In a delta or mesh connection the dissimilar ends of the three-phase windings are joined together i.e., the ‘starting end of one phase is joined to the ‘finishing end of the other phase and so
on as shown in Fig. 17. In other words, the three windings are joined in series to form a closed mesh.
Three leads are taken out from the junctions as shown and outward directions are taken as positive.
(a) Relation between line voltages and phase voltages:
Since in delta or mesh connected system, only one phase is included between any pair of line
outers, therefore potential difference between the line outers, called the line voltage, is equal to phase
i.e., Line voltage, EL = phase voltage, Eph‘
(b) Relation between line currents and phase currents:
From Fig. 17 it is obvious that line current is the vector difference of phase currents of two phases concerned.
Thus, line current, IR = IYR – IRB (Vector difference)
= IYR + (-IRB) (Vector sum)
Similarly, IY = IBY – IYR and IB = IRB – IBY
Fig. 17. Delia or mesh connected diagram.
Refer Fig. 18. Since phase angle between phase current IYR and – IRB is 60°,
Fig. 18. Vector diagram for delta connected net work.
Assuming the delta connected system or network be balanced, the phase current in (In(‘h
winding is equal and lot each be equal to Iph (i.e., IYR = lBY = IRB = Iph) …(5)
Hence, in terms of line values, the above expression for power becomes
where Ф = the phase power factor angle.
(Apparent power = )
In case of delta or mesh connected system the following points are worthnoting:
(i) Line voltages are equal to phase voltages.
(ii) Line currents are times phase currents.
(iii) Line currents are 120o apart.
(iv) Line currents are 30o behind their respective phase currents.
(v) The angle between line currents and corresponding line voltages is (300 ± Ф) as in the star system.
(vi) True power = , where Ф is the phase angle between respective phase current and phase voltage.
(vii) Apparent power =
(viii) In balanced system, the resultant e.m.f. in the closed circuit will be zero.
i.e., ERY+EYB+EBR = 0.
Hence, there will no circulating current in the mesh if no-load is connected to the lines.
Example 11. Three identical coils connected in delta across 400 V, 50 Hz, 3-phase supply take a line current of 15 A at a power factor 0.8 lagging. Calculate:
(i) The phase current, and
(ii) The impedance, resistance and inductance of each winding.
Solution. Line voltage, EL = 400 V
Line current, IL = 15 A
Power factor, cos = 0.8 lagging
Iph ; Zph ; Rph ; L :
Phase voltage, Eph = EL = 400 V
(i) Phase current, 〖 I〗_ph = I_L/√3= 15/√3=8.66A. (Ans.)
(ii) Impedance of each phase,〖 Z〗_ph= E_ph/I_ph = 400/8.66=46.19 Ω.(Ans.)
Resistance of each phase, R_ph=Z_ph cosΦ=46.19 ×0.8=36.95 Ω.(Ans.)
Reactance of each phase, X_ph=Z_ph sinΦ=46.19 √((1-cos^2 Φ))
=46.19 √((1-〖(0.8)〗^2 )=27.71 Ω
∴Inductance, L= X_ph/2πf= 27.71/(2π ×50)=0.088 H.(Ans.)
Example 12. A 220 V; 3-phase voltage is applied to a balanced delta-connected 3-phase load
of phase impedance (6 + j8).
(i) Find the phasor current in each line.
(ii) What is the power consumed per phase?
(iii) What is the phasor sum of the three line currents? What does it have this value?
Solution. Refer Fig. 19.
Resistance per phase, R_ph=6 Ω
Reactance per phase, X_ph=8 Ω
Impedance per phase, Z_ph=√(R_(〖ph〗^2 )+ X_(〖ph〗^2 ) )= √(6^2+ 8^2 )=10 Ω
(i) Phase current, I_ph=E_ph/Z_ph = 220/10=22 A
∴ Line current, I_L=√3 ×22=38.1 A (Ans.)
(ii) Power consumed per phase,
P_ph=I_(〖ph〗^2 ) × R_ph= 〖22〗^6 ×6=2204 W. (Ans)
(iii) Phasor sum would be zero because the three currents are equal in magnitudes and have a mutual phase difference of 120°.
Solution by Symbolic Notation. Let ERY is taken as a reference vector (Fig, 20).
I_RY=E_RY/Z= (220∠0^o)/(10∠〖53〗^o 8′)= 22∠〖-53〗^o 8^’,=(13-j17.6) A
I_YB=E_YB/Z= (220∠〖-120〗^o)/(10∠〖53〗^o 8′)= 22∠〖-173〗^o 8^’,=(-21.84-j2.63)
I_BR=E_BR/Z= (220∠〖120〗^o)/(10∠〖53〗^o 8′)= 22∠〖66〗^o 52′,=(8.64+j20.23)
(i) Current in each line:
I_R=I_RY- I_BR=(13.22-j17.6)- (8.64+j20.23)=4.58-j37.83 = 38.1 ∠83.1°Ans.
I_Y=I_YB- I_RY=(- 21.84 -j2.63) – (13.22 – j17.6)
=21.84 – j2.63 – 13.22 +j17.60 = – 35.06 + j14.97 =38.12∠〖156.8〗^o.(Ans.)
(ii) Power consumed per phase:
Using conjugate of voltage, we get for R-phase
PVA = ERY.IRY = (220 – j0)(13.22 – j17.6) = (2908.4 – j3872) volt ampere
True power per phase = 2.908 kW. (Ans.)
(iii) Phase sum of the three line currents
= (4.58 – j37.83) + (- 35.06 + j14.96) + (30.48 + j22.86) = 0
Hence, the phasor sum of three line currents drawn by a ‘balanced load’ is zero. (Ans.)
Example 13. A delta-connected balanced 3-phase load is supplied from a 3-phase, 400 V supply. The line current is 30 A and the power taken by the load is 12 k W. Find:
(i) Impedance in each branch; and
(ii) The line current, power factor and power consumed if the same load is connected is star.
Eph = EL = 400 V
∴ I_ph=I_L/√3= 30/√3=17.32 A
(i) Impedance per phase
Z_ph=E_ph/I_ph = 400/17.32=23.09 Ω. (Ans)
Now P=√3 E_L I_L cosΦ
or cos〖Φ (power factor)= 〗 12000/(√3 ×400×30)=0.577
E_ph=E_L/√3= 400/√3=231 V
I_L=I_ph E_ph/Z_ph = 231/23.09=10 A (Ans.)
Power factor, cos ϕ = 0.577 (since impedance is same)
Power consumed = √3 E_L I_L cos〖Φ=√3〗×400×10×0.577=3997.6W (Ans)
Example 14. Three 50 n non-inductive resistances are connected in (i) star, (ii) delta across a 400 V, 50 Hz., 3-phase mains. Calculate the power taken from the supply system in each case. In the event of one of the three resistances getting opened, what would be the value of the total power taken from the mains in each of the two cases.
Solution. Star connection:
Phase voltage, E_ph=E_L/√3=400/√3=231 V
Phase current, I_ph=E_ph/R_ph =231/50=4.62 A
Power consumed, P = 3I_(〖ph〗^2 ) R_ph= 3 × 〖4.62〗^2 × 50 = 3200 W.(Ans.)
[ or P = √3 E_L I_L cosΦ= √3 × 400 × 4.62 ×1 = 3200 W.] Delta connection :
Phase voltage, E_ph=E_L=400 V
Phase current, I_ph=E_ph/R_ph =400/50=8 A
Power consumed, P = 3I_(〖ph〗^2 ) R_ph= 3 × 8^2 × 50 = 9600 W.(Ans.)
When one of the resistances is disconnected:
(i) Star connection. Refer Fig. 21.
When one of the resistances is disconnected, the circuit is no longer 3-phase but converted into single-phase circuit, having two resistances each of 50 ohm connected in series across supply of 400V.
Hence line current, I_L=E_L/〖2R〗_ph =400/(2×50)=4 A
Power consumed, P = 42 (50 + 50) = 1600 W. (Ans.)
[or P = VI cos Φ= 400×4×1 = 1600 W].
(ii) Delta connection. Refer Fig. 22.
Potential difference across each resistance, EL = 400 V
Current in each resistance =400/50=8 A
Power consumed in both resistances = 2 × 82 × 50 = 6400 W. (Ans.)
[or P = 2×E_ph I_ph cos Φ= 2×400×8×1 = 6400 W].
Example 15. The secondary of’a 3-phase star-connected transformer, which has a phase voltage of 230 V feeds a 3-phase delta connected load; each phase of which has a resistance of 30 Ω
and an inductive reactance of 40 Ω. Draw the circuit diagram of the system and calculate:
(i) The voltage across each phase of load,
(ii) The current in each phase of load,
(iii) The current in the transformer secondary windings, and
(iv) The total power taken from the supply and its power factor.
Solution. Refer Fig.23.
Resistance per phase, R_ph=20 Ω
Reactance per phase, X_ph=40 Ω
Phase voltage across transformer secondary,
Line voltage across delta connected load = line voltage across transformer secondry
√3 ×230=400 V (app.)
(i) Voltage across each phase of the load,
(ii) Current in each phase of the load,
I_ph=E_ph/Z_ph =400/√(R_(〖ph〗^2 )+X_(〖ph〗^2 ) )=400/√(〖30〗^2+〖40〗^2 )=400/50=8 A.(Ans.)
(iii) Current in the transformer secondary
=line current of load=√3×I_ph=√3×8=13.86 A.(Ans.)
(iv) Power factor, cosΦ=R_ph/Z_ph =30/50=0.6 (Ans.)
Total Power consumed P=√3 E_L I_L cosΦ=√3×400×13.86×0.6=5761.5 W.(Ans)
Example 16. Three non-inductive resistances, each of 50 ohms are connected in star to a
3-phase 415 volts supply. Three inductive coils of 100 ohms each connected in delta fashion are also connected to the same supply. Calculate:
(i) Line current
(ii) Power factor of the system.
(AMIE Winter, 1999)
Solution. The connection diagram and the vector diagram of the star and delta connected
impedances are shown in Fig. 24 (a) and (b).
The voltage E10 between the line 1 and neutral is taken along the X-axis. Since the load is
balanced, it will suffice to determine the current in one line only.
Applying Kirchhoff’s law to junction 1, we have
Now E_10=415/√3=239.6 V
Now V_12=E_10+E_02=E_10+(-E_20 )=E_10-E_20
(i) Line current:
∴ I_(〖11〗^’ )= I_10+I_12+I_13
Hence line current =8.639 A. (Ans.)
Power factor =〖cos56.3〗^o=0.5548. (Ans.)
Example 17. (Balanced parallel loads) Three star-connected impedances Z1 = (20 + j37.7) Ω per phase are in parallel with three delta-connected impedances Z2 = (30 – j159.3) Ω per phase. The line voltage is 398 V. Find the line current, power and reactive volt-amperes taken by the combination.
(AMIE Winter, 1998)
Solution. Any balanced star-connected system can be completely replaced by an equivalent delta-connected system or vice-versa because of their relationship between phase and line voltages and currents. For example, a balanced star-connected load having an impedance of magnitude Z with a power factor cos ɸ (or Z ∠ɸ in each phase can be replaced by an equivalent
delta-connected load having an impedance of magnitude 3Z and power factor ɸ (i.e., 3Z ∠ɸ) in each phase.
This may be established as follows:
For a balanced star-connected load, let the line voltage be VL, line current be IL and impedance per phase be Z.
Phase voltage, V_ph= V_ph/√3
Phase current, I_ph= I_L
∴Impedance per phase,Z_star= V_ph/I_ph =(V_L √3)/I_L =1/√3=V_L/I_L ….(i)
Now in equivalent delta-connected system for the same line values of voltage and current as in case of star-connected systems,
Phase voltage, V_ph= V_L
Phase current, I_ph= I_L/√3
∴Impedance per phase,Z_delta= V_ph/I_ph =V_L/(I_L √3)=√3 V_L/I_L ….(ii)
Comparing (i) and (ii), we have
Z_star= 1/3× Z_(delta ) or Z_delta= 〖3Z〗_star
The equivalent delta load of the star-connected load is given as 3Z, i.e., (60 + j113.I) Ω
Line voltage, VL = 398 V
Phase voltage, Vph = 398 V in case of delta-connections.
Taking phase voltage as reference phasor
〖 V〗_ph=398 ∠0^o V
Phase current for Impedance〖 Z_1, I〗_ph1=V_ph/Z_1 =(398 ∠0^o)/((60+j113.1))
=(398 ∠0^o)/(128 ∠〖62.05〗^o )=3.109∠〖-62.05〗^o A or (1.46-j2.746)A
Phase current for Impedance〖 Z_2, I〗_ph2=(398 ∠0^o)/((30+j159.3))
=(398 ∠0^o)/(162.1 ∠〖-79.33〗^o )
=2.455∠〖79.33〗^o A or (0.4545+j2.412)A
Total phase current I_ph=I_ph1+I_ph2
=(1.914-j0.334)A or 1.943 ∠〖-9.9〗^o
Total line current, 〖 I〗_L=√3 I_ph √3×1.943=3.36A.(Ans.)
Power, P=√3 V_L I_L cosΦ
Reactive Power, √3 V_L I_L sinΦ
=√3×398×3.36×〖sin 〗(〖-9.9〗^o )
=-398.2 VAR or 398.2 VAR (inductive). (Ans.)