Codes, Acoustics, Economics & Electrical

10% of exam

Codes and standards, sound and vibration control (NC, dBA, isolation), life-cycle and payback economics, and motor power, amperage, and heat output.

6 concepts

Exam areas

A. Codes and Standards

Codes and Standards: ASHRAE 90.1 / 62.1 / 15 / 34, IMC, IECC

What ASHRAE 90.1, 62.1, 15, and 34, the IMC, and the IECC each govern, and the engineer's role in demonstrating compliance.

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B. Acoustics and Vibration Control

Sound Control: dB Addition, NC and Attenuation

Add sound levels logarithmically, separate sound power from pressure, rate rooms with NC/RC, and balance duct attenuation against the regenerated-noise floor.

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Vibration Isolation: Transmissibility and Isolator Selection

Set the forcing-to-natural frequency ratio for isolation, size isolators by static deflection, and choose between pads and springs.

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C. Economic Analysis

Life-Cycle Cost, Simple and Discounted Payback

Compare HVAC alternatives on a present-worth basis with the interest factors, and judge energy retrofits with simple and discounted payback.

On the PE Mechanical exam the economics is rarely hard finance — it is bookkeeping in time. A chiller that costs more up front but less to run, a premium-efficiency motor, a heat-recovery wheel: each is a stream of cash flows at different dates, and you cannot add dollars from different years until you slide them all to one date with an interest factor. Points are lost not in the algebra but in the setup — discounting an annual stream as if it were a single payment, mixing nominal and effective rates, or counting the salvage twice. The NCEES PE Mechanical Reference Handbook §1.6 Economic Analysis gives you the factor formulas and the factor tables; your job is to know which factor moves which cash flow, and in which direction.

The time value of money and the six factors

Every factor is built from one idea: a dollar today grows to

(1+i)^n

dollars after

n

periods at interest

i

per period. The single-payment compound-amount factor

(F/P,i,n)=(1+i)^n

moves a present sum forward; its reciprocal

(P/F,i,n)=(1+i)^{-n}

discounts a future sum back. For a level annual series

A

— the way energy and maintenance costs are modeled — the uniform-series present-worth factor

(P/A,i,n)

collapses the whole stream to one present number, and its reciprocal, the capital-recovery factor

(A/P,i,n)

, spreads a present cost into equal annual payments. Read the factor name as 'find-given':

(P/A)

means find

P

given

A

. Two more round out the set you may need: the uniform-series compound-amount factor

(F/A)

grows a level series to a future lump (its reciprocal

(A/F)

is the sinking fund that sets aside equal deposits toward a future replacement), and the uniform-gradient factor

(P/G)

handles a cost that climbs by a constant amount each year, such as a maintenance budget that escalates as equipment ages.

(P/A,i,n)=\dfrac{(1+i)^n-1}{i\,(1+i)^n},\qquad (A/P,i,n)=\dfrac{i\,(1+i)^n}{(1+i)^n-1}

Nominal versus effective interest

Problems quote a nominal annual rate

r

but compound it

m

times a year. You must convert to the effective rate that actually governs growth before you use any factor whose period matches the compounding. The annual effective rate is

i_e=(1+r/m)^m-1

; a nominal

8\%

compounded monthly is

i_e=(1+0.08/12)^{12}-1=8.30\%

. When cash flows are annual and compounding is annual,

r

and

i

coincide and no conversion is needed — but read for the word 'compounded'.

i_e=\left(1+\dfrac{r}{m}\right)^{m}-1

Life-cycle cost as a present worth

Life-cycle cost (LCC) is the single present-worth number that captures the whole ownership story: first cost now, plus the present worth of every recurring cost over the study life, minus the present worth of the salvage value at the end. Recurring annual costs use

(P/A)

; a one-time future replacement or salvage uses

(P/F)

. The alternative with the lowest LCC wins, even if it has the highest first cost — that is the entire point of the method, and the reason a high-efficiency machine usually beats a cheap one over a 20-year life.

LCC = C_0 + A_{\text{O\&M}}\,(P/A,i,n) - S_n\,(P/F,i,n)

Simple payback — fast, blunt, undiscounted

Simple payback is the screening tool everyone reaches for first: the years it takes the annual savings to repay the extra first cost, with no discounting. It is honest only as a rough filter — it ignores the time value of money, anything that happens after payback, and salvage. A 5-year simple payback on an energy retrofit tells you the order of magnitude, nothing more, and it always looks better than the truth because future dollars are counted at full value.

n_{\text{simple}} = \dfrac{\text{Initial investment} - \text{Salvage}}{\text{Net annual savings}}

Discounted payback

Discounted payback fixes the headline flaw by discounting each year's savings before accumulating them, so it always lands later than simple payback. Setting the present worth of the savings equal to the investment, and writing

CRF

as the ratio of annual savings to initial cost (

A/P

), the handbook's improved-payback relation solves directly for the period. If the savings rate

CRF

does not exceed the interest rate

i

, the project never pays back — the logarithm argument goes non-positive, which is the math telling you the discounted stream can never catch the investment.

n = \dfrac{\ln\!\left[\dfrac{CRF}{CRF-i}\right]}{\ln(1+i)},\qquad CRF=\dfrac{\text{annual savings}}{\text{initial investment}}

Energy-savings economics

Most HVAC economics questions are an energy retrofit dressed up: an efficiency measure costs an increment

\Delta C_0

now and returns annual energy savings

A=\dot{E}\cdot(\text{rate})

. Convert the energy saved to dollars per year, take the present worth with

(P/A)

, and compare to the increment — a positive net present worth means the measure pays. Annual worth is the mirror image: spread

\Delta C_0

with

(A/P)

and compare the equivalent annual cost of capital to the annual savings. The two methods always agree on the decision.

Exam strategy

Draw the cash-flow diagram first — arrows up for income/savings, down for costs, time on the axis — then label each arrow with the factor that moves it to year 0. Use the supplied factor tables when

i

and

n

are tabulated (they almost always are); only fall back to the closed-form factors for odd rates. Keep three discriminators straight: simple payback ignores

i

and always reads optimistically; discounted payback is later; LCC and net present worth are the defensible answers. If two alternatives have different lives, compare equal annual worth, not raw present worth. Watch the sign on salvage — it is a benefit, so it subtracts from LCC.

Key equations

Single-payment compound amount (F/P)

(F/P,i,n)=(1+i)^{n}

Moves a present sum $P$ forward $n$ periods to its future worth $F$ at periodic rate $i$ .

Single-payment present worth (P/F)

(P/F,i,n)=(1+i)^{-n}

Discounts a single future sum $F$ (e.g., salvage, a future replacement) back to year 0.

Uniform-series present worth (P/A)

(P/A,i,n)=\dfrac{(1+i)^{n}-1}{i\,(1+i)^{n}}

Collapses a level annual series $A$ (energy, O&M) over $n$ years into one present worth $P$ .

Capital recovery factor (A/P)

(A/P,i,n)=\dfrac{i\,(1+i)^{n}}{(1+i)^{n}-1}

Spreads a present cost into $n$ equal annual payments; the reciprocal of $(P/A)$ . Used for equivalent annual cost.

Uniform-series compound amount (F/A)

(F/A,i,n)=\dfrac{(1+i)^{n}-1}{i}

Future worth of a level annual series; $(A/F)=i/[(1+i)^n-1]$ is its reciprocal (sinking fund).

Effective interest rate

i_e=\left(1+\dfrac{r}{m}\right)^{m}-1

$r$ = nominal annual rate, $m$ = compounding periods per year. Convert before using period factors.

Life-cycle cost

LCC = C_0 + A_{\text{O\&M}}\,(P/A,i,n) - S_n\,(P/F,i,n)

Present-worth total of ownership: first cost $C_0$ , annual costs, less salvage $S_n$ . Lowest LCC wins.

Simple payback

n_{\text{simple}}=\dfrac{C_0-S}{A_{\text{savings}}}

Years for undiscounted annual savings to repay first cost; screening only — ignores $i$ and post-payback value.

Discounted payback

n=\dfrac{\ln[\,CRF/(CRF-i)\,]}{\ln(1+i)},\quad CRF=\dfrac{A_{\text{savings}}}{C_0}

Handbook improved-payback period. Requires $CRF>i$ or the project never pays back.

Net present worth of a measure

NPW = A_{\text{savings}}\,(P/A,i,n) - \Delta C_0

Decision metric for an efficiency upgrade; $NPW>0$ accept. $\Delta C_0$ = incremental first cost.

Uniform gradient present worth (P/G)

(P/G,i,n)=\dfrac{(1+i)^{n}-i n-1}{i^{2}(1+i)^{n}}

Present worth of a cost that increases by a constant amount $G$ each year (e.g., escalating maintenance).

Worked examples

Life-cycle cost of two chillers

Problem. Chiller X costs $120,000 installed and $18,000/yr to run and maintain. Chiller Y costs $160,000 installed and $12,000/yr. Use a $20$ -yr study life, $i=6\%$ , and negligible salvage. Which has the lower life-cycle cost?

Solution. The annual streams are level, so use

(P/A,6\%,20)

(P/A,6\%,20)=\frac{1.06^{20}-1}{0.06\,(1.06^{20})}=\frac{3.2071-1}{0.06(3.2071)}=11.47.

Chiller X:

LCC = 120{,}000 + 18{,}000(11.47) = 120{,}000+206{,}459 = \$326{,}000

. Chiller Y:

LCC = 160{,}000 + 12{,}000(11.47) = 160{,}000+137{,}639 = \$298{,}000

. Chiller Y wins by about $28,800 over the life despite costing $40,000 more up front. Sanity: the $6,000/yr operating saving, worth

6{,}000\times11.47=\$68{,}800

present, exceeds the $40,000 first-cost premium — consistent.

LCC_Y-LCC_X = 40{,}000 - 6{,}000\,(11.47) = -\$28{,}800

Simple versus discounted payback

Problem. A controls retrofit costs $45,000 and saves $9,000/yr in energy with no salvage. Find the simple payback, and the discounted payback at $i=8\%$ .

Solution. Simple payback:

n=\dfrac{45{,}000}{9{,}000}=5.00\text{ yr}

. Discounted:

CRF=\dfrac{9{,}000}{45{,}000}=0.20

, which exceeds

i=0.08

, so it does pay back.

n=\frac{\ln[\,0.20/(0.20-0.08)\,]}{\ln(1.08)}=\frac{\ln(1.667)}{\ln(1.08)}=\frac{0.5108}{0.07696}=6.64\text{ yr}.

The discounted payback (6.64 yr) is later than the simple payback (5.00 yr), as it must be. Sanity: the present worth of 5 years of savings is only

9{,}000(P/A,8\%,5)=9{,}000(3.993)=\$35{,}900<\$45{,}000

, confirming repayment needs more than 5 yr.

n=\dfrac{\ln[\,CRF/(CRF-i)\,]}{\ln(1+i)}=6.64\ \text{yr}

Net present worth of a VFD upgrade

Problem. Adding a variable-frequency drive to a pump costs $8,000 more than the across-the-line starter and saves $14{,}500\,\text{kWh/yr}$ at $0.11/kWh. Over $15$ yr at $i=7\%$ , is the upgrade justified?

Solution. Annual saving:

A = 14{,}500\times 0.11 = \$1{,}595/\text{yr}

(P/A,7\%,15)=\dfrac{1.07^{15}-1}{0.07(1.07^{15})}=\dfrac{2.759-1}{0.07(2.759)}=9.108.

Present worth of savings:

1{,}595\times 9.108 = \$14{,}530

NPW = 14{,}530 - 8{,}000 = +\$6{,}530 > 0

, so the VFD is justified. Sanity: simple payback is

8{,}000/1{,}595 = 5.0\,\text{yr}

, well inside the 15-yr life, so a positive NPW is expected.

NPW = 1{,}595\,(P/A,7\%,15) - 8{,}000 = +\$6{,}530

Common pitfalls

•Discounting an annual stream with $(P/F)$ instead of $(P/A)$ : a level yearly cost is a series — use $(P/A)$ once, not $(P/F)$ on each year unless the amounts differ.
•Mixing nominal and effective rates: if interest is 'compounded monthly', convert $r$ to $i_e$ before using annual factors, or work in monthly periods throughout.
•Treating simple and discounted payback as interchangeable. Simple payback ignores $i$ and always reads earlier; the exam often asks for the discounted value specifically.
•Adding salvage to LCC instead of subtracting it — salvage is a benefit received at year $n$ , so it lowers life-cycle cost via $-S_n(P/F)$ .
•Comparing alternatives with different service lives on raw present worth. Equalize the horizon or compare equivalent annual worth, $(A/P)$ of each LCC.
•Forgetting that a project with $CRF \le i$ never reaches discounted payback — the log argument goes non-positive; do not report a spurious number.
•Confusing the factor direction: $(A/P)$ spreads a present cost into payments; $(P/A)$ collapses payments to a present sum. Reading the name 'find/given' prevents inverting it.

References

NCEES PE Mechanical Reference Handbook — §1.6 Economic Analysis (factor formulas and factor tables)
ASHRAE Handbook—HVAC Applications, Owning and Operating Costs (life-cycle cost method)
NIST Handbook 135 — Life-Cycle Costing Manual for the Federal Energy Management Program — background on LCC, SIR, and discounted payback conventions

Building Energy Usage: EUI/CUI and Benchmarking

Compute energy and cost use intensity, estimate annual energy with degree-day and bin methods, and normalize and benchmark across buildings.

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D. Electrical Concepts

Motor Power, Amperage and Heat Output

Compute single- and three-phase motor power and full-load amperage, apply efficiency and service factor, and find the heat a motor releases to a conditioned space.

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