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Words 6124

Pages 25

MECH 421

Huzeyfe SHAHIN

Gokce SAGIR

Usman Arshad SHAH

28th December 2015

Table of Contents

1.

ABSTRACT .............................................................................................................................................. 3

2.

INTRODUCTION ..................................................................................................................................... 3

3.

LITERATURE REVIEW ............................................................................................................................. 3

4.

SIMPLE DESIGN ..................................................................................................................................... 3

4.1 Evaporator at 800 kPa (Counter Flow) ................................................................................................ 4

4.1.1 Preheater ..................................................................................................................................... 4

4.1.2 Boiler ............................................................................................................................................ 4

4.1.3 Superheater ................................................................................................................................. 5

4.2

Evaporator at 800 kPa (Shell & Tube) ........................................................................................... 5

4.3 Evaporator at 650 kPa (Counter Flow) ................................................................................................ 7

4.3.1 Preheater ..................................................................................................................................... 7

4.3.2 Boiler ............................................................................................................................................ 7

4.3.3 Superheater ................................................................................................................................. 8

4.4 Evaporator at 650 kPa (Shell & Tube) ................................................................................................. 8

4.5 Condenser (Counter Flow) .................................................................................................................. 9

4.6 Condenser (Shell & Tube) ................................................................................................................. 11

5.

RATING ................................................................................................................................................ 12

5.1 Evaporator at 800 kPa (Counter Flow) .............................................................................................. 12

5.2 Evaporator at 650 kPa (Counter Flow) .............................................................................................. 14

5.3. Condenser (Counter Flow) ............................................................................................................... 16

5.4. Conclusion ........................................................................................................................................ 17

6.

REFERENCES ........................................................................................................................................ 18

1. ABSTRACT

The report focuses on a superheated ORC for the power plant capacity of 50 to 100 kW using R245fa. Design of all heat exchangers for the cycle are made. The report is divided into simple design and rating to identify the design conditions (diameter, length etc.). The rating part shows the effect on output temperatures, pressure drops and work outputs by changing the mass flow rate of the working and the outer fluid. Furthermore simple design of both counter flow and shell & tube type heat exchangers have been discussed.

2.

INTRODUCTION

Today, 69.3% of the electricity is generated by different types of steam generators, all of which rely on the application of Heat

Exchangers[5]. The aim of this project is to take a closer look at these heat exchangers, and accordingly, design a super-heated

Rankine cycle with a capacity of 50 to 100 kW. According to the literature review, it is found that the most suitable system to fit into this capacity range is Organic Rankine

Cycle (ORC).

3.

LITERATURE REVIEW

Organic Rankine Cycle is a well-known and widely spread thermodynamic cycle that converts heat into work. The heat is applied externally on a closed loop that uses one of the organic fluids such as R11, R123, R113,

R114, R245fa, R245ca, R236fa, and R134a as working fluid [4].

For the capacity range of this project, geothermal sourced Organic Rankine cycle is the optimum solution. Also, it is beneficial for environment since a boiler is not needed in this process and there will be no emission formed.

In geothermal resourced ORC, hot water vapor with temperature between 150°C and 230°Cis used to heat the working fluid to obtain energy, and the condensing fluid with approximately

10°C inlet temperature is used to condense the working fluid coming out of the turbine [1].

According to the performance test for a 100 kW

Organic Rankine Cycle using low temperature geothermal resources, the refrigerant mass flow rate should be 5 kg/s for an ORC designed for

113.1 kW with 9.74 cycle efficiency. This test appeared that the maximum power output for such a system was 91.22 kW with the cycle efficiency of 7.14 % and the measured refrigerant mass flow rate was 4.17 kg/s. The working fluid used for this test was R245fa [1].

On the other hand, one American based company (Infinity Turbine) claims that for a 50 kW Organic Rankine Cycle the mass flow rate of working fluid (R245fa) should be between

7.57 kg/s and 9.46 kg/s. Also, the mass flow rate of cooling fluid should be above 22.7 kg/s. The company also indicates that the dimensions for the cycle should be 92 cm x 152 cm x 183 cm

[2].

Furthermore, for an ORC with a pump efficiency of 40 %, aerodynamic turbine efficiency of 80 %, mechanical turbine efficiency 95% and electrical generator efficiency of 96%, the acceptable or expected pressure drop is 2% [3].

4. SIMPLE DESIGN

Organic Rankine Cycle running with geothermal resource: This part of the project focuses on the size of the heat exchanger. It is decided to have two different heat exchanger designs for two different pressure values (i.e. Counter flow at

800 kPa and 650kPa). The later part consists of the design of the condenser using counter flow type heat exchanger.

(Refrigerant: R245fa)

4.1 Evaporatorat 800 kPa(Counter Flow)

For the geothermal side Nusselt number calculation, the Dittus-Boelter equation for cooling was used from which ho was obtained as: ho =88862.6 W/m2K

Q = 313.592 kW ε= Q/Qmax = 0.91

NTU =

By using the Dittus-Boelter equation again for heating this time, Nu was calculated for a phase changing fluid (R-245fa at 800kPa) from which hi was obtained as:

A=

d

ln ( do ) ∗ di

1 1

1

i

= + R f,ref +

+ R f,wat +

U hi

2k al ho U=513.21 W/m2K

Th,i=Twat,i: 85.54°C Th,o=Twat,o: 77.962°C

Tc,i=Tref,i: 30°C Tc,o=Tref,o: ? mh: 9.9 kg/s

CP,h: 4.180 kJ/kg.K

mc: 4.17 kg/s

CP,c: 1.488 kJ/kg.K

Ch = mh * CP,h = (9.9 kg/s)*(4.180 kJ/kg.K)

= 41.382 kW/K

Cc = mc * CP,c = (4.17 kg/s)*(1.488 kJ/kg.K)

= 6.205 kW/K

Cmin = Cc = 6.205 kW/K

Cmax = Ch = 41.382 kW/K

C* = Cmin/Cmax = 0.15

Qmax= Cmin(Th,i - Tc,i) = 344.6 kW

Q = Qc = Qh = mc* CP,c(Tc,o - Tc,i ) = mh *

CP,h (Th,i - Th,o)

Tc,o = 80.54°C

Q = Qc = Qh = mh * CP,h (Th,i - Th,o)

NTU ∗ Cmin

U

A = 32.16 m2

And then the required L and ΔPlosscan be calculated as:

L=

hi = 618.28 W/m2K

Then the overall heat transfer coefficient can be calculated as:

1 ε − 1) ln ( ∗

)

−1

C ε−1

NTU = 2.66

P:800 kPa di:100 mm do: 110 mm D:130mm kaluminium:240W/mK Rf,ref:0.0002 Rf,wat:0.0001

4.1.1 Preheater

∗

A

= 102.38 m πdi 1

∆ = ( ) ( 2 ) = 0.418

2

4.1.2 Boiler

By using the Kandlikar correlations, Nu was calculated for a phase changing fluid (R-245fa at

800kPa) from which hi was obtained as: hi = 2322.16 W/m2K

Then the overall heat transfer coefficient can be calculated as: d ln ( do ) ∗ di

1 1

1

i

= + R f,ref +

+ R f,wat +

U hi

2k al ho U=1312.78 W/m2K

Th,i=Twat,i: 95.87°C Th,o=Twat,o: 85.54 °C

Tc,i=Tref,i: 80.54°C Tc,o=Tref,o: 80.54°C mh: 9.9 kg/s

mc: 4.17 kg/s

hfg,ref= 152.060 kJ/kg

CP,h: 4.180 kJ/kg.K

CP,c: ∞

Ch = mh * CP,h = (9.9 kg/s)*(4.180 kJ/kg.K)

= 41.382 kW/K

Cc = ∞

Cmin = Ch = 41.382 kW/K

Cmax = Cc = ∞

C* = Cmin/Cmax = 0

Qmax= Cmin(Th,i - Tc,I) = 634.09 kW

Cc = mc * CP,c = (4.17 kg/s)*(1.208 kJ/kg.K)

= 5.035 kW/K

Cmin = Cc = 5.035 kW/K

Q =Qc=Qh=mc(hfg)=mh*CP,h (Th,i - Th,o)

Cmax = Ch = 41.382 kW/K

Q = 427.282 kW

C* = Cmin/Cmax = 0.12

ε= Q/Qmax = 0.674

1

ε − 1)

NTU = ∗ ln ( ∗

)

−1

C ε−1

NTU = 1.12

Qmax= Cmin(Th,i - Tc,i) = 828.683 kW

Q = Qc = Qh = mc* CP,c(Tc,o - Tc,i ) = mh *

CP,h (Th,i - Th,o)

Tc,o = 89.87°C

NTU ∗ Cmin

A=

U

Q = Qc = Qh = mh * CP,h (Th,i - Th,o)

Q = 46.956 kW

A = 35.318 m2

And then the required L and ΔPlosscan be calculated as:

A

L=

= 112.42 m πdi ∆

ε= Q/Qmax = 0.57

NTU =

By using the Dittus-Boelter equation for heating this time, Nu was calculated for a phase changing fluid (R-245fa at 800kPa) from which hi was obtained as:

A=

d

NTU ∗ Cmin

U

A = 7.15 m2

And then the required L and ΔPlosscan be calculated as:

L=

hi = 769.87 W/m2K

Then the overall heat transfer coefficient can be calculated as:

1 ε − 1) ln ( ∗

)

−1

C ε−1

NTU = 0.871

1

= ( ) ( 2 ) = 0.477

2

4.1.3 Superheater

∗

A

= 7.15 m πdi 1

∆ = ( ) ( 2 ) = 2.428

2

ln ( do ) ∗ di

1 1

1

i

= + R f,ref +

+ R f,wat +

U hi

2k al ho The total values are as follows:

U=613.49 W/m2K

L= 237.54 m

Th,i=Twat,i: 97°C Th,o=Twat,o: 95.87 °C Tc,i=Tref,i:

80.54°C Tc,o=Tref,o: ? mh: 9.9 kg/s

CP,h: 4.180 kJ/kg.K

mc: 4.17 kg/s

CP,c: 1.208 kJ/kg.K

Ch = mh * CP,h = (9.9 kg/s)*(4.180 kJ/kg.K)

= 41.382 kW/K

Q= 787.83 kW

ΔPloss= 3.32 kPa

4.2 Evaporator at 800 kPa (Shell & Tube)

Shell and tube heat exchangers are widely used types of heat exchangers due to the flexibility they provide to the designer for a wide range of pressure and temperature, and also, for limited

space that the heat exchanger is supposed to fit for some applications.

=

The correlations used for shell and tube heat exchanger design are as follows; = (0.4) = 0.543 = − ,0 = 0.028 . . =

= 0.1472

ℎ

= 67.25/2

=

4[

=

2

2 √3

− 8,0 ]

4

(. ,0 /2)

= 0.0795

Where dt,0 represents the outer diameter of the tube, PT represents tube pitch and equal to

1.25dt,0, C represents clearance, As represents bundle cross-flow area, DS represents inside diameter of the shell, GS represents equivalent shell diameter and B is baffle spacing.

ℎ

. 0.55 . 1/3 0.14

)

(

) ( )

( )

2

= 749.4/

= 0.36 (

Where ho is heat transfer coefficient for heating water, µ is dynamic viscosity of the water Cp is the specific heat of water, k is the thermal conductivity of the water.

The heat exchanger is designed according the triangular pitch and single pass tubes. Therefore;

=

()2 ,0

= (0.637) (

)(

)

2

= (785) (

2

)(

) = 79

()2 2

,

Where PR is the pitch ratio, △Tlm,cf is the log mean temperature difference, △Tm mean temperature difference, Af is the fouled area, Ac is clean area, and Nt is the number of tubes.

For the design of the shell and tube heat exchanger, three parts, pre-heater, boiler and super-heater, are examined separately By fixing overall length and shell diameter ratio, and by iterating the shell diameter, optimum length values are obtained, for each of the parts and overall length is fixed as 6.6 m for the assumed value of Ds as 1.328 m. Also, surface over design is found for all parts by Af/Ac as 1

Pressure drop for shell side: 0.16 kPa

Pressure drop for tube side: 4.74 kPa

The following formulas was used to calculate pressure drop formulas for shell side and tube side. = exp(0.576 − 0.19 ∗ ln()) = ⁄ =

CTP = 0.90

∆, =

,0

∆ =

∆1 − ∆2

∆

ln (∆1 )

2

= 0.9 => ∆ = (0.9)(∆, )

∆

2

CL = 0.87

=

. ∆

∆ = (

−1

∗ 2 ∗ ( + 1) ∗

2 ∗ ∗ ∗ ∅

4 ∗ ∗ ∗ ∗ 2

+ 4 ∗ ) ∗

2

4.3Evaporatorat 650 kPa(Counter Flow)

Q = Qc = Qh = mh * CP,h (Th,i–Th,o)

For the geothermal side Nusselt number calculation, the Dittus-Boelter equation for cooling was used from which ho was obtained as: Q = 257.447 kW

ho =88862.6 W/m2K

P:650kPa di:100 mm do: 110 mm D:130mm kaluminium:240W/mK Rf,ref:0.0002 Rf,wat:0.0001

ε= Q/Qmax = 0.89

NTU =

∗

NTU = 2.47

A=

4.3.1 Preheater

By using the Dittus-Boelter equation again for heating this time, Nu was calculated for a phase changing fluid (R-245fa at 800kPa) from which hi was obtained as:

d

ln ( o ) ∗ di

1 1

1

di

= + R f,ref +

+ R f,wat +

U hi

2k al ho U=502.42W/m2K

Th,i=Twat,i: 77.42°C Th,o=Twat,o: 71.19°C

Tc,i=Tref,i: 30°C Tc,o=Tref,o: ? mh: 9.9 kg/s

CP,h: 4.180 kJ/kg.K

mc: 4.17 kg/s

CP,c: 1.455 kJ/kg.K

Ch = mh * CP,h = (9.9 kg/s)*(4.180 kJ/kg.K)

= 41.382 kW/K

Cc = mc * CP,c = (4.17 kg/s)*(1.455 kJ/kg.K)

= 6.069 kW/K

Cmin = Cc = 6.069 kW/K

Cmax = Ch = 41.382 kW/K

C* = Cmin/Cmax = 0.15

Qmax= Cmin(Th,i–Tc,i) = 287.8 kW

Q = Qc = Qh = mc* CP,c(Tc,o–Tc,i ) = mh * CP,h

(Th,i–Th,o)

Tc,o = 85.60°C

NTU ∗ Cmin

U

A = 29.85 m2

And then the required L and ΔPlosscan be calculated as:

hi =602.67 W/m2K

Then the overall heat transfer coefficient can be calculated as:

1 ε − 1) ln ( ∗

)

−1

C ε−1

L=

A

= 95.028 m πdi 1

∆ = ( ) ( 2 ) = 0.378

2

4.3.2 Boiler

By using the Kandlikar correlations, Nu was calculated for a phase changing fluid (R-245fa at

800kPa) from which hi was obtained as: hi = 2293.34 W/m2K

Then the overall heat transfer coefficient can be calculated as: d ln ( do ) ∗ di

1 1

1

i

= + R f,ref +

+ R f,wat +

U hi

2k al ho U=1309.97 W/m2K

Th,i=Twat,i: 88.39°C Th,o=Twat,o: 77.42 °C

Tc,i=Tref,i: 72.42°C Tc,o=Tref,o:72.42°C mh: 9.9 kg/s

mc: 4.17 kg/s

hfg,ref= 158.540 kJ/kg

CP,h: 4.180 kJ/kg.K

CP,c: ∞

Ch = mh * CP,h = (9.9 kg/s)*(4.180 kJ/kg.K)

= 41.382 kW/K

Cc = ∞

Cmin = Ch = 41.382 kW/K

Cmax = Cc = ∞

C* = Cmin/Cmax = 0

Qmax= Cmin(Th,i–Tc,I) = 661.11 kW

Ch = mh * CP,h = (9.9 kg/s)*(4.180 kJ/kg.K)

= 41.382 kW/K

Cc = mc * CP,c = (4.17 kg/s)*(1.208 kJ/kg.K)

= 5.035 kW/K

Q =Qc=Qh=mc(hfg)=mh*CP,h (Th,i–Th,o)

Cmin = Cc = 5.035 kW/K

Q = 454.201 kW

Cmax = Ch = 41.382 kW/K

ε= Q/Qmax = 0.687

NTU =

∗

1 ε − 1) ln ( ∗

)

−1

C ε−1

NTU = 1.16

A=

NTU ∗ Cmin

U

A =36.696 m2

And then the required L and ΔPlosscan be calculated as:

L=

A

= 116.81 m πdi 1

∆ = ( ) ( 2 ) = 0.479

2

C* = Cmin/Cmax = 0.12

Qmax= Cmin(Th,i–Tc,i) = 88.52kW

Q = Qc = Qh = mc* CP,c(Tc,o–Tc,i ) = mh * CP,h

(Th,i–Th,o)

Tc,o = 85.6°C

Q = Qc = Qh = mh * CP,h (Th,i–Th,o)

Q = 66.38 kW ε= Q/Qmax = 0.75

NTU =

∗

NTU = 1.469

A=

4.3.3 Superheater

By using the Dittus-Boelter equation for heating this time, Nu was calculated for a phase changing fluid (R-245fa at 800kPa) from which hi was obtained as:

And then the required L and ΔPlosscan be calculated as:

L=

d

NTU ∗ Cmin

U

A = 12.38 m2

hi = 745.18 W/m2K

Then the overall heat transfer coefficient can be calculated as:

1 ε − 1) ln ( ∗

)

−1

C ε−1

A

= 39.39 m πdi 1

∆ = ( ) ( 2 ) = 5.2

2

ln ( do ) ∗ di

1 1

1

i

= + R f,ref +

+ R f,wat +

U hi

2k al ho The total values are as follows:

U=597.7 W/m2K

Q= 778.03 kW

Th,i=Twat,i: 90°C Th,o=Twat,o: 88.39 °C Tc,i=Tref,i:

72.42°C Tc,o=Tref,o: ? mh: 9.9 kg/s

CP,h: 4.180 kJ/kg.K

mc: 4.17 kg/s

CP,c: 1.208 kJ/kg.K

L= 251.23 m

ΔPloss= 6.07kPa

4.4 Evaporator at 650 kPa (Shell & Tube)

Similarly like before in order to decrease the length of the evaporator, Shell & Tube must be used. ℎ

1/3

0.55

.

)

= 749.4/2

= 0.36 (

(

.

)

(

0.14

)

.(

)

The following formulas was used to calculate pressure drop formulas for shell side and tube side. = exp(0.576 − 0.19 ∗ ln())

The heat exchanger is designed according the triangular pitch and single pass tubes. Therefore;

= ⁄

CL = 0.87

=

CTP = 0.90 =

∆, =

∆ =

,0

∆1 − ∆2

4 ∗ ∗ ∗ ∗ 2

+ 4 ∗ ) ∗

2

∆1

)

∆2

= . ∆

∆

()2 ,0

= (0.637) (

)(

)

2

2

= (785) (

∗ 2 ∗ ( + 1) ∗

2 ∗ ∗ ∗ ∅

ln (

= 0.9 => ∆ = (0.9)(∆, )

=

∆ = (

−1

2

)(

) = 82

()2 2

,

Where PR is the pitch ratio, △Tlm,cf is the log mean temperature difference, △Tm mean temperature difference, Af is the fouled area, Ac is clean area, and Nt is the number of tubes.

For the design of the shell and tube heat exchanger, three parts, pre-heater, boiler and super-heater, are examined separately By fixing overall length and shell diameter ratio, and by iterating the shell diameter, optimum length values are obtained, for each of the parts and overall length is fixed as 6.8m for the assumed value of Ds as 1.357 m. Also, surface over design is found for all parts by Af/Ac as 1%.

Pressure drop for shell side: 0.15 kPa

Pressure drop for tube side:13.16 kPa

4.5Condenser (Counter Flow)

The flow that comes from the turbine to the condenser is still in vapor form, 150KPa and

40°C. To have the perfect situation, fully liquid flow at saturation temperature, mass flow rate of the coolant water has to be calculated. For this purpose, whole process divided into two parts, first one is cooling through saturation temperature, and second one is phase change process. The regions and the pinch point temperature difference can be seen in figure1. In addition to that, PPTD was assumed to be 5°C.

By knowing the inlet and outlet temperatures of the cooling water during phase change operation and the heat rejected during phase change from refrigerant fluid, mass flow rate of the cooling water for this perfect case can be determined.

During the whole design procedure, wall thickness assumed to be very small and effects of fouling was neglected. By using try-error method, at the end of the calculations diameters of the tubes was determined to have 2-3 percent pressure loss during whole condensation process. UA2(cooling region)=5,08 KW/K;

To find the required lengths, U has to be determined for both regions. For outer flow

(cooling water) by using Dittus-Boelter equation it was found as; ho =19.3 KW/m2K

“hi” was calculated by using the Dittus-Boelter formula for cooling region; for phase change region following equation was used. Nusselt number for phase change region;

Figure I di=120mm; do=130mm; mref=4,17kg/s;

PPTD=5°C;

P=150KPa;

Twat,in=10°C;

Tref,in=40°C;

Tsat=25.1°C; hfg,ref=160kj/kg; cwat=4,182 kj/(kg*K); cref,g=0,86 kj/(kg*K);

From equality of [mref* hfg,ref= mwat * cwat*(Tsatref-pptd-Twatin)]; mwat found as

15,79 kg/s.

Now, by knowing the mass flowrates, c values, it is possible to calculate UA values for first and second region to calculate total length and total pressure loss. Total heat rejected from the cycle can also be calculated at that step. To determine

UA values, E-NTU method was used just like the boiler design.

Q1= mref* cref,g*(Tref,in-Tref,sat)

Q2=mref* hfg,ref

Qtot,out=Q1+Q2=720KW

For a cycle with 50-100 KW output, this amount of heat rejection seems to be not acceptable but after subtracting this value from Qtot,in; it will give an acceptable work output.

While using the E-NTU method, in the first region Cr will be zero and cmin will equal to mwat*cwat. For the second region, cooling region, mref*cref will be cmin. This mref*cref will be equal to infinity during phase change process. For the formulas that was given in the previous parts for counter flow;

UA1(phase change region)=73KW/K;

Surface temperature was assumed to be constant and it equals to average cooling water temperature during phase region. hi,phase =0,86 KW/m2K hi,cooling =0,45 KW/m2K

U1=0,82KW/K(phase change)

U2=0,44KW/K(cooling)

Length of cooling section is 30,44m;

Length of phase change section is 234,84m;

Total length is 265,30 meters

Also, pressure drop for the tube that refrigerant flows, has to be calculated separately for phase change and cooling parts. 1

∆ = ( ) ( 2 ) 2

This formula can be directly used for cooling section and it says that pressure loss during cooling section equals to 3,46KPa.

For simple design part, at the entrance point of the phase change quality was 1 and at the exit quality was zero. So average quality was calculated and all the values was calculated by using it. Density, viscosity, average velocity etc.

= ∗ ∗ ∗

For phase change region pressure loss was calculated as 0,73KPa.

Then total pressure loss equals to 4,18KPa which meets with our limit.

= (1.58 ∗ ln() − 3.28)−2

=

4.6 Condenser (Shell & Tube)

In this part, two pass shell and tube heat exchanger was designed by using the same conditions with first counter flow type heat exchanger for condenser. The process was divided into two parts; cooling and phase change part. They were thought like two different heat exchangers. For the cooling part, U was assumed initially as

U=0.44 kW/(m2K), which was calculated previously at simple desgn of counter flow type hex. Then, UA was found by using ε-NTU method. After that, area and initial length was calculated. Using the initial length, and the formulas which can be found at the following used to calculate htube (hi) and hshell (ho) values. 0.5 ∗ ∗ ∗

2⁄

3

(1.07 + 12.7 ∗ √(0.5 ∗ ) ∗ (()

ℎ =

− 1))

∗

With these values, new U was calculated and new UA is also availible. Using an iterative process; number of tubes, length and pressure drops were calculated. Restictions were obeyed during this iterative process. The inputs and outputs are the followings;

Tube diameter: 29mm ϕ=1(assumed, which was used at calculation of pressure drop of shell side)

Number of tubes: 20

Length:2.87meters

Shell diameter:190mm

Pressure drop for shell side: 673kPa

Pressure drop for tube side:231kPa

=

∗ ∗ 2 ∗ 2 = 0.637 ∗ (

)

∗ = 0.4 ∗ = 4 ∗ ( 2 − ∗

2

1

)∗

4

∗

= ∗ ( − ) ∗ =

ℎ =

∗ ∗

0.36 ∗ ∗ 0.55 ∗ 0.33

4 ∗ = ∗ 2 ∗ ∗

For phase change part, the procedure is nearly same. U was assumed initially as U=0.90

Kw/(m2*K). For htube (hi) value, nusselt number correlation for phase change was used which is availible in counterflow hex simple design part. For the tube side during phase change, averages of each variables were taken

(average value of liquid viscosity and gas viscosity for example) as an assumption. The inputs and outputs are the followings.

Tube diameter: 60mm ϕ=1(assumed, which was used at calculation of pressure drop of shell side)

Number of tubes: 86

Length:9.67meters

Shell diameter:830mm

Pressure drop for shell side: 0.03kPa

Pressure drop for tube side:1.3Mpa

The following formulas was used to calculate pressure drop formulas for shell side and tube side.

= exp(0.576 − 0.19 ∗ ln())

= ⁄

=

−1

∗ 2 ∗ ( + 1) ∗

∆ =

2 ∗ ∗ ∗ ∅

∆ = (

4 ∗ ∗ ∗ ∗ 2

+ 4 ∗ ) ∗

2

To conclude, length seems to be reasonable for each heat exchanger. However, pressure drops are very high for shell side. Also, the detailed calculations can be provided if required in which all the matlab codes are availible for

“counterflow hex simple design for condenser”,

“Rating of counterflow hex for condenser” and

“simple design of shell and tube hex for condenser”. For this project, the desired work output for group E is between 50 and 100 kW. The work output for the simple design with 800kPa was calculated as 67 kPa and 58 kPa for the 650kPa evaporator which is in the desired range of the

ORC.

5. RATING

This part of the assignment requires changing the mass flow rates of all the fluids by 20 %.

The results found were graphed to show the trend of how changing the mass flow rate affects outlet temperatures and the work output of the overall Organic Rankine Cycle.

In the rating part area was kept constant which was taken from simple design. The approach roughly was as follows for each mass flow rate

(which may change slightly for evaporator or condenser):

Determine overall heat transfer coefficient as to obtain the UA value

Calculate NTU=UA/Cmin

Calculate the effectiveness ɛ(C*, NTU) using the tables

Since Cc and Ch are known Qmax can be determined using the inlet and exit temperatures Q= ε*Qmax may then be found

Finally outlet temperatres can be determined (for the condenser inlet temperature of refrigerant is calculated whereas outlet is kept constant, see part

4. for details.)

5.1 Evaporator at 800 kPa (Counter

Flow)

It is observed that increasing the mass flow rate of the refrigerant decreases both the outlet temperatures (water(hot) and refrigerant(cold)) but has an obvious effect on the outlet temperature of the geothermal source (i.e. the outer and hotter fluid) whereas not a noticeable effect on the outlet temperature of the R245fa refrigerant (Fig 1). This is what was expected from the results. This further shows that increasing mass flow rate on the refrigerant side will reduce the overall heat transfer between the two fluids.

On the other hand increasing the water mass flow rate would increase the outlet temperature of the refrigerant and decrease it for the water but only a minor change is observed (Fig 2). It can also be seen that the increase and decrease in outlet temperatures of the refrigerant and the water respectively starts decreasing, and eventually a mass flow rate will be achieved where a further increase in mass flow rate will not affect the temperature outlet. This implies that increasing mass flow rate on the water side will increase the overall heat transfer between the two fluids to a certain value after which it will remain constant.

900.00

92

850.00

88

Qtot,in (kW)

Temperature (celcius)

90

86

84

82

800.00

750.00

700.00

80

650.00

78

76

600.00

74

3.00

3.00

3.40

3.80

4.20

4.60

5.00

5.40

3.40

3.80

4.20

4.60

5.00

5.40

Mass flowrate of ref (kg/s)

mass flowrate of ref (kg/s)

mref vs Qtot,in mref vs Tho

mref vs Tco

Figure 4

Figure 2

860.00

94

840.00

93

820.00

Qtot,in (KW)

Temperatures(celcius)

93.5

92.5

92

91.5

91

800.00

780.00

760.00

90.5

740.00

90

720.00

89.5

7

8

9

10

11

12

mass flowrate of water mwat vs Tho

mwat vs Tco

Figure 3

The heat input of the cycle increases with increase in the mass flow rate of the refrigerant

(Fig 3) and also with the mass flow rate of the water (Fig 4).

13

7.5

8.5

9.5

10.5

11.5

12.5

mass flowrate of water(kg/s) mwat vs Qtot,in

Figure 5

When the refrigerant side mass flow rate is increased the total pressure losses increase (Fig

5) as it obvious from the formula used for pressure loss. It depends on the square of the mass flow rate and hence increases the pressure losses. increase in mass flow rate will not affect the temperature outlet. This implies that increasing mass flow rate on the water side will increase the overall heat transfer between the two fluids to a certain value after which it will remain constant. 5.00

4.60

3.80

3.40

3.00

2.60

2.20

1.80

1.40

1.00

3.00

3.40

3.80

4.20

4.60

5.00

mass flowrate (kg/s) mref vs Pressure loss

5.40

Temperature (celcius)

Pressure loss (KPa)

4.20

90.00

85.00

80.00

75.00

70.00

65.00

60.00

55.00

50.00

3.00 3.40 3.80 4.20 4.60 5.00 5.40

Figure 6

mass flowrate (kg/s)

The work output graphs are shown and discussed in the conclusion (Fig 19 & 20).All the calculations done using EXCEL 2013 for the evaporator can be made available if required.

mref vs Tho

Figure 7

5.2Evaporator at 650 kPa (Counter Flow)

On the other hand increasing the water mass flow rate would increase both of the outlet temperature of the refrigerant and the water but only a minor change is observed for the refrigerant outlet temperature (Fig 8). It can also be seen that the increase and decrease in outlet temperatures of the refrigerant and the water respectively starts decreasing, and eventually a mass flow rate will be achieved where a further

86

Temperatures(celcius)

It is observed that increasing the mass flow rate of the refrigerant decreases both the outlet temperatures (water(hot) and refrigerant(cold)) but has an obvious effect on the outlet temperature of the geothermal source (i.e. the outer and hotter fluid) whereas not a noticeable effect on the outlet temperature of the R245fa refrigerant (Fig 7). This is what was expected from the results. This further shows that increasing mass flow rate on the refrigerant side will reduce the overall heat transfer between the two fluids.

mref vs Tco

85.5

85

84.5

84

83.5

83

7

9

11

13

mass flowrate of water mwat vs Tho

mwat vs Tco

Figure 8

The heat input of the cycle increases with increase in the mass flow rate of the refrigerant

(Fig 9) and also with the mass flow rate of the water (Fig 10).

900.00

Pressure loss (KPa)

Qtot,in (KW)

850.00

800.00

750.00

700.00

650.00

600.00

3.00

3.40

3.80

4.20

4.60

5.00

5.40

9.80

9.00

8.20

7.40

6.60

5.80

5.00

4.20

3.40

2.60

1.80

1.00

3.00

Mass flowrate (kg/s)

3.80

4.20

4.60

5.00

5.40

mass flowrate (kg/s)

mref vs Qtot,in

mref vs Pressure loss

Figure 9

Figure 11

The work output of the cycle decreases when the mass flow rate of the refrigerant is increased (Fig 12) whereas the vice versa is observed when outer fluid’s mass flow rate is increased (Fig 13). The work output with this evaporator is obviously lower than the work output with evaporator at 800 kPa (Fig 19 &

20). For this reason the efficiency of the cycle is also lower, 7.33 %, compared to the other one

(see conclusion).

840

820

Qtot,in (KW)

3.40

800

780

760

740

720

700

7.5

8.5

9.5

10.5

11.5

12.5

120.00

mass flowrate of water(kg/s)

100.00

Figure 10

When the refrigerant side mass flow rate is increased the total pressure losses increase (Fig

11) as it obvious from the formula used for pressure loss. It depends on the square of the mass flow rate and hence increases the pressure losses. Woutput [kW]

mwat vs Qtot,in

80.00

60.00

40.00

20.00

0.00

3.00

3.50

4.00

4.50

mref [kg/s]

Figure 12

5.00

5.50

60

-30

-20

-10

50

40

Temp [C]

Woutput [kW

100

90

80

70

60

50

40

30

20

10

0

30

20

10

0

10

20

30

change in mouterfluid [%]

0

3.00

3.50

4.00

change in % of mouter fluid

4.50

5.00

5.50

mref [kg/s] mref vs Trefin

Figure 13

All the calculations done using EXCEL 2013 for the evaporator can be made available if required.

5.3. Condenser (Counter Flow)

Changing the mass flow rate of the refrigerant on the condenser increases the inlet temperature and decreases the outlet temperature of the coolant (water) if the outer refrigerant temperature is kept constant (Fig 14). This was done to avoid the outlet refrigerant temperature to reach the subcooled region in order to simplify the calculations. On the other side the temperature outlet of the coolant will decrease when mass flow rate of water is increased (Fig

15). This also corresponds to a lower heat transfer. mref vs Tcoolant,o

Figure 15

The increase in mass flow rate of the refrigerant almost gives a linear increase in the Qout from the condenser (Fig 16). This is obvious since increase in this mass flow rate has a proportional effect on the heat transfer rate. The increase in the coolant mass flow rate also corresponds to a higher Qout of the system but at a decreasing rate (Fig 17). A value will be reached when

Qout does not increase with further increase in mass flow rate of water.

900.0

850.0

800.0

Qout [kW]

Temp [C]

50

40

30

20

750.0

700.0

650.0

600.0

10

12.00

550.0

14.00

16.00

18.00

20.00

mwater [kg/s] mwater vs Tref,i

mwater vs Tcoolant,o

500.0

3.00

3.50

4.00

4.50

mref [kg/s] mref vs Qout

Figure 14

Figure 16

5.00

5.50

5.4. Conclusion

750.0

Qout [kW]

740.0

730.0

720.0

710.0

700.0

690.0

12.00

14.00

16.00

18.00

20.00

mwater [kg/s]

For this project, the desired work output for group E is between 50 and 100 kW. The work output for the simple design was calculated as

67.21 kPa which is in the desired range of the

ORC. The efficiency of the cycle is calculated as

8.53 % which is acceptable as it should be close to 10 % for an ORC [1]. It can be seen that the work output decreases with increasing the mass flow rate of the refrigerant in the ORC. It can be observed from that by decreasing the mass flow rate of refrigerant up to 20%, optimum work output can be achieved.

120.00

mwater vs Qout

100.00

Woutput [kW]

Figure 17

When the refrigerant side mass flow rate is increased the total pressure losses increase (Fig

18) as it obvious from the formula used for pressure loss. It depends on the square of the mass flow rate and hence increases the pressure losses. 80.00

60.00

40.00

20.00

0.00

3.00

3.50

6

4.00

4.50

5.00

5.50

20

30

mref [kg/s]

5.5

Figure 19

4.5

120.00

4

3.5

100.00

Woutput [kW

dPloss [kPa]

5

3

2.5

2

3.00

3.50

4.00

4.50

5.00

80.00

60.00

40.00

20.00

5.50

mref [kg/s] mref vs total dPloss

0.00

-30

-20

-10

0

10

change in mouterfluid [%] change in % of mouter fluid

Figure 18

All the calculations done using MATLAB

R2013a for the condenser will be available in the Appendix of the Final report prepared for this project.

Figure 20

6.

REFERENCES

1. Lee D. H. (et al.), “Development and

Test of a 100 kW Class ORC PowerGenerator for Low Temperature

Geothermal Applications”, 2015.

2. Infinityturbine.com, ‘IT50 ORC Infinity

Turbine', 2015. [Online]. Available: http://www.infinityturbine.com/. [Accessed: Oct-2015].

3. Brasz L. J. , Bilbow W. M., “Ranking of

Working Fluids for Organic Rankine

Cycle Applications”, 2004.

4. Wang E.H. (et al.), “Study of working fluid selection of organic Rankine cycle

(ORC) for engine waste heat recovery”,2011. 5. Instituteforenergyresearch.org,

“Electricity Distribution”,1998. http://instituteforenergyresearch.org/elec tricity-distribution/. [Accessed: Oct2015].

6. S. Kakac and H. Liu, “Heat Exchangers:

Selection, Rating and Thermal Design”,

2002.

7. I. Frank, (et al.), “Principles of Heat and

Mass transfer”, 2013

8. S. Ramesh and S. Dusan,

“Fundamentals of Heat Echanger

Design”, 2003.

9. Y. Almila,”ME 421 Heat Exchanger

Design”, 2014

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