Question 11.AE.1: A very good example for energy consumption reduction of a re......

The floor-heating of the house maintained the bedroom temperatures at 17-20°C and all other rooms at 20-22°C. The residence is equipped with a 7.64 kW ground-water heat pump [16]. During summer the cooling of the house is via floor-cooling using ground-water (e.g., 8-11°C) maintaining room temperatures at less than 24°C. The above described generation (EI_i), energy fed-into grid (EG_i), energy requirements for both heating (including hot water heating during entire year for kitchen and baths) and cooling modes (EHC_i) as well as the energy supplied by the utility (ELA_i) are reflected in Figure E11.1.3 and Table E11.1.1.

Note that E I=7 \cdot \sum_{i=1}^{52} E I_i+E I_{7 / 14 / 2014}=6577 \mathrm{kWh} / \text { year, } E G=7 \cdot \sum_{i=1}^{52} E G_i+E G_{7 / 4 / 2014} =6088 \mathrm{kWh} / \text { year, } \quad E H C=7 \cdot \sum_{i=1}^{52} E H C_i+E H C_{7 / 14 / 2014}=3376 \mathrm{kWh} / \text { year, } and E L A=7 \cdot \sum_{i=1}^{52} E L A_i+E L A_{7 / 14 / 2014}=593 \mathrm{kWh} / \text { year }

The predicted value of 18 kWh/(m² -a) can be checked using some of the values (EHC_i) related to either Figure E11.1.3 or Table E11.1.1 as follows:

with \left.7 \cdot\left[\sum_{i=1}^{52} E H C_i\right]+E H C_{7 / 14 / 2014}+7 \cdot\left[\sum_{i=1}^{52} E L A_i\right]+E L A_{7 / 14 / 2014}\right]=(3376+593) \mathrm{kWh} / \text { year }=3969 \mathrm{kWh} / \text { year, }

energy consumption in house without PV system

\begin{aligned} & =\left\{\begin{array}{c} \text { energy input or energy consumption in house } \\ \text { on } 14^{\text {th }} \text { July } 2014(E H C+E L A)-\text { energy input or energy } \\ \text { consumption in house on } 14^{\text {th }} \text { July } 2013(E H C+E L A) \end{array}\right\} /\left(220 \mathrm{~m}^2 \cdot \mathrm{a}\right) \\ & =\{5982 \mathrm{kWh}-2013 \mathrm{kWh}\} / 220 \mathrm{~m}^2=3969 \mathrm{kWh} / 220 \mathrm{~m}^2=18.04 \mathrm{kWh} /\left(\mathrm{m}^2 \cdot \mathrm{a}\right) . \end{aligned}                        (E11.1-1a)

This measured value is about the same as predicted by the energy disclosure (18 kWh/ (m² ·a) of Figure E11.1.2 based on calculations of Konig [ € 15]. If the generation of the PV system is taken into account via net metering –discussed later– then the energy coefficient of the house will be greatly reduced. Figure E11.1.3 (EI_i) depicts the generation of the PV system as a function of time from 14 July 2013 to 14 July 2014. The CO_2=15 (kg-force)/ (m² ·a) emission (Figure E11.1.2) will be greatly reduced as well, and will approach zero.
The total generated electric energy EI is for this time span 6577 kWh/year taking the inverter efficiency of 0.96 into account one obtains the generated energy available at the meter EI_{meter}=EI_{out}=0.96·6577 kWh/year¼6313.9 kWh/year. Therefore, one obtains:

energy consumption in house taking PV generation into account
\begin{aligned} & =\left\{\begin{array}{c} \text { energy consumption on } 14 \text { July } 2014(E H C+E L A)-\text { energy consumption } \\ \text { on } 14 \text { July } 2013(E H C+E L A)-\text { generation of PV system }\left(E I_{\text {out }}\right) \end{array}\right\} /\left(220 \mathrm{~m}^2 \cdot \mathrm{a}\right) \\ & =(5982 \mathrm{kWh}-2013 \mathrm{kWh}-6313.9 \mathrm{kWh}) /\left(220 \mathrm{~m}^2 \cdot \mathrm{a}\right)=-10.66 \mathrm{kWh} /\left(\mathrm{m}^2 \cdot \mathrm{a}\right) \end{aligned}                  (E11.1-1b)

This modified energy coefficient corresponds to that of the “passive house” construction [17], and it is negative which is even better. Because of utility policy, the PV system does not supply the water-to-water heat pump system with electricity.
The average coefficient of performance [1] of the heat pump is from April 2013 to November 2014 with the measured heat energy Q_H =25,070 kWh as provided by the heat pump and \left(P_{\text {compressor }}+P_{\text {water handlens }}\right)=E H C=\left(P_{\text {compressor }}+P_{\text {water handlers }}\right)= 6,299 kWh

\mathrm{COP}_{\mathrm{H}}=\mathrm{Q}_{\mathrm{H}} /\left(P_{\text {compressor }}+P_{\text {witer handlers }}\right)=25,070 / 6,299=3.98          (E11.1-2a)

corresponding to a seasonal energy efficiency ratio (SEER) of

\mathrm{SEER}=3.413 \cdot \mathrm{COP}_{\mathrm{H}}=13.58                (E11.1-2b)

which corresponds to a low-efficiency heat pump. However, if the P_{water\ handlers} required for cooling is neglected in (E11.1-2a) during the summer season then the COP_H will be somewhat above 4. High-efficiency [1] heat pumps have CO_H =5.57 or SEER=19. This means there is still room for efficiency increase, requiring higher investment costs. The feed-in compensation of the PV power fed-into the public utility system is 0.18 Euros/kWh¼US$ 0.25/ kWh: of this amount a value-added tax of 19% must be paid in Germany. The feed-in arrangement is shown in Figure E11.1.4 where the supply ELA from utility is the all-time available electricity, EHC includes the highand low-tariff electricity available from utility only at certain times during working days, nights, weekends and holidays.

The ELA is about 1.65 kWh/day and in addition about 0.6 kWh/day are supplied by the PV plant, that is, self-consumption (e.g., lights, appliances, outlets, electronic equipment) is about 2.25 kWh/day, which amounts to about 13% of the total generated PV power 0.96·EI=6313.9 kWh/year. Comparing the graphs of Figure E11.1.3, one notes that energy consumption EHC is highest and energy generation (EI or EG) is lowest during winter. Correspondingly, during summer, energy consumption is lowest –even if cooling is employed– and energy generation is highest. This oppositeness calls for long-term storage facilities such as hydro-pump [18,19] and compressed-air [20,21]. According to Figure E11.1.3 from October to February (EHC+ELA) is larger than {EG+(ELAR=365days·0.6 kWh/day=219 kWh/year)} requiring storage plant at the residence. New hydro pump-storage plants are planned in the Black Forest [22] and the Bavarian alpine regions of Germany [23, 24]. As expected in both cases recreationists oppose these new necessary storage plants.
Comparing the annual energy yield of the 6.15 kW_p PV plant with one inverter in Boulder, Colorado [1] with the 6.48 kW_p PV plant with two inverters in Munich, Germany, one concludes that for the same array conditions

(energy yield in Munich)/(energy yield in Boulder)_{measured}=6242:1=7953 = 0:785,          (E11.1-3a)

where the measured energy yield EI in Boulder [1] during one year (12/31/2008 to 12/ 31/2009) for the 6.15kW_p plant=18370 kWh-10417 kWh=7953kWh/year, and the measured energy yield EI in Munich (7/14/2013 to 7/14/2014) for an equivalent 6.15 kW_p plant is (6.15 kW_p/6.48 kW_p)· 6577 kWh/year=6242.1 kWh/year. The ratio (E11.1-3a) is approximately confirmed through calculations using National Renewable Energy (NREL) software [25] resulting in

(energy yield in Munich)/(energy yield in Boulder)_{predicted\ by\ NREL\ software} ≈ 0.65.                (E11.1-3b)

The above significant difference (17.2%) may lie in the use of two inverters in the Munich plant minimizing cloud influence as compared to using one inverter in Boulder, whereby in the Munich 6.48 kWp residential PV plant two inverters each process approximately half the rated power. In addition the environmental conditions (temperature, snow coverage, clouds) might be somewhat different during 2008/2009 and 2013/ 2014 time frames. Inverters have an efficiency of about 96%. This was measured when there was no electrical consumption in the house during a sunny day (7/21/2013) in Munich when the generated power was EI_{7/21/2013}=37.57 kWh and the corresponding delivered power to the grid was EG_{7/21/2013}=36.14 kWh resulting in a total inverter efficiency of η_{inverter}=36.14/37.57≈0.96.