ap-4-10.dvi Acta Polytechnica Vol. 50 No. 4/2010 Nonlinear Aspects of Heat Pump Utilization R. Najman Abstract This work attempts to answer the question: How much can we believe that the coefficient of performance provided by the manufacturer is correct, when a heat pump is required to face the real load coming from changes of temperature? The paper summarizes some basics of heat pump theory and describes the results of numerical models. Keywords: heat pumps, thermal mass of an object, coefficient of performance. 1 Introduction The idea of a heat pump is quite old, but nowadayswe are confronted with the perspective of a potential en- ergy crisis in the coming decades; people are starting to look for ways to lower their expenditure on domes- tic heating. A few years ago, when we started studies of this topic, only a few thousand heat pumps were installed in the Czech Republic, but the number has grown by 20–50% every year. Houses with additional thermal insulation and new insulation windows have spread evenmore. These installations havea profound impact on the performance of heat pumps. 1.1 Basic principles of heat pumps Heat pumps use energy from a colder source and re- lease it into a warmer ambience. Almost every refrig- eratorhasone, sowehavebeen livingwithheatpumps for a long time now. Legend 1 Condenser coil (hot side heat exchanger) 2 Expansion valve (gas expands, cools and liquifies) 3 Evaporator coil (cold side heat exchanger) 4 Compressor Red=Gas at high pressure and temperature Pink=Gas at high pressure and reduced temperature Blue=Liquid at low pressure and greatly reduced temper- ature Light Blue=Gas at low pressure and warmer temperature Fig. 1: Basic idea of a heat pump [*1] To describe a heat pump correctlywe need the fol- lowing data: COP BheatingB, Source of heat, Target medium of heat transition COP: (efficiency) COPheating = Q P . (1) Q Amount of heat transferred to hot reservoir [W] COPheating Coefficient of performance (for heating purposes) [–] P Dissipated work of the compressor [W] 1.2 Source and target of heat The three typical sources are air, water and soil. The targets are air or water. The systems are therefore re- ferred to as air/air, air/water,water/air, water/water, soil/air and soil/water. For most technical and eco- nomic evaluation purposes, the target medium itself does not matter, only its temperature is important. 1.2.1 Sources Air is the cheapest source for initial investment, but if you do not possess a source with stable temperature (suchaswarmair fromsometechnologicalprocess) the COP is quite low, especially when temperatures out- side hit minus values (◦C), or when you need output temperature above 40◦C. Water is the “golden” middle way. It is a cheaper source than soil and its temperature is quite steady through the year, so achievable COP is quite good. Of course, use can be limited by unavailability of a usable water source. In most cases, the water source must be approved by the local authorities. Soil is the most expensive, but surely the best source forCOP.There aremany technologicalways to obtain heat from soil. The most common way is from drill holes, or from ground collectors. For new build- 60 Acta Polytechnica Vol. 50 No. 4/2010 ings, a quite cheap and effective way is to use energy pilots (pilots of building foundations with integrated heat collectors). Fig. 2: Air, water, soil to water heat pumps [*2] 2 Input values and equations 2.1 Subject for testing our mathematical model Our test subject is a house with a surface area of 400 m2 (surfaces with applicable thermal insulation). Our model heat pump is a water/water type, source water 7◦C (from a well), output 35◦C for a screed floor. After making the first calculation of en- ergy losses, the Stiebel Eltron WPW 7 heat pump seems to be right choice, in combination with some thermal insulation. With no thermal insulation we should use some larger model. Technical data of WPW7: 6.9 kW heat output at 35◦C water and COP 5.2. 2.2 Required heat output The heat output heat source is calculated according to the following model: The indoor temperature Tv is maintained by regu- lating temperature T1. T1 highest heating temperature [ ◦C] Twd lowest heating temperature [ ◦C] Tv indoor temperature [ ◦C] Kheating constant representing efficiency of heat transfer [W/K] Qven heat losses through ventilation [W] Qiz. heat losses through walls (insulated) [W] Qok. heat losses through windows [W] QT UV heat consumed on supply water [W] Fig. 3: Heat losses diagram To calculate the heat losses we need to establish the thermal resistances Rthv = 1 αv , Rthz = d λz , Rthiz = diz 1000 · λiz (2) Rtha = 1 αa , Rtho = 1 λo Where: Rthv indoor convection thermal resistance [ K ·m2 W ] αv indoor heat transfer coefficient [ W K ·m2 ] Rthz thermal resistance of wall [ K ·m2 W ] d wall width [m] λz thermal conductivity of walls [ W K ·m ] Rthiz thermal resistance of walls [ K ·m2 W ] λiz thermal conductivity of thermal insulation [ W K ·m ] diz thermal insulation width [mm] Rtha outdoor convection thermal resistance [ K ·m2 W ] αa outdoor heat transfer coefficient [ W K ·m2 ] Rtho thermal resistance of windows [ K ·m2 W ] λo thermal conductivity of windows [ W K ·m ] Qok = Sok · Tv − Ta Rthv + Rtho + Rtha (3) Qiz = S · Tv − Ta Rthv + Rthz + Rthiz + Rtha (4) Qven = 10 · (Tv − Ta) · os – empiric value (5) When using a heat pump, the consumption is givenby ČSN as follows: (Heating to 35◦C, and then applying an additional source of heat. With other heat sources than the heat pumps it can be done in one step.) QT UV = os ·82 · cp · (35−7) 3600 · 24 + pom (6) Additional heating up to 55◦C: pom = os · 82 · cp · (55−35) 3600 ·24 (7) os number of people [–] cp specific heat capacity of water [J/kg · K] 61 Acta Polytechnica Vol. 50 No. 4/2010 2.3 Heat pump in a thermal circuit QH2O−in heat taken from the source [W] Pel-HP B electricity consumption of HP [W] m2 mass flow the in heating circuit [kg/s] Twh max. temp. of water in the HP [ ◦C] Pel-B electricity consumption of a boiler [W] (for others, see before) Fig. 4: Thermal circuit In the heat pump, thewater is heated toTwh (ap- proximately 35◦C), then it passes through the water boiler, where it can be heated to T1 (if necessary). In the screed floor, the temperature of the water goes down to Twd, and the cycle is repeated. 2.4 Equations of the system COP · Pel-HP = m2 · cp · (Twh − Twd) (8) Pel-B = m2 · cp · (T1 − Twh) (9) Qcelk = m2 · cp · (T1 − Twd) (10) Qcelk = Kheating · ( T1 + Twd 2 − Tv ) (11) Qcelk sum of all heat losses of the object [W] This system of equations can be solved. The m2 is chosen from the manufacturer’s catalogue. The so- lutions are as functions of diz and Ta (from heat loss formulas). 2.5 Thermal mass Thenonlinearbehavior of aheat pump(COP is a non- linear function of the source and target temperature) indicates that the thermal mass of the object must be taken into account. Here I present the algorithm that allows us to estimate the impact of thermal masses on a defined thermal circuit. The first step is to define the command function, which makes it easy for us to input the desired indoor temperature: Ptop heating output [W] Fig. 5: Heating regulation The following equations describe the thermal sta- tus of the system: The wall: ρs · cps · ∂Ts(x, t) ∂t = λs · ∂2Ts(x, t) ∂x2 (12) With this initial condition Ts(x,0)= 0 (13) and border conditions 1 1 αa + diz λiz ·(Ta(t)−Ts(0, t))= −λs · ∂Ts(x, t) ∂x |x=0 (14) αv · (Ts(d, t) − Tv(d, t))= −λs · ∂Ts(x, t) ∂x |x=d (15) Where: ρs density [kg/m 3] cps specific heat capacity [J/kg · K] Ts temperature in the wall [ ◦C] λs thermal conductivity of wall [W/K · m] (for others, see above) Equations for air, adapted to fit better into Math- ematica software: ρv · cpv · ∂Tv(x, t) ∂t =100 · ∂2Tv(x, t) ∂x2 + Ptop(Tv(x, t)) − S · αv · (Tv(x, t) − Ts(x, t)) V − (16) ρv · cpv · (Tv(x, t) − Ta) tv With the initial condition Tv(x,0)= 0 (17) and border conditions: ∂Tv(x, t) ∂x = 0|x→0 (18) ∂Tv(x, t) ∂x = 0|x→d (19) 62 Acta Polytechnica Vol. 50 No. 4/2010 Where: Tv air temperature [ ◦C] ρv air density [kg/m 3] cpv air specific heat capacity [J/kg · K] S surface of walls [S] V indoor volume of air [m3] Ptop heating output [W] tv time constant of ventilation [s] The solutions to the equations give heating output as a function of time. With sinusoid outdoor temper- ature in the graph below: Fig. 6: Heating output From the part of the solution where transient ef- fects no longer apply, we extract a period of one day Ptop = a + b · sin(ωt + ϕ). If such heat loses are implemented into the heat pump model described above, we can obtain the de- pendencies of the variables needed to evaluate the ef- fects of the thermal masses on the heat pump. Where “red” is the power input of the heat pump, and “black” is the power input of water boiler. Variant A: outdoor temperature −10 to +10◦C, sinus, without thermal mass Fig. 7: Power inputs, variant A Variant B: outdoor temperature −10 to +10◦C, sinus, with thermal mass Fig. 8: Power inputs, variant B Variant C: Outdoor temperature 0◦C Fig. 9: Power inputs, variant C If thepower inputs are integrated, there is little dif- ference between Var. B (reality) and Var. C. (5.8 %). However, since integrating the full thermal mass in- cludingmodel into our heat pump circuitmodelwould increase the numerical flaws and greatly reduce the stability and reliability of the outcome, we chose from the models according to Var. A (no thermal mass) or Var. C (the thermal mass is so great, that it would negate all changes of temperature during one day). I have chosen to use a curve of input temperatures that it will simulate a system equivalent to variant C. It is very close to reality, and can be solved more precisely with numerical methods in our system of equations. 3 Output values of the heat pump circuit model (for diz =50 mm) If the outdoor temperature is stable, the input powers of the heat pump (red) and the water boiler (black) are as follows: Fig. 10: Power inputs in stable conditions For choosing the heating period of the year we take the long-term average temperature and compare it with 15◦C. Fig. 11: Average temperature 63 Acta Polytechnica Vol. 50 No. 4/2010 For the real effects on a heat pump, we add some oscillations to the outdoor temperature. Fig. 12: Average temperature with oscillations This givesus a gooddisplayof the consumedpower of the heat pump and water boiler in the course of a year. It shows us that the heat pump (red) covers al- most all heat losses. The water boiler (black) needs to be switched on only when there is a long period of cold weather. Fig. 13: Power consumptions in the course of a model year We also obtain a very significantCOP value (red): Fig. 14: COP in the course of model year Results With our input parameters the heat pump worked throughout the year with COP 4.82. The heat pump covers almost 100 % of heat consumption with 5900 kWh per year consumed for heating. The simplified models give us different results: The“merchant”model,which is oftenusedbyheat pump sellers gives usCOP 5.2, and energy consump- tion of 5 400 kWh. It calculates with an average tem- perature during the heating season. The “normative” model, which uses Ta = −15◦C for the working parameters for heating gives us COP 2.5 and consumed power of circa 11000 kWh per year. This shows that it isworthwhile to investigateheat pumps utilization more deeply. Appropriate use of heat pumps should be considered on the basis of as much factual information as possible. Acknowledgement The research described in this paper was supervised by Doc. Dr. Jan Kyncl. References [1] http://en.wikipedia.org/wiki/Heat pump [2] Stiebel, E.: Heat Pump catalogue 2008. Richard Najman E-mail: najmaric@.fel.cvut.cz Dept. of Electroenergetics Czech Technical University Technická 2, 166 27 Praha, Czech Republic 64