What this formula explains
Rectifier calculations are used to evaluate AC input power, DC output power, rectifier efficiency,
operating time, downtime, meter constant, and energy use. These relationships are common in CP field
work when checking whether a rectifier has been operating properly, estimating costs, or troubleshooting
abnormal performance.
Reference formulas
1) AC input power (watts)
\[
P_{ac} = \frac{3600 \,\times\, K_h \,\times\, N}{t}
\]
\(P_{ac}\) = AC input power (W)
\(K_h\) = meter constant (Wh/rev)
\(N\) = number of revolutions (rev)
\(t\) = time for \(N\) revolutions (s)
2) DC output power (watts)
\[
P_{dc} = V_{dc} \,\times\, I_{dc}
\]
\(P_{dc}\) = DC output power (W)
3) Rectifier efficiency (%)
\[
\eta(\%) = \left(\frac{P_{dc}}{P_{ac}}\right) \,\times\, 100
\]
\(\eta\) = efficiency (%)
\(P_{dc}\) = DC output power (W)
\(P_{ac}\) = AC input power (W)
4) Operating time (hours ON)
\[
H_{on} = \frac{E_{kWh}}{P_{ac}(kW)}
\]
\(H_{on}\) = hours ON (h)
\(E_{kWh}\) = energy used during period (kWh)
\(P_{ac}(kW)\) = AC input power in kW
5) Downtime (hours OFF)
\[
H_{off} = H_{total} - H_{on}
\]
\(H_{off}\) = hours OFF (h)
\(H_{total}\) = total hours in billing period (h)
\(H_{on}\) = hours ON (h)
6) Meter constant (solve for \(K_h\))
\[
K_h = \frac{P_{ac} \,\times\, t}{3600 \,\times\, N}
\]
\(K_h\) = meter constant (Wh/rev)
\(P_{ac}\) = AC input power (W)
\(t\) = time for \(N\) revolutions (s)
\(N\) = revolutions (rev)
Problem 1 — Downtime calculation (days)
A rectifier has a DC output of 40 V and 20 A. Calculate the downtime based on the following data:
- Meter constant \((K_h)\) = 2.0
- Revolutions \((N)\) = 5
- Time \((t)\) = 20 seconds
- Billing period = 30 days
- Total consumption = 864 kWh
How many days was the rectifier OFF?
Reveal solution
\[
\begin{array}{c}
P_{ac}
=
\frac{3600 \,\times\, 2.0 \,\times\, 5}{20}
=
1800\ \text{W}
=
1.8\ \text{kW} \\[10pt]
H_{on}
=
\frac{864}{1.8}
=
480\ \text{h} \\[10pt]
H_{total}
=
30 \,\times\, 24
=
720\ \text{h} \\[10pt]
H_{off}
=
720 - 480
=
240\ \text{h} \\[10pt]
\text{Days OFF}
=
\frac{240}{24}
=
10\ \text{days}
\end{array}
\]
Answer: 10 Days OFF
Problem 2 — Efficiency calculation
You are performing an annual survey. Calculate the rectifier efficiency.
- DC output = 50 V, 20 A
- Meter constant \((K_h)\) = 5.0
- Revolutions \((N)\) = 2
- Time \((t)\) = 30 seconds
What is the rectifier efficiency?
Reveal solution
\[
\begin{array}{c}
P_{dc}
=
50 \,\times\, 20
=
1000\ \text{W} \\[10pt]
P_{ac}
=
\frac{3600 \,\times\, 5.0 \,\times\, 2}{30}
=
1200\ \text{W} \\[10pt]
\eta
=
\left(\frac{1000}{1200}\right) \,\times\, 100
=
83.33\%
\end{array}
\]
Answer: 83.3% Efficiency
Problem 3 — Exam sample problem (downtime)
(From Appendix F of Exam Guide) How long has this rectifier been OFF during this 30-day period given the following data?
- DC output = 15 V and 10 A
- Meter data = 4 revolutions per minute \((K_h = 1.0)\)
- Consumption = 120 kWh for 30 days
How many days was the unit OFF?
Reveal solution
\[
\begin{array}{c}
P_{ac}
=
\frac{3600 \,\times\, 1.0 \,\times\, 4}{60}
=
240\ \text{W}
=
0.24\ \text{kW} \\[10pt]
H_{on}
=
\frac{120}{0.24}
=
500\ \text{h} \\[10pt]
H_{total}
=
30 \,\times\, 24
=
720\ \text{h} \\[10pt]
H_{off}
=
720 - 500
=
220\ \text{h} \\[10pt]
\text{Days OFF}
=
\frac{220}{24}
=
9.16\ \text{days}
\end{array}
\]
Answer: ~9 Days OFF
Problem 4 — Downtime (low consumption)
Determine the hours of downtime for a rectifier with the following characteristics:
- Meter constant \((K_h)\) = 3.0
- Revolutions \((N)\) = 2
- Time \((t)\) = 30 seconds
- Period = 30 days
- Consumption = 360 kWh
How many hours was the rectifier OFF?
Reveal solution
\[
\begin{array}{c}
P_{ac}
=
\frac{3600 \,\times\, 3.0 \,\times\, 2}{30}
=
720\ \text{W}
=
0.72\ \text{kW} \\[10pt]
H_{on}
=
\frac{360}{0.72}
=
500\ \text{h} \\[10pt]
H_{off}
=
(30 \,\times\, 24) - 500
=
720 - 500
=
220\ \text{h}
\end{array}
\]
Answer: 220 Hours OFF
Problem 5 — Finding meter constant (\(K_h\))
You are troubleshooting a rectifier installation. You measure the AC input with a clamp-on wattmeter and find it is drawing 2,000 W. You time the utility meter disk and it makes 5 revolutions in 18 seconds.
What is the meter constant \((K_h)\) of the utility meter?
Reveal solution
\[
\begin{array}{c}
K_h
=
\frac{P_{ac} \,\times\, t}{3600 \,\times\, N} \\[10pt]
K_h
=
\frac{2000 \,\times\, 18}{3600 \,\times\, 5} \\[10pt]
K_h
=
\frac{36000}{18000}
=
2.0
\end{array}
\]
Answer: \(K_h = 2.0\)
Problem 6 — Power cost estimation
A rectifier operates at 60 V and 40 A with an efficiency of 80%. The local utility rate is $0.15 per kWh. The unit runs continuously with 0 days downtime.
What is the estimated monthly 30-day power bill?
Reveal solution
\[
\begin{array}{c}
P_{dc}
=
60 \,\times\, 40
=
2400\ \text{W} \\[10pt]
P_{ac}
=
\frac{2400}{0.80}
=
3000\ \text{W}
=
3.0\ \text{kW} \\[10pt]
E
=
3.0 \,\times\, 24 \,\times\, 30
=
2160\ \text{kWh} \\[10pt]
\text{Cost}
=
2160 \,\times\, 0.15
=
324.00
\end{array}
\]
Answer: $324.00
Problem 7 — Full operation verification
Determine if the rectifier has been running continuously based on the monthly bill.
- Meter constant \((K_h)\) = 1.5
- Revolutions \((N)\) = 10
- Time \((t)\) = 60 seconds
- Period = 30 days
- Consumption = 648 kWh
How long has the rectifier been OFF?
Reveal solution
\[
\begin{array}{c}
P_{ac}
=
\frac{3600 \,\times\, 1.5 \,\times\, 10}{60}
=
900\ \text{W}
=
0.9\ \text{kW} \\[10pt]
H_{on}
=
\frac{648}{0.9}
=
720\ \text{h} \\[10pt]
H_{total}
=
30 \,\times\, 24
=
720\ \text{h} \\[10pt]
H_{off}
=
720 - 720
=
0\ \text{h}
\end{array}
\]
Answer: 0 Hours OFF
Problem 8 — Troubleshooting low efficiency
You find an older air-cooled rectifier. You measure the output at 20 V and 10 A. You time the meter \((K_h = 3.0)\) at 2 revolutions in 40 seconds.
What is the efficiency, and what does it indicate?
Reveal solution
\[
\begin{array}{c}
P_{dc}
=
20 \,\times\, 10
=
200\ \text{W} \\[10pt]
P_{ac}
=
\frac{3600 \,\times\, 3.0 \,\times\, 2}{40}
=
540\ \text{W} \\[10pt]
\eta
=
\left(\frac{200}{540}\right) \,\times\, 100
=
37\%
\end{array}
\]
37% is very low compared with typical values around 60% to 80%, which indicates probable aging or failure of selenium stacks.
Answer: 37% (Stack Failure)
You need to size a breaker for a single-phase rectifier. The unit outputs 50 V and 50 A. The efficiency is 85%. The AC supply voltage is 120 V.
What is the AC input current?
Reveal solution
\[
\begin{array}{c}
P_{dc}
=
50 \,\times\, 50
=
2500\ \text{W} \\[10pt]
P_{ac}
=
\frac{2500}{0.85}
=
2941\ \text{W} \\[10pt]
I_{ac}
=
\frac{2941}{120}
=
24.5\ \text{A}
\end{array}
\]
Answer: 24.5 Amps AC
Problem 10 — Downtime (slow meter/long cycle)
A small rectifier is on a bi-monthly 60-day reading cycle. Calculate the downtime.
- Meter constant \((K_h)\) = 1.0
- Revolutions \((N)\) = 6
- Time \((t)\) = 60 seconds
- Period = 60 days
- Consumption = 200 kWh
How many days was the unit OFF?
Reveal solution
\[
\begin{array}{c}
P_{ac}
=
\frac{3600 \,\times\, 1.0 \,\times\, 6}{60}
=
360\ \text{W}
=
0.36\ \text{kW} \\[10pt]
H_{on}
=
\frac{200}{0.36}
=
555.55\ \text{h} \\[10pt]
H_{total}
=
60 \,\times\, 24
=
1440\ \text{h} \\[10pt]
H_{off}
=
1440 - 555.55
=
884.45\ \text{h} \\[10pt]
\text{Days OFF}
=
\frac{884.45}{24}
=
36.85\ \text{days}
\end{array}
\]
Answer: ~36.9 Days OFF
