Tutorial-OCProt_2015
Q1. A 380V 3φ 60 kW water pump motor is supplied by a 400 mm 2 4-core PVC/SWA/PVC copper cable. The cable is mounted on a perforated cable tray with 2 other similar cables touching. HRC fuses to BS88 are used in the switch board. Length of the cable is 50 m from the switchboard to the Direct-On-Line motor starter, which is adjacent to the motor. The following information may be used in the calculation. (State clearly all assumptions made in all your calculations.)
Design ambient temperature: 40?C Motor starting time: 4 s Starting current: 6.25 times the full-load current Power factor of the motor: 90% Efficiency of the motor: 89%
Earth fault loop impedance external to the installation at the switchboard: 0.1 Ω Circuit protective conductor: Steel wire armour of the cable
Determine the appropriate cable size to comply with all relevant protective requirements of the CoP and IET on Wiring Regulations.
(a) Cable size
Grouping for 3 circuits and T a = 40?C
Starting current I st = 6.25 ? I b ; where I b is full load design current Starting time = 4s Power factor pf = 80%
Design full load current I b
= 114 A
I st = 6.25 ?
Fuse to protect cable: I b ≤ I N
Correction factors, for grouping C g; 3 = 0.81 (Table 4B1);
M
Motor
DOL starter
50 m
Switchboard
BS88 Rating = ?
4/c PVC/SWA/PVC Cu on tray, touching
SWA used as CPC
Ze=0.1Ω
I, A
114 A 712 A
for ambient temperature C a; 40?C = 0.87 (Table 4C1)
Fuses are used for both overload protection and fault protection. The fuse rating of 160A is used for cable sizing.
Tabulated current, I t =
160
0.810.87
?= 227 A Form Table 4D4A col. 5, Select 95 mm 2 PVC cable, I ta = 251A Therefore I z = 251 ? 0.87 ? 0.81 = 177A > In of fuses
Check for voltage drop. (2.5% for final circuit in practice) VD ta,70 = 0.41 (for r); 0.135 (for x); 0.43(for z) (in mV ?A -1?m -1)
VD Ib=114A = 0.43 ? 10-3 ? 50 ? 114 = 2.45V (0.7% of 380V) satisfied! Ploss = 3*Ifl*Ifl*R = 3*114*114*0.41/1.732*50 = 585W Ploss (%) = 585/67527 = 0.87% <1.0% ( OK)
(b) Earth fault
SWA is used as CPC for the circuit made by the 95 mm 2 pvc/swa/pvc Cu cable, R 1 + R 2 = 1.719 m Ω?m -1 (Schedule 2.8B 4/c cable Col. ?) The cross-sectional-area (CSA) of SWA A swa = 160 mm 2 Earth fault loop impedance
Z s = Z E + (R 1 + R 2) = 0.1 + 1.719? 10-3 ? 50 = 0.186 Ω
Prospective earth fault current I EF
I EF = s
Z U 0 = 2200.186 = 1183A
Disconnecting time read from characteristics curve for 160 A fuses to BS88, t dis = 2.8 sec. <5 sec. (OK)
Assume the setting current Ir is equal to the full load current. Ir=35A;
Iz=Ita*Ca*Cg*Ci = 36A * 1 * 1 *1 = 36A
Therefore,Ib<=Ir<=Iz (OK)
(c) Part C (3/c 16mm2 Cable)
In=63A;
Iz=Ita*Ca*Cg*Ci = 83A * 1 * 1 *1 = 83A
Therefore,Ib<=In<=Iz (OK)
Ist=6*Ifl=6*35= 210A
t= 20 sec. for the 63A fuse > 3sec. starting time (OK)
5. IDMT Relay Settings
Q1.
Select relay type and settings for a 2500A MICB in the circuit supplied by a 1500kVA
Solution:
(1) Overcurrent protection – O/C relays (PS = 50% -- 200%) ? Circuit Break and CTs
I fl (Tx) = 1500kVA/sqrt(3)/380 = 2279A
I b is normally less than the transformer full-load current (e.g., 90% of I fl ) =2051A ACB rating I n >I b , therefore, I n =2500A
CT rating is normally equal to ACB rating, but we need to follow the requirement of power companies, that is, CT rating = 2250:5A ? Relay Type:
For fast operation, EI relays are selected for O/C protection ? Relay Settings
General speaking, b n z I I I ≤≤
For IDMT relays, ,b setting p z I I I ≤≤ or b z I CT Rating PS I ≤?≤
PS >= 2279*90%/2250=0.91
Plug setting =100% or Current setting (S) = 5A or Current setting (P) = 2250A
Time setting is determined in such a way that (1) relay operating time is as short as possible, and (2) the O/C protection requirement from the power company shall be fulfilled (discrimination). In other words, under all fault currents specified in the figure (or I f =7kA – 40kA) the operating time of the O/C relays is shorter than the time given in the figure (e.g., t<=74ms at I f =40kA, and t<=1.4s at I f =7kA).
(a) If=40kA
? PSM= I f /CT rating (P) *PS = 40kA/2250A*100% = 17.8
? 1
PSM 80
t 2-=*TM = 80/(17.8^2-1)*TM =0.253*TM <0.074 s
? TM <0.29 TM = 0.1 (not 0.2)
(b) If=7kA
? PSM= If/CT rating (P) *PS = 7kA/2250A*100% = 3.11
? 1
PSM 80
t 2-=*TM = 80/(3.11^2-1)*0.1 =9.22*TM <1.4 s
? TM < 0.15 s
To meet the requirements in (a) and (b) together, TM is selected to be 0.1. Therefore, we use EI IDMT O/C relays with PS=100% and TS= 0.1
(2) Earth-fault protection – E/F relays (PS = 10% -- 40%) given by I e/f =500A
Relay type and relay settings are determined in such a way that under earth fault the circuit is disconnected effectively within 5 sec.
We use 3.0sec. standard inverse IDMT relays for earth fault protection.
PSM=I e/f / PS * CT Rating >1.3 for effective operation under the fault (would be better if PSM>2)
CT ratio is given in Part (1), that is, 2250:5,
Therefore, PS < 500/(2250)/2 = 11.1% PS = 10%
Therefore, PSM = I e/f / plug setting * CT rating = 500/2250*10% = 2.22
1PSM 14.0t 02.0-=
*TM = 122.214
.002.0-*TM =8.71*TM < 5 sec. TM < 0.574 or TM =0.5
Therefore, we use SI IDMT E/F relays with PS=10% and TS= 0.5
Q2. Determine rating of CTs, and type and settings of IDMT relays for a 1600A ACB in an outgoing circuit (Ib=1500A). Check whether discrimination between MICB and ACB is achieved or not. The grading time is 0.1s. The settings of the relays for MICB are given in Q1. It is assumed the maximum fault current the ACB may experience is 25kA, and the minimum and maximum earth-fault currents are equal to 500A and 10kA.
Solution:
O/C protection If=25kA
(i) MICBB Relay R2 (PS=100%, TM=0.1)
? PSM= I f /CT rating (P) *PS = 25kA/2250A*100% = 11.1
? 22
80
1
t PSM =-*TM = 80/(11.1^2-1)*0.1 =0.652*0.1 <=0.0652 s (ii) ACB Relay R1
ACB rating =1600A, CT=1600:5A PS >= 1.06*1500/1600=0.994
Plug setting =100% or Current setting (S) = 5A or Current setting (P) = 1600A ? PSM= I f /CT rating (P) *PS = 25kA/1600A*100% = 15.6
? 12
80
1
t PSM =-*TM = 80/(15.6^2-1)*0.1 =0.329*0.1 <=0.0329 s t2-t1=0.032s < 0.1s (not OK)
Therefore, it is not possible to achieve the discrimination between MICB and ACB. In other words, both MICB and ACB will trip simultaneously if there is a fault at the outgoing circuit.
E/F protection
(a) Min. earth fault current Ie/f =500A (i) MICB Relay R2 (PS=10%, TM=0.5)
PSM = I e/f / PS * CT Rating = 500/2250*10% = 2.22
20.02
0.14
1t PSM =-*0.5 = 122.214.002.0-*0.5 =8.71*0.5 = 4.36 sec. t1 (ii) ACB Relay R1 with Ie/f =500A We use 3.0sec. standard inverse IDMT relays for earth fault protection. PSM=I e/f / PS * CT Rating > 2 PS < 500/(1600)/2 = 31.25% PS = 10% PSM = I e/f / PS * CT rating = 500/1600*10% = 3.1 1PSM 14.0t 02.0-=*TM = 0.02 0.14 3.1251-*TM =6.074*TM < t2-0.1sec = 4.26 sec. TM < 0.701 or TM =0.4 Therefore, we use SI IDMT E/F relays with PS=10% and TS= 0.4 (b) Max. earth fault current Ie/f =10kA (i) MICB Relay R2 (PS=10%, TM=0.5) Ie/f =10kA PSM = I e/f / PS * CT Rating = 10kA/2250*10% = 44.4 T2=2s*0.5 = 1sec. t1 (ii) ACB Relay R1 (PS=10%, TM=0.4) Ie/f =10kA PSM = I e/f / PS * CT rating = 10kA/1600*10% = 62.5 t1=2sec*0.4 = 0.8sec < 0.9 sec. Therefore, we use SI IDMT E/F relays with PS=10% and TS= 0.4