22-Mec-B2 Environmental Control in Buildings · December 2017

Question 8 of 8: Seasonal gas consumption of an R-2000 house and the value of night setback

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada national examination 16-Mec-B2 Environmental Control in Buildings, December 2017, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. ASHRAE Psychrometric Chart No. 1 (SI and inch-pound) and a pressure–enthalpy diagram for R-717 are appended to the paper as pages 6–8.

All eight problems are worked here. The paper mixes SI and inch-pound units deliberately: Problems 1, 3 and 4 are SI, Problems 2, 6 and 7 are inch-pound, and Problem 8 is SI with a Canadian climate. Each solution is worked in the units the question uses, as the cover-page instructions require.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry and the psychrometric chart), Ch. 6 (air-conditioning plant cycles), Ch. 7 (the cooling coil, apparatus dew point and by-pass factor), Ch. 9 (cooling towers), Ch. 15 (fans).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12 (fans and duct design).
ASHRAE Handbook — Fundamentals — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 21 (duct design), Ch. 25–27 (thermal and moisture performance of the building envelope), Ch. 30 (fenestration).
Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., McGraw-Hill — Ch. 10–12 (vapour-compression cycle, compressors, condensers and evaporators); ASHRAE Handbook — Refrigeration for ammonia plant practice.
National Building Code of Canada and the National Energy Code of Canada for Buildings (NRC), Appendix C climatic data; CSA and Canada Green Building Council material for Problem 5.

Property basis used throughout. Moist-air properties are computed from the ASHRAE Fundamentals ideal-moist-air relations, so every state quoted here can be read back off the psychrometric chart supplied with the paper:

$$W=\frac{0.621945\,p_w}{p-p_w},\qquad h_{\text{SI}}=1.006\,t+W\,(2501+1.86\,t),\qquad h_{\text{IP}}=0.240\,t+W\,(1061+0.444\,t)$$

with $h$ in $\text{kJ/kg}$ of dry air for $t$ in $\,{}^{\circ}\text{C}$ and in $\text{Btu/lb}$ of dry air for $t$ in $\,{}^{\circ}\text{F}$. Ammonia properties are quoted on the same datum as the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ and $s_f=1.0\ \text{kJ/(kg}\cdot\text{K)}$ for saturated liquid at $0\,{}^{\circ}\text{C}$); only differences enter the answers, so any consistent chart or table gives the same duties.

Question 8: Seasonal gas consumption of an R-2000 house and the value of night setback (20 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Design heat loss	$\dot q_{design}$	$10\ \text{kW}$
Indoor / outdoor design temperature	$t_i,\ t_o$	$22\,{}^{\circ}\text{C}$ / $-25\,{}^{\circ}\text{C}$
Furnace efficiency	$\eta_f$	$0.85$
Heating value of natural gas	$HV$	$37\ \text{MJ/m}^3$
Ottawa heating degree-days below $18\,{}^{\circ}\text{C}$	$\text{HDD}_{18}$	$4500\ {}^{\circ}\text{C}\cdot\text{day}$ (NBC Appendix C)
Degree-day correction factor	$C_D$	$0.65$
Setback	$\Delta t_{sb}$, hours	$4\ \text{K}$ for $7\ \text{h}$ of every $24$

Find. The annual natural-gas consumption, and the saving that results from a $4\ \text{K}$ night setback held for seven hours a day.

Approach. Convert the design heat loss into a building loss coefficient, apply the modified degree-day method with Ottawa climatic data to get the seasonal heat requirement, and divide by the furnace efficiency and the heating value. For the setback, treat the seven setback hours as a season at a lower base temperature and re-sum.

Building loss coefficient. The design heat loss is the product of the coefficient and the design temperature difference, so $$UA=\frac{\dot q_{design}}{t_i-t_o}=\frac{10}{22-(-25)}=\frac{10}{47} =0.2128\ \text{kW/K}$$ This single number carries the whole envelope — transmission plus infiltration — and it is all the degree-day method needs.
Climatic data. Appendix C of the National Building Code gives Ottawa a January $2.5\%$ design temperature of $-25\,{}^{\circ}\text{C}$ — which is exactly the value the question uses, confirming the data set — and $4500\ {}^{\circ}\text{C}\cdot\text{day}$ of heating degree-days below $18\,{}^{\circ}\text{C}$.
Part (a) — seasonal heat requirement. The modified degree-day method scales the design loss by the accumulated temperature deficit and applies an empirical correction: $$Q_{annual}=24\;UA\;\text{HDD}_{18}\;C_D =24\times0.2128\times4500\times0.65$$ $$Q_{annual}=14{,}940\ \text{kWh}=53{,}770\ \text{MJ}$$ The factor $C_D$ — here taken as $0.65$ — corrects for the two effects the raw degree-day sum ignores: internal gains from occupants, lighting and appliances, which are relatively large in a well-sealed R-2000 house, and the fact that a furnace runs less efficiently at part load than at its rated output. Values between $0.60$ and $0.70$ are standard for a modern Canadian house.
Part (a) — gas volume. Dividing by the furnace efficiency and the heating value: $$V_{gas}=\frac{Q_{annual}}{\eta_f\,HV}=\frac{53{,}770}{0.85\times37} =\boxed{1710\ \text{m}^3/\text{yr}}$$ about $4.7\ \text{m}^3$ a day averaged over the year, or $63\ \text{GJ}$ of purchased gas. Sanity-check it as equivalent full-load hours: $$\text{EFLH}=\frac{24\,\text{HDD}\,C_D}{t_i-t_o} =\frac{24\times4500\times0.65}{47}=1494\ \text{h}$$ so the furnace runs the equivalent of about $1500$ hours a year at full output, which is the right order for southern Ontario and confirms the arithmetic.
Part (b) — how a setback actually saves energy. The saving is not simply the fraction of time multiplied by the fraction of temperature difference. Lowering the thermostat by $4\ \text{K}$ lowers the balance temperature of the house by the same $4\ \text{K}$, so during the setback hours the house is losing heat against a base of $14\,{}^{\circ}\text{C}$ rather than $18\,{}^{\circ}\text{C}$ — and a degree-day total measured to a lower base is smaller, because the hours when the outdoor temperature lies between the two bases now contribute nothing at all. The seasonal heat is therefore re-summed with the day split between the two bases: $$Q_{sb}=UA\,C_D\big[(24-7)\,\text{HDD}_{18}+7\,\text{HDD}_{14}\big]$$
Part (b) — the degree-days at the lower base. Near a base of $18\,{}^{\circ}\text{C}$ the sensitivity of the degree-day total to the base temperature is, by definition, the number of days a year colder than that base — about $295\ {}^{\circ}\text{C}\cdot\text{day}$ per kelvin for Ottawa. Hence $$\text{HDD}_{14}=4500-4\times295=3320\ {}^{\circ}\text{C}\cdot\text{day}$$ $$Q_{sb}=0.2128\times0.65\times\big[17(4500)+7(3320)\big] =0.1383\times99{,}740=13{,}790\ \text{kWh}$$
Part (b) — the saving. Comparing with the $14{,}940\ \text{kWh}$ of part (a): $$\Delta Q=14{,}940-13{,}790=1150\ \text{kWh}\ \text{per year}$$ $$\text{saving}=\frac{1150}{14{,}940}=\boxed{7.6\%}$$ $$\Delta V_{gas}=\frac{1150\times3.6}{0.85\times37} =\boxed{131\ \text{m}^3/\text{yr}}$$ leaving an annual consumption of about $1580\ \text{m}^3$. At a delivered gas price of roughly $\$0.40$ per cubic metre this is on the order of $\$50$ a year — real, but modest, and worth stating plainly to a client who has been promised more.
Why the saving is smaller than intuition suggests, and what limits it. A naive calculation — $7/24$ of the time at $4/47$ less temperature difference — gives only $2.5\%$, because it wrongly scales against the design difference rather than the seasonal average. The correct degree-day treatment gives $7.6\%$. Two practical qualifications cut into even that. First, thermal mass: the house does not reach $18\,{}^{\circ}\text{C}$ the moment the thermostat drops, so the effective setback period is shorter than seven hours — in a heavy house perhaps four or five. Second, the recovery penalty: at the end of the setback the furnace must supply the steady loss plus the energy to re-warm the structure. The extra steady capacity needed is only $UA\,\Delta t_{sb}=0.2128\times4=0.85\ \text{kW}$ against $9.15\ \text{kW}$ of spare capacity at design conditions, so the furnace can recover even on the coldest night; but the recovery is at full fire, and an over-aggressive setback in a house with an undersized furnace will simply fail to recover by morning. The realistic expectation is therefore $5$–$7\%$, and the setback should be ramped back an hour before occupancy rather than stepped.

The setback schedule. Because the balance temperature falls with the thermostat, the seven setback hours must be summed against a 14 °C degree-day base and the remaining seventeen against 18 °C.

Final results.

Quantity	Result
Building loss coefficient	$UA=0.213\ \text{kW/K}$
Seasonal heat requirement	$14{,}940\ \text{kWh}=53{,}770\ \text{MJ}$
(a) Annual natural gas	$\mathbf{1710\ \text{m}^3/\text{yr}}$ ($63\ \text{GJ}$)
Equivalent full-load hours	$1494\ \text{h}$
Degree-days at the $14\,{}^{\circ}\text{C}$ base	$3320\ {}^{\circ}\text{C}\cdot\text{day}$
Seasonal heat with setback	$13{,}790\ \text{kWh}$
(b) Saving	$\mathbf{7.6\%}$, i.e. $1150\ \text{kWh}$ or $\mathbf{131\ \text{m}^3}$ of gas a year
Consumption with setback	$1580\ \text{m}^3/\text{yr}$
Extra capacity needed for recovery	$0.85\ \text{kW}$ against $9.15\ \text{kW}$ spare

Check: assumptions declared under cover-page instruction 1. The question supplies no climatic data, so two values must be brought in and stated. (i) $\text{HDD}_{18}=4500\ {}^{\circ}\text{C}\cdot\text{day}$ for Ottawa, from Appendix C of the National Building Code; the same table gives the $-25\,{}^{\circ}\text{C}$ design temperature the question uses, which confirms the source. Published Ottawa values range from about $4400$ to $4700$ depending on the station and the normals period, so the gas figure carries roughly $\pm5\%$ on that account alone. (ii) $C_D=0.65$; at $0.60$ the answer becomes $1580\ \text{m}^3$ and at $0.70$ it becomes $1840\ \text{m}^3$, so this is the largest single uncertainty in part (a) and the answer should be quoted as "about $1700\ \text{m}^3$". The percentage saving in part (b) is far more robust, because $C_D$ and $UA$ cancel out of the ratio — only the degree-day sensitivity of $295\ {}^{\circ}\text{C}\cdot\text{day}$ per kelvin affects it. (iii) Domestic hot water is excluded throughout: the question asks for the gas required to heat the house, and water heating would add roughly another $1000\ \text{m}^3$ a year to a real bill.

Topic. The degree-day method and the balance-point basis of setback savings.

Key relations. $UA=\dot q_{design}/(t_i-t_o)$; modified degree-day method $Q=24\,UA\,\text{HDD}_{base}\,C_D$; fuel volume $V=Q/(\eta_f\,HV)$; equivalent full-load hours $\text{EFLH}=24\,\text{HDD}\,C_D/(t_i-t_o)$. Balance point $t_{bal}=t_i-\dot q_{int}/UA$, and $\text{d(HDD)}/\text{d}t_{base}$ = the number of days a year colder than the base.

Why this works. The degree-day method works because a building's heat loss is very nearly proportional to the indoor–outdoor temperature difference, so summing that difference over the year and multiplying by $UA$ gives the seasonal energy directly. The subtlety that this question turns on is that the base temperature is not the thermostat setting: the conventional $18\,{}^{\circ}\text{C}$ base is a stand-in for a house held near $21\,{}^{\circ}\text{C}$ with about $3\ \text{K}$ worth of internal gain, so gains are already half counted in the base. Setting the thermostat back lowers the balance point by the same amount, which is why the correct treatment is to re-sum the degree-days to a lower base rather than to scale the answer by a temperature ratio. It also explains why setback saves less than people expect: the coldest hours contribute the most degree-days and are least affected by a $4\ \text{K}$ shift, and the mildest hours — the ones a lower base removes entirely — were contributing little to begin with.

Common pitfall. Subtracting the internal gain from a base-$18$ result. The base-$18$ convention already embeds an allowance for internal gains, so deducting them again double-counts and understates the heating requirement. The related error in part (b) is scaling the saving against the design temperature difference, which gives about $2.5\%$; the seasonal average difference is much smaller than the design difference, so the true saving is roughly three times that.

Source. McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning, 6th ed., Ch. 8 (energy estimating: degree-day and bin methods); ASHRAE Handbook — Fundamentals, Ch. 14 (climatic design information) and Ch. 19; National Building Code of Canada, Appendix C.

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