 0.114◦ C. The numerical problem is constrained by one inequality constraint, which is necessary to impose the boundary conditions of the system. The con- straint is needed to impose that the total surface area of the heat exchanger doesn’t vary from the design to the off-design scenario and this is true because we didn’t change the device: |Aof f B T OT−A T OT| ≤1E−3(constraint area B) Here in the next page we represented theT−˙ Qdiagram comparison (see Figure7) between the design analysis (continuous lines) and the off-design scenario B (dashed lines). We plotted also the saturated curves and we can clearly see that off-design one is just scaled to the left with the same reduction of mass flow rate of the water. Both the cold streams end on the saturated vapor front. We can clearly see that the slope of the of the cold stream changed because of the change of its mass flow rate, with the reduction, we decrease the heat capacity and, as consequence, we increase the line slope: ˙mof f B C ∂ T∂ ˙ Q CFigure 7:Off-Design scenario B: Result B T-Q Diagram 14 Energy Systems- Report GA1 - Groups xx and xx In fact, if we consider just the water stream and we plot them in aT−∆hdiagram (see Figure8) by dividing the heat powers for the corresponding mass flow rate, the curves will perfectly overlap.∂ T∂∆h  C,of f B= ∂ T∂∆h  CFigure 8:Off-Design scenario B: Result B T-∆h Comparison Diagram 5.Further considerations between all scenarios Here in the Figure we can compare the Pre-heating and Evaporation section areas distribution between all case scenarios. In the picture we can clearly see that the evaporation surface (top portion) are in both off-design cases is higher than the design case and it is highest in the off-design scenario B.Figure 9:Pre-heating (bottom) and Evaporation (top) section areas distribution comparison We also provided a fictitious representation of the temperature gradient in each heat exchanger using a color scale bar. This does not represent the actual temperature distribution, since the relationship betweenTand a generic spatial variablexis not linear, as it is in theT−˙ Qdiagram. It is merely a graphical illustration. 1525°C 98°C 133°C 260°C 210°C 130°C Design Off-Design A Off-Design B Energy Systems- Report GA1 - Groups xx and xx We also wrote the parametric relation between the effectiveness in pre-heating and evaporation starting from the expressions (10.1A) and (10.1B) and we plotted the parametric curves for each scenario with the values obtained. εP RE=T sat−T in,C(1−ε E V A)·T in,H+ε E V A·T sat−T in,CFigure 10:ε P RE−ε E V Acomparison The effectiveness values (for both the pre-heating and evaporation sections) increase progressively along the sequence: off-design scenario A, design case, and off-design scenario B. This behavior is due to the progressively smaller temperature difference between the hot and cold streams. In theory, a heat exchanger with unit effectiveness could be achieved ifTsat=T mid,Hbut, it would require and infinite heat transfer surface. GA AssessmentGroupMember NameEvaluation Dario Adella 3 out of 3 14 Andrea Davide Delana 3 out of 3Vijay Mani 3 out of 3 46 Fabio Santoro 3 out of 3References [1]Cüneyt Ezgi.Basic Design Methods of Heat Exchanger. 04 2017. [2]Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, and Margaret B. Bailey.Fundamentalsof Engineering Thermodynamics. Wiley, 10 edition, 2025. [3]Keith Woodbury, Bob Taylor, Joseph Chappell, Kenny Mahan, Jesse Huguet, and Troy Dent. Ther-modynamics.https://excelinme.net/thermodynamics/, 2025. Excel in Mechanical Engineering Research Team. 16