Electrical and Electronic Measurements Test 4

Concept:

Energy consumption is, E = V × I × cos ϕ × t × 10^-3 kWh

And error A_m - A_t

Where A_m = measured value

A_t = true value

Calculation:

V = 230, I = 10, cos ϕ = 0.8, t = 30 min

(Energy consumption)_t = V × I × cos ϕ × t × 10^-3 kW

\(= 230 \times 10 \times 0.8 \times \left( {\frac{{30}}{{60}}} \right) \times {10^{ - 3}}\)

(E.C)_t = 0.92 kWh

(E.C)_m = 58.25 – 57.35 = 0.90 kWh

Error = A_m - A_t = 0.90 – 0.92 = -0.02 kWh

Speed of the energy meter is given by,

\(N \propto \frac{r}{{{\phi ^2}R}}\)

Where r is resistance

ϕ is magnetic flux

R is radius

Speed of disc is inversely proportional to radius.

The brake magnet is shifted from its position and moved a small distance towards the edge of the disc. So, the radius of disc increases and hence speed decreases.

10 revolutions in 90 sec.

⇒ 400 revolutions in 3600 sec.

⇒ 400 revolutions in 1 hour

Meter constant is 2400 rev/kwh.

Voltage (V) = 240 V

Current (I) = 20 A

At half load, speed of disc = 32 rpm

Energy consumed in one minute at half load and unity power factor is,

E = VI cos ϕ × t × 10^-3

\(= 240 \times \frac{{20}}{2} \times 1 \times \frac{1}{{60}} \times {10^{ - 3}}\)

= 0.04 kWh

Meter constant \(= \frac{{revolutions}}{{energy}}\)

Energy actually consumed = VI cosϕ × t × 10^-3 = 220 × 50 × 0.5 × 1 × 10^-3 = 5.5 kWh

The meter makes 120 revolutions for one unit ie., one kwh of energy consumed

∴ Number of revolutions made = 120 × 5.5 = 660

In case the meter makes 600 revolutions,

Energy consumed in kWh \(= {V_I}\cos \phi \times t \times {10^{ - 3}}\)

Energy recorded \( = \frac{{2400}}{{400}} = 6\ kWh\)

Since no information is given about error, we assume the meter as error free.

\(\Rightarrow {V_{}}I\cos \phi \times t \times {10^{ - 3}} = 6\)

240 × 12 cos ϕ × 8 × 10^-3 = 6

\(\begin{gathered} \cos \phi = \frac{{6000}}{{240 \times 12 \times 8}} \\ \cos \phi = 0.26\\ \end{gathered} \)

Concept:

The energy consumed in 1 hr is

E = VI cos ϕ × t × 10^-3 kWh

And the percentage error \(= \frac{{{{\left( {No.\;of\;revolutions} \right)}_m} - {{\left( {No.\;of\;revolutions} \right)}_t}}}{{{{\left( {No.\;of\;revolutions} \right)}_t}}}\)

Where, m denotes measured value

t denotes measure true value

Calculation:

K = 80 rev/kwH which is a meter constant

V = 230 V, I = 30 A, cos ϕ = 0.6

Energy consumed in one hour is

E = VI cos ϕ × t × 10^-3 kwh

= 230 × 30 × 0.6 × 1 × 10^-3

= 4.14 kWh

No. of revolutions = Energy consumed × K (Meter constant)

= 4.14 × 80 = 331.2

Meter actually makes 330 revolutions.

So, the percentage error \(= \frac{{340 - 331.2}}{{331.2}} \times 100\)

= 2.6%

The positive sign indicates that the meter is running fast.

Concept:

Percentage Error = \(\frac{{{E_m} - {E_t}}}{{{E_t}}} \times 100\)

Where E_m is measured value

E_t­ is true value

Calculation:

Energy consumed by 10 lamps of 100 W each = 10 × 100 × 1 = 1000 Wh

Energy consumed by 3 fans of 60 W each = 3 × 60 × 1 = 180 Wh

Energy consumed by 2 heaters of 1 kW each = 2 × 1000 × 1 = 2000 Wh

Total energy consumed by the domestic load = 1000 + 180 + 2000 = 3180 Wh = 3.180 kWh

Meter constant = 1500 revolutions per kWh

The number revolutions as per meter constant = 3.18 × 1500 = 4770

Energy meter reading = 4500 revolutions

Percentage Error = \(\frac{{4500 - 4770}}{{4770}} \times 100 = - 5.66\;\% \)

Concept:

Energy consumed in t time is,

E = V × I × cos ϕ × t × 10^-3 kWh

And the percentage error \(= \frac{{Measured\;angle - True\;angle}}{{True\;angle}}\)

Revolutions/min = kwh × k

Calculation:

V = 240 V, I = 10 A, cos ϕ = 0.8,

t = 1 minute, k = 600

Now, energy consumed in 1 minute with rated current and 0.8 pf lagging is

= V × I × cos ϕ × t × 10^-3 kWh

\(= 240 \times 10 \times 0.8 \times \left( {\frac{1}{{60}}} \right) \times {10^{ - 3}}\;kWh\)

= 0.032 kWh      

Revolution/min = Energy consumed × Meter constant

= 0.032 × 600

= 19.2 rpm

Now, In the energy meter due to the Induction effect we assume, there are two fluxes; one is from top of the disc which rotates and another at the bottom, with a phase difference of β angle. Due to these interactions of fluxes, the disc rotates.

Where \({T_d} = {T_{{d_1}}} \sim {T_{{d_2}}}\), where T_d = driving force

And T_d α ϕ₁ ϕ₂ sin β [as α ≈ 0]

To convert T_d α ϕ₁ϕ₂ sin β to V_LI_L cos ϕ, we have to make β angle (90 – ϕ).

So, % Error introduced \(= \frac{{\sin \left( {{\rm{\Delta }} - \phi } \right) - \sin \left( {90 - \phi } \right)}}{{\sin \left( {90 - \phi } \right)}}\)

Where, Δ = 86° and p.f = 0.5 lag

∴ ϕ = 60°

Error \(= \frac{{\sin \left( {86^\circ - 60^\circ } \right) - \cos \left( {60^\circ \;} \right)}}{{\cos \left( {60^\circ } \right)}} \times 100\)

Error = -12.3%