SYSTEM FOR ELECTRICITY GENERATION USING HEAT PUMP AND HYDRO TURBINES

Abstract

A System for Electricity Generation Using Heat Pump and Hydro Turbine, having very high efficiency is disclosed herein. The said system comprises Circular pipes arranged in rectangular pattern to form close circuit of water (10), water pump (1)attached to first end of said closed circuit (10), heat pump, hydraulic turbines (T1(3), T2(4) and T3(5)) placed at other three corners of the closed circuit (10), heater (2). Present invention uses pressure raised after low temperature heating of water in closed circuit (10) for energy generation. The hydraulic turbines are used to capture pressure generated in the closed circuit (10) and to convert it in mechanical work. As system works on low temperature it avoids huge losses of heat energy occurring at currently available mechanisms. The system is simple in construction, compact, low in capital cost, suitable for operation with local water and over a wide range of climatic conditions.

Claims

1. A System for Electricity Generation Using Heat Pump and Hydro Turbine comprising: Circular pipes arranged in rectangular pattern to form close circuit of water (10); water pump (1)attached to first end of said closed circuit (10), heat pump; hydraulic turbines (T1(3), T2(4) and T3(5)) placed at three corners of the closed circuit (10); characterized in that to generate electricity, said system uses pressure raised, by low temperature heating of water in its liquid state in closed circuit (10)

2. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 1 wherein for water heating system uses heat sourse.

3. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 2 wherein the heat source used is external heater (2)

4. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 2 wherein the heat absorbed by the heat absorber (11) from river water is used as heat source for the system

5. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 1 wherein the ratio of cross-section area of the first pipe to other three pipes is 2:1

6. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 1 wherein: the system uses Heat pump cycle to recover the remaining heat in closed circuit (10) after the pressure energy generated in the said circuit has been used by all three turbines; the evaporator (6), conderser (8), compressor (7) and expansion tube (9) are the componets of heat pump;

7. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 1 wherein the water pump is used to feed the water in the system; to provide the one directional low velocity to the water in the system

8. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 1 wherein hydralic turbines T1(3), T2(4) and T3(5) convert the mechanical energy generated due to high pressure condition into the work which is then converted into electricity through generators G1, G2 and G3 respectively.

9. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 8 wherein the hydraulic turbines used in the system is water reaction turbine

10. The System for Electricity Generation Using Heat Pump and Hydro Turbine as claimed in claim 4 wherein the heat absorber (11) is used to absorb the heat from water taken from natural source such as river, lake etc. and no heater (2) is used to provide thermal energy to the system

Description

BRIEF DESCRIPTION OF THE DRAWINGS

[0040] FIG. 1 shows top view of system according to present invention with external heat/thermal source

[0041] FIG. 2 shows the solid view of the present invention with external thermal source FIG. 3 shows the isometric view of the present invention with external thermal source

[0042] FIG. 4 shows top view of system according to present invention which uses river water and heat absorber for electricity generation

[0043] FIG. 5 shows solid top view of the present invention uses river water and heat absorber for electricity generation

[0044] FIG. 6 shows the solid isometric view of the present invention uses river water and heat absorber for electricity generation

[0045] FIG. 7 shows the top solid view of components of heat pump when external heat/thermal source is used in the system

[0046] FIG. 8 shows the isometric solid view of heat pump components when external heat/thermal source is used in the system

[0047] FIG. 9 shows top solid view of components of heat pump when no external heat/thermal source is used in the system

[0048] FIG. 10 shows the isometric solid view of heat pump components when no external heat/thermal source is used in the system

[0049] FIG. 11 shows graphical representation of the entropy in the present invention with external heat source

[0050] FIG. 12 shows graphical representation of the entropy in the present invention uses river water and heat absorber for electricity generation

[0051] As illustrated in FIG. 1 to FIG. 4, present invention comprises a water pump (1), heat pump, pipes arranged in rectangular pattern to form close circuit of water (10), hydraulic reaction turbines (T1(3), T2(4), T3(5)) placed at three corners/end of the closed circuit. The water pump (1) is attached to last end of the close circuit (10). The evaporator (6), compressor (7), condenser (8) and expansion tube (9) are the components of heat pump. Generators G1, G2 and G3 are attached to turbines T1, T2 and T3 respectively.

[0052] In one embodiment, the system may use tap water and external heater (2) to generate electricity in the rectangular shape closed circuit of pipes (10) as shown in FIG. 1 FIG. 2 and FIG. 3.

[0053] In other embodiment, the system may use river water and heat absorber (11) to generate electricity as shown in FIG. 4, FIG. 5 and FIG. 6. In this case heat pump components are used to remove heat from lake /river water and to recover remaining heat in the closed circuit (10) after water has passed through all three turbines.

Working of the System

[0054] As illustrated in FIG. 1 Water at room temperature is allowed to flow through the closed circuit (10). Heater/Boiler (2) is used to provide thermal energy to water. Or as shown in FIG. 3, Heat absorbed from river water trough heat absorber (11) is used to provide thermal energy to the system in FIG. 4.

[0055] The water is heated at low temperature up to 150° C. As the system is closed circuit, water keeps on circulating in the loop and passes through hydraulic turbines T1(3), T2(4) and T3(5). A hydraulic turbine takes water in one direction and gives water out in 90 degree from inlet.

[0056] The Heat pump is divided into two parts. The upper section works on the heat recovery system which takes heat from evaporator (6) and other section which provides the heat from heater/boiler (2). The water in the closed circuit (10) is heated by the heater (2). As the heating chamber is completely filled with water, rise in temperature of water leads to increase in volume thereby increase in pressure. The turbines convert the mechanical energy generated due to high pressure condition into the work which is then converted into electricity through generators.

[0057] The water in the Heat pump is forced to flow in one direction by pump (1). The power required to pump (1) the water is recovered by the generators as energy given by the pump (1) to water is kinetic or the pressure energy. At the end the kinetic energy gets converted into pressure energy which is absorb by the pressure head of the hydraulic turbines.

[0058] Initially as water is not uniformly heated, the first turbine T1(3) absorbs less amount of power from closed circuit (10). After water passes through hydraulic turbine T1(3), heat gets uniformly distributed in the closed circuit (10) due to orderly distribution of water through the turbine's fins. The water now has high pressure due to uniform heating. The second turbine T2(4) absorbs large amount of pressure to generate power. The third turbine T3(5) absorbs the remaining pressure so as to match the inlet pressure of the pump (1). The pressure created due to loss in kinetic energy of water at corners is absorb by the third generator T3(5). The outlet of third turbine T3(5) is connected to evaporator (6) to reuse heat by Heat Pump which is then used for heating by condenser (8). Water vapour is converted to water in the condenser (8) and then that water is fed into the expansion tube (9). In best mode of the system, the water is heated up to 150° C. temperature and 10 bar base pressure. When temperature is rose by 1° C. i.e. when water is heated from 150° C. to 151° C. the system gets maximum increase in pressure i.e. 19 bar which is then absorb by the hydraulic turbines to generate electricity.

[0059] Following calculations shows maximum pressure generated through the system when water temperature is increased by 1° C. temperature from 150° C. to 151° C. by way of example:

[0060] A. When System Uses External Heat Source and Tap Water: [0061] Case 1: Water of 1 m.sup.3 was passed through in closed circuit of pipes and was heated to rose its temperature from 1500 C to 151° C. the total pressure rose was calculated in following ways:

[0062] Water at 150° C. has density of 917 Kg/m.sup.3, Specific heat is 4.3 KJ/Kg and volume of 0.001091 m.sup.3/Kg. Also, its isothermal compressibility is 0.0006204/MPa.

[0063] As Bulk Modules=1/compressibility=1611.86 MPa=16118.6 bar

[0064] To Calculate volumetric expansion of water after 150° C. for 1° C. rise.

[0065] Standard values from Steam Table:

[0066] State 1: Pressure=4.8 bar

[0067] Temperature=150.31° C.

[00001]$\mathrm{Volume}=0.001091m^3/\mathrm{Kg}\mathrm{Density}=916.59{\text{Kg/m}}^3$

[0068] State 2: Pressure=5 bar

[0069] Temperature=151.84° C. Volume=0.001093 m.sup.3 /Kg Density=914.91 Kg/m.sup.3

[0070] Let consider 1 kg of water so total temperature change and volume change in above state is

[00002]$T0=151.84-15031=1.53°C.V0=0.001093-0.001091=0.000002m^3\mathrm{Therefore},\mathrm{Volume}\mathrm{change}\mathrm{per}\mathrm{degree}\mathrm{Celsius}=V0/T0=0.000002/1.53=1.307\times {10}^{-6}{m}^3$

[0071] So volumetric expansion of water after 150° C. for 1° C. rise is 1.307×10.sup.−6 m.sup.3/Kg

[0072] Hence if the water of 1 m.sup.3 was placed in closed circuit and was heated to rise its temperature from 150° C. to 151° C. the total pressure rise was:

[00003]$\mathrm{Rise}\mathrm{in}\mathrm{Total}\mathrm{Volume}\mathrm{if}\mathrm{allowed}\mathrm{to}\mathrm{expand}=\mathrm{Density}\times \mathrm{Expansion}\mathrm{per}\mathrm{Kg}=917\times 1.307\times {10}^{-6}=1.1985\times {10}^{-3}{m}^3$$\mathrm{As}\mathrm{bulk}\mathrm{modules}=\left(P1-P0\right)/\left(\left(V1-V0\right)/V0\right)$$\mathrm{Where}P0=\mathrm{Initial}\mathrm{Pressure}$$P1=\mathrm{Final}\mathrm{Pressure}$$V0=\mathrm{Initial}\mathrm{Volume}$$V1=\mathrm{Final}\mathrm{Pressure}$

[0073] Hence here the volume taken is 1 m.sup.3, therefore V0=1 m.sup.3

[0074] V1−V0 is the total volume change.

[0075] Here considers the initial pressure is 5 bar, therefore P0=5 bar P1−P0 is the total pressure rise.

[0076] The Bulk modules is found to be 16118.6 bar

[0077] Therefore, 16118.6=Pressure rise/1.1985×10.sup.−3

[0078] Pressure rise=19.31 bar

[0079] That's is approximately 19 bars [0080] Case 2: Supply total heat to the 5 Kg of water instead of total water in the 1 m.sup.3

[0081] The specific heat of water at 150° C. is 4.3 KJ /Kg. Therefore, for 1 m.sup.3 of water to get 1° C. rise the total heat required is Q=4.3×Density=4.3×917=3943.1 KJ

[0082] Supply total heat to the 5 Kg of water instead of total water in the 1 m.sup.3 So heat given to each Kg of water is

[00004]$Q0=3943.1/5=788.62\mathrm{KJ}$

[0083] The Enthalpy of water at 150° C. is 633.5 KJ/Kg (H1)

[0084] Hence the total resultant Enthalpy is

[00005]$H=H1+Q0=633.5+788.62=1422.12\mathrm{KJ}$

[0085] The Properties of water with 1422.12 KJ/Kg according to the steam table are 103.37 bar, 313.267° C. and volume 0.00146546 m.sup.3 /Kg.

[00006]$\mathrm{Total}\mathrm{Volume}\mathrm{Increase}=5\mathrm{Kg}\times \left(0.00146546-0.001091\right)=1.8723\times {10}^{-3}$

[0086] Hence the pressure rise will be

[00007]$P=\mathrm{Total}\mathrm{Volume}\mathrm{Increase}X\mathrm{Bulk}\mathrm{Modules}=1.8723\times {10}^{-3}\times 16118.6=30.17\mathrm{bar}$

[0087] Hence the Approximate Pressure rise is 30 bars. In Case 1, pressure head has 19 bar pressure In Case 2, pressure head has 30 bar pressure.

[0088] The 1 bar pressure head for 1 m.sup.3 of water gives power as 11.11 m depth gives 1 bar pressure at 917 Kg/m.sup.3 density

[0089] Therefore

[0090] For 1 bar Pressure head, Power=mass*gravitational acceleration*height=917×9.81×11.11=99943.0 J=99.94 KJ

[0091] Hence the Approximate Power is 100 KJ

[00008]$\mathrm{Therefore},\mathrm{in}\mathrm{Case}1\mathrm{Power}\mathrm{Generated}=\mathrm{Total}\mathrm{Pressure}\mathrm{Head}X\mathrm{Power}\mathrm{per}\mathrm{head}=19\times 100\mathrm{KJ}=1900\mathrm{KJ}=1.9{\text{MJ/m}}^3$$\mathrm{In}\mathrm{case}2\mathrm{Power}\mathrm{Generated}=\mathrm{Total}\mathrm{Pressure}\mathrm{Head}X\mathrm{Power}\mathrm{per}\mathrm{head}=30\times 100\mathrm{KJ}=3{\text{MJ/m}}^3$$\mathrm{Total}\mathrm{heat}\mathrm{input}\mathrm{for}1m^3\mathrm{is}Q=\mathrm{Specific}\mathrm{heat}\times \mathrm{Total}\mathrm{mass}=4.3\times 917=3.9{\text{MJ/m}}^3$

[0092] Therefore, efficiency of the system is derived as

[00009]$\mathrm{Case}1=1.9/3.9=48.7\%$$\mathrm{Case}2=3/3.9=76.92\%$

[0093] Heat Pump Calculations

[0094] Remaining heat can be recovered by the heat pump

[0095] The heat pump with water as working fluid has a COP of 7

[0096] So small part of output power can be used to run a heat pump to recover total heat, therefore overall efficiency of thermal power plant in both cases is 100%.

[0097] The heat pump will work in the range of 123.27° C. to 175° C. so as shown in

[0098] FIG. 11 From the T-s Diagram for water, it is found out that

[00010]$H1\mathrm{enthalpy}\mathrm{before}\mathrm{compression}=2491.3\text{KJ/Kg}—\hspace{1em}\left(\mathrm{Vapour}\left(90\%\right)+\mathrm{Liquid}\left(10\%\right)\mathrm{at}123.27°C.\mathrm{at}2.1\mathrm{bar}\right)H2\mathrm{enthalpy}\mathrm{after}\mathrm{compression}=2772.1\text{KJ/Kg}—\left(\mathrm{Vapour}\mathrm{at}175°C.\right)H3\mathrm{enthalpy}\mathrm{after}\mathrm{condensation}\mathrm{of}\mathrm{condenser}=2772.1\text{KJ/Kg}—\left(\mathrm{Liquid}\mathrm{at}175°C.\mathrm{at}9\mathrm{bar}\right)H4\mathrm{enthalpy}\mathrm{after}\mathrm{expansion}=H3—\left(\mathrm{Liquid}\left(90\%\right)+\mathrm{Vapour}\left(10\%\right)\mathrm{at}123.27°C.\right)\mathrm{COP}\mathrm{of}\mathrm{heat}\mathrm{pump}\mathrm{is}=H2-H3/H2-H1=\left(2772.1-742.6\right)/\left(2772.1-2491.3\right)=2029.5/280.8=7.2\mathrm{COP}$

[0099] Hence the Approximate COP is 7 For system with 100 MW plant [0100] (1) Condenser converts 40% of heat to work (pressure).—(Case 1) [0101] (2) Boiler/heater converts 50% of heat to work (pressure).—(Case 2)

[0102] B. When River Water and Heat Absorber (11) is Used in System for Electricity Generation.

[0103] In this design the heat is not given by external heat source but it is taken from the water of nearby lake or river. Heat absorber (11) has water of 7° C. which absorbs heat from river (lake) and converts some part of 7° C. water to 7° C. of vapour. Then the heat absorber absorbs remaining unused heat from system which is 152° C. water at 10 bar and converts in to 150° C. which is then pass to pump to initial cycle. The water in evaporator is now at 123° C. vapour from 7° C. water and is now pass to compressor.

Heat Pump Calculations

[0104] Remaining heat can be recovered by the heat pump

[0105] The heat pump with water as working fluid has a COP of 7

[0106] The heat pump will work in the range of 7° C. to 175° C. so as shown in

[0107] FIG. 12, From the T-s Diagram for water, we found out that

[00011]$H1\mathrm{enthalpy}\mathrm{before}\mathrm{compression}=2491.3\text{KJ/Kg}—\hspace{1em}\left(\mathrm{Vapour}\left(90\%\right)+\mathrm{Liquid}\left(10\%\right)\mathrm{at}123.27°C.\mathrm{at}2.1\mathrm{bar}\right)H2\mathrm{enthalpy}\mathrm{after}\mathrm{compression}=2772.1\text{KJ/Kg}—\left(\mathrm{Vapour}\mathrm{at}175°C.\right)H3\mathrm{enthalpy}\mathrm{after}\mathrm{condensation}\mathrm{of}\mathrm{condenser}=2772.1\text{KJ/Kg}—\left(\mathrm{Liquid}\mathrm{at}175°C.\mathrm{at}9\mathrm{bar}\right)H4\mathrm{enthalpy}\mathrm{after}\mathrm{expansion}=H3—\left(\mathrm{Liquid}\left(90\%\right)+\mathrm{Vapour}\left(10\%\right)\mathrm{at}123.27°C.\right)\mathrm{COP}\mathrm{of}\mathrm{heat}\mathrm{pump}\mathrm{is}=H2-H3/H2-H1=\left(2772.1-742.6\right)/\left(2772.1-2491.3\right)=2029.5/280.8=7.2\mathrm{COP}$

[0108] Hence the Approximate COP is 7

Best Mode of the System:

[0109] To generate electricity of 100MW where heat in system is provided through heater (2):

[0110] System specifications: [0111] 1. Closed circuit of pipes (10): [0112] Length of all pipes: 100 m, Diameter of first pipe: 6 m, Diameter of remaining three pipes: 4 m [0113] 2. Water pump (1) attached to the first end of closed circuit [0114] 3. Heater (2) with 100 MW heating capacity [0115] 4. Three Francis turbines (T1(3), T2(4), and (T3(5)) attached at three corners of the closed circuit [0116] 5. Heat Pump components such as evaporator (6), condenser (8), compressor (7) and expansion tube (9).

[0117] Water which was at room temperature was passed through rectangular closed circuit of pipes (10). To provide velocity and one directional flow for water in system, pump (1) was used. Volume flow rate in system was 25.64 m.sup.3/s and water pass at velocity of 2.04 m/s through 4 m diameter pipes. Heater (2) was used to provide heat input to the closed circuit (10).

[0118] Water was heated to up to 150° C. by the heater (2). Once the temperature is at 150 ° C. now we can heat the water in boiler (heater) in two different ways

[0119] Case-1. We can heat water uniformly (by 1° C. rise) to generate pressure about 19 bar

[00012]$\left(\mathrm{Total}\mathrm{pressure}=19+10=29\mathrm{bar}\right)$

[0120] Case-2. We can heat small amount of water, e.g. 5 Kg of water per 1 m.sup.3 volume. Which generates greater expansion and also greater pressure about 35 bar

[00013]$\left(\mathrm{Total}\mathrm{pressure}=35+10=45\mathrm{bar}\right)$

[0121] This pressure developed in the system by Case-2 is 16 bar more than Case-1. So, the 16-bar work is need to be use by turbine (1) to generate work. Because after passing through turbine (1) the water will get uniformly heated and get convert to Case-1.

[0122] The remaining work of 19 bar will be converted in to work by turbine (2). So, the remaining heat in water and base pressure of 10 bar is present in water is than pass to evaporator.

[0123] To remove pressure generated in the closed circuit (10) and to convert it in to mechanical work (for generating electricity) three Francis turbines were used in the system. When water was heated from 150° C. to 151° C. in water state in closed circuit (10), the pressure exerted by the water was 19 bar. The pressure generated by the heated water was absorbed by the hydraulic turbines to generate electricity. The outlet of third turbine T3(5) was connected to evaporator (6) to reuse heat by Heat Pump; further this heat was used by condenser (8) for heating. Water vapours were converted to water in the condenser (8) and then that water was fed into expansion tube (9). Thus, Heat pump cycle continued along with the main cycle of the closed circuit (10) to generate electricity by heating water at very low temperature.

Calculations for Heat Pump Cycle in Above System:

[0124] 1. Evaporator—123° C. water is flowing through it which absorbs remaining heat from water in closed circuit which was not use by the circuit; this unused water gets converted in to 123° C. vapour.

[0125] The length of evaporator is 100 m and width is 4 m and 4.2 m height

[0126] It has heat 200 plates along the length and 100 plates along width. Total 20,000 plates.

[0127] Each plate has 0.2 m length, 4 m height and 1.6 cm thick. With 1 cm hollow space in it to allow refrigerant water at 123° C. to flow.

[00014]$\mathrm{Total}\mathrm{heat}\mathrm{area}\mathrm{of}\mathrm{plates}\mathrm{is}=2*4*0.2*20000=32000m^2$$\mathrm{Total}\mathrm{heat}\mathrm{transfer}=340*25*32000=270\mathrm{MW}$

[0128] (Where 340 W/m.sup.2 ° C. is overall heat transfer coefficient for Water-Mild Steel-Water in contact and 25° C. is temperature difference)

[0129] Heat absorb by evaporator was 176 MW

[0130] 2. Compressor—converts the 123° C. vapour in to 175° C. vapour by use of mechanical power. The mechanical input to the compressor is 34 MW. The flow rate of refrigerant is 0.1199 m.sup.3/s or (110 kg/sec). The diameter of pipe for refrigerant is 0.5 m and 0.61 m/s velocity.

[0131] 3. Condenser—Has 175° C. vapour flowing through it which gives heat to water which is use to generate pressure and converted in to 175° C. water.

[0132] The heat given by condenser is 210 MW

[0133] The length of condenser is 100 m and width of 2.8 m and 3 m height

[0134] It has heat 300 plates along the length and 100 plates along width. Total 30,000 plates

[0135] Each plate has 0.2 m length, 2.8 m height and 1.6 cm thick. With 1 cm hollow space in it to allow refrigerant vapour at 175° C. to flow.

[00015]$\mathrm{Total}\mathrm{heat}\mathrm{area}\mathrm{of}\mathrm{plates}\mathrm{is}=2*2.8*0.2*300000=33600m^2$$\mathrm{Total}\mathrm{heat}\mathrm{transfer}=340*25*33600=285\mathrm{MW}$

[0136] 4. Expansion tube—The expansion tube reduces pressure of 175° C. water and boils till it drop its temperature to 123° C. water and some vapour. The diameter of this expansion tube is 0.