Answer / gajanan bharad

I am too seeking right answer, but I know that the calculations above are incorrect. We have to consider the heat generated by BTSs and not the power consumed by BTSs. We have to take into account the heat generated/ rate of heat generation in Watts by all the equipment inside the shelter. There are 48V SMPS, Battery sets, BTSs, Microwave equipment, lights and assume 2 persons working. Heat generated by the SMPS equals input power minus the output power (i.e. loss of energy in terms of heat). But for rest items need the information on heat generated. Then only we can arrive at correct air conditioner capacity requirement.

Answer / suraj.dahiya

Genarally 1 Ton capacity means---The cooling produced by
melting of 1 ton of ice in a day is.taken as unit of
refrigeration (in tons). Now unit of refrigeration used in
S.I. System is KW. 1 ton of refrigeration = mass of ice x
latent heat of ice / 24 hrs i.e
= 900 kg x 335 KJ/Kg ( one short ton = 900 kgs )
---------------------
24 x 3600 secs

= 3.4895 KJ/Sec Now we should convert KJ to KW i.e 1KJ=1KW
so 3.5 KJ / Sec = 3.5 KW
that is 1 ton of refrigeration = 3.5 KW
Always in telecom sheleters we shold take the load of BTS
For Ex: In shelter there are 4Nos of BTS,If each BTS is
consuming 20A the total Amps for 4BTS is 80A.
So The total Load of the BTS is P=VI i.e
P=230*80=18400Watts i.e 18.4KW Now each 1Ton =3.5KW we can
suggest for 5Ton of AC for 18.4 KW
plus surface area of BTS shelter suppose (9feet*6 feet)=54
sqft area.
required temp =54*25=1.350 kw
so total kw = 18.4+1.350=19.7KW
total Tr. required to cool the shelter @25 C =19.7/3.5
=5.6 Tr. AC required

Answer / chandrashekar

Genarally 1 Ton capacity means---The cooling produced by
melting of 1 ton of ice in a day is.taken as unit of
refrigeration (in tons). Now unit of refrigeration used in
S.I. System is KW. 1 ton of refrigeration = mass of ice x
latent heat of ice / 24 hrs i.e
= 900 kg x 335 KJ/Kg ( one short ton = 900 kgs )
---------------------
24 x 3600 secs

= 3.4895 KJ/Sec Now we should convert KJ to KW i.e 1KJ=1KW
so 3.5 KJ / Sec = 3.5 KW
that is 1 ton of refrigeration = 3.5 KW
Always in telecom sheleters we shold take the load of BTS
For Ex: In shelter there are 4Nos of BTS,If each BTS is
consuming 20A the total Amps for 4BTS is 80A.
So The total Load of the BTS is P=VI i.e
P=230*80=18400Watts i.e 18.4KW Now each 1Ton =3.5KW we can
suggest for 5Ton of AC for 18.4 KW

Capacitor panel CT is normally fixed on Y phase of the main panel.why???

3 Answers  

---

please, Send me last technical test question paper in my email id..

0 Answers  

---

what is bausbar?

4 Answers   Essar,

---

Why we use a shunt capacitor or synchronizing Motor To Improve a Power factor pls reply me in this mail dinesh_elect_eng@yahoo.com

4 Answers  

---

What will happen if the 50HZ Transformer is connected with 60HZ power Supply

2 Answers  

---

where to use CT and PT?

4 Answers   CEB, Reliance,

---

What is meant by sotf,lbb,power swing,pole discrepency,auto reclose?

1 Answers   Bunge, Electrical, Omega, TCS, TNEB Tamil Nadu Electricity Board,

---

Is any AC drive(VFD) had a feature of DC injection braking? if available please mention the detail. i.e drive can supply braking torque when the motor in idle state.

0 Answers  

---

can we use ordinary transformer in place of autotransformer?give reasons also

1 Answers  

---

sourse is 66 KV / 350 MVA, transfarmer is 1000 KVA 6.6 / 0.415 kv, iF fault is 28.75 MVA on 0.415 KV Bus then calculat the Transfarmer Impidance

0 Answers  

---

What is PLL? what are it's advantages?

1 Answers   BREB,

---

RELAY 1. REF RELAY : • OPERATION WITH CT CURRENT FLOW DIRECTION. • WHY REF RELAY WHEN DIFFERENTIAL RELAY IS THERE. • REF RELAY STABILITY CHECK BY PRIMARY INJECTION – WHICH POINT FOR CUR RENT INJECTION. • WHY REF RELAY NEED STABILISING RESISTOR. IF RESISTOR SHORTED OR BYPASSED, WHAT WILL HAPPEN. • WHICH TYPE OF RELAY FOR REF, HIGH IMPEDANCE OR LOW IMPEDANCE. • IF NEUTRAL CT IS NOT CONNECTED IN THE CIRCUIT, WHAT WILL HAPPEN. • REF PROTECTION ACTS ON WHICH OF FOLLOWING: SINGLE PH. TO E/F; PH-PH TO E/F; THREE PH. TO E/F. 2. DISTANCE RELAY : • BASIC PRINCIPLE AND TYPES • SWITCHED MODE PROTECTION – WHY IS IT CALLED SO. • SCHEMES USED • WHY ZONE-1 SET AT 80% • SETTING RANGE FOR ZONE-2 AND ZONE-3 • HOW SINGLE PHASE TO GROUND FAULT MEASUREMENT IS DONE. • 3. WHAT IS POLARITY. CT LOCATION IF INTERCHANGED WHAT WILL HAPPEN. 4. SOLKOR RELAY: 1. WHY SOLKOR IS USED. 2. CIRCUIT DIAGRAM OF SOLKOR RELAY 3. DIFFERENT PARTS OF RELAY 4. HOW IT WORKS. 5. EXPLAIN BY VECTOR DIGRAM. 6. WHAT ADVANTAGE FOR USING SUMMATION TR. 7. DIFFERENCE BETWEEN R & RF WITH CIRCUIT. 8. ADJUSTMENT OF PADDING RESISTANCE. 9. WHAT IS LEAKAGE CURRENT AND HOW TO MEASURE. 10. WHAT HAPPEN IF LINK IS LEFT OPEN. 11. 5KV/15KV INSULATION TR. ARE USED IN WHICH SYSTEM. 12. WHAT PRECAUTION TAKEN DURING TESTING. Relay Page 1 of 5 13. PROCEDURE BEFORE TESTING AND AFTER NORMALISATION. 14. HOW RELAY IS TESTED IN LIVE CONDITION. 15. RANGE OF SPILL CURRENT ANS SETTING OF RELAY 16. IF MEASURED CURRENT IS 5MILLI AMPS, CAN RELAY OPERATE. 17. EACH FAULT APPROXIMATE VALUE OF PICK UP. 18. WHAT IS THE REASON SOLKOR-R MODE PILOT SHORTED CONDITION RELAY CURRENT PRESENT. 19. AFTER SHUTDOWN, WHY NEED TO OPEN SOLKOR PILOT, IF NOT WHAT WILL HAPPEN. 20. AFTER NORMALISING WHAT CURRENT TO BE MEASURED, WHAT IS ACCEPTABLE LIMIT. 21. WHAT HAPPENS IF PILOT OPENS FROM RELAY, WHEN FEEDER IS LOADED. 22. CAN WE USE THIS RELAY AS O/C OR E/F RELAY. 23. ANY RELAY OTHER THAN PILOT CABLE TYPE. 24. IF FEEDER TRIPS ON SOLKOR, WHAT TO DO. 25. WHAT HAPPENS IF PILOT POLARITY REVERSE. 26. WHAT HAPPENS WHEN CT SHORTED AT ONE END. 27. WHAT HAPPENS IF PILOT SHORTED – STATUS OF RELAY 28. WHAT HAPPENS TO RELAY ,IF POWER CABLE GETS OPEN. 29. WHAT HAPPEN IF PADDING RESISTOR IS SHORTED AT ONE END. 5. TR. DIFFERENTIAL: 1. CIRCUIT DIAGRAM 2. DIRECTION OF FAULT CURRENT. 3. WHEN THE FAULT IS IN ZONE AND OUT OF ZONE. 4. CT CONNECTION IN PY. AND SY. SIDE. 5. BIAS SETTING AND BIAS CURRENT. 6. WHICH BIAS SETTING USED IN TRS. FOR HIGHER TAPS. 7. PURPOSE OF RESTRAINING BIAS COIL. 8. WHY BIASING REQUIRED. 9. EXPLAIN IPCT, FOR EXAMPLE, DYN10 VECTOR GROUP. WHY MATCHING TR. USED. 10. WHY NEED 2ND HARMONIC BLOCKING DURING INRUSH. 11. WHICH HARMONIC IS MORE 2ND OR 3RD. 12. HOW TO AVOID FALSE TRIPPING IN DIFF. PROTECTION WHEN INRUSH CURRENT FLOW DURING TR. IS ENERGISED. Relay Page2 of5 6. DIRECTIONAL RELAY: 1. CIRCUIT FOR USE OF DIRECTIONAL PROTECTION SYSTEM. 2. WORKING ZONE / NON-WORKING ZONE. 3. HOW THE CORRECT DIRECTION OF FLOW OF CURRENT IS DETERMINED. 4. CT & PT CONNECTION DIAGRAM. 5. SETTING OF PHASE & EARTH FAULT ANGLE AND OPERATING REGION. 6. WHICH DIRECTION IT IS PROTECTING. 7. WHAT PROTECTION AVAILABLE FOR POWER TR. 8. WHY CT CONNECTION IS STAR, IF TR. WINDING IS DELTA. 9. DIFFERENCE BETWEEN REF AND DIFF. RELAY 10. WHERE IS O/C & E/F RELAY CONNECTED EITHER HV SIDE OR LV SIDE AND HOW E/F RELAY WORKS IF TR. WINDING IS DELTA CONNECTED. 11. IN DELTA WINDING OF TR., IF ONE WINDING IS CUT AND NOT EARTHED, WHICH PROTECTION WILL ACT. 12. IF DIFF. & REF RELAY ARE NOT WORKING, WHAT IS BACK UP PROTECTION IN TR. 13. WHAT IS PURPOSE OF NGR AND HOW STANDBY E/F RELAY WORKING, TESTING PRECAUTIONS IF COMMON NGR IS USED. 14. WHAT IS PROTECTIVE ZONE OF STANDBY E/F RELAY. 15. WHAT ARE PROTECTION AVAILABLE FOR FEEDER. 16. AFTER TR. IS NORMALISED, WHAT CURRENT TO BE MEASURED. 17. BUSBAR PROTECTION: a. BASIC PRINCIPLE AND DIAGRAM WITH CURRENT DIRECTION. b. WHY OVER LAPPING CT’S REQUIRED, IF NOT WHAT WILL BE THE EFFECT. c. WHY CT SUPERVISION REQUIRED, HOW TRIPPING IS AVOIDED, WHAT IS CT SUPERVISION SETTING VALUE. d. HIGH IMPEDANCE & LOW IMPEDANCE TYPE e. TEST PROCEDURE(STEP BY STEP) f. AFTER NORMALISING, SPILL CURRENT VALUE AND LIMIT. Relay Page3of5 18. CIRCUIT BREAKER FAILURE – FUNCTION, WHY NEEDED, HOW TRIPPING ARRANGEMENT. 19. WHY INSTANTANEOUS ELEMENT IS NOT USED IN THE NETWORK. 20. WHEN FEEDER IS LIVE AND BY MISTAKE LINE EARTH SWITCH IS CLOSED, WHICH RELAY OPERATES – O/C OR E/F. 21. DURING TR. SWITCHING ON, IF DIFF. RELAY IS BLOCKED BY 2ND HARMONICS, WHAT ARE THE OTHER PROTECTION IN SERVICE. 22. WHICH OF FOLLOWING PROTECTION OPERATES FAST, WHEN 11KV CABLE COMPARTMENT GET FLASH OVER: CABLE DIFFERENTIAL; OVERCURRENT & EARTH FAULT; ARC PROTECTION. 23. WHICH PROTECTION OPERATES WHEN ANY ONE OF THE OHL PHASE IS CUT AND HANGING, NOT TOUCHED WITH EARTH. 24. WHAT ARE DIFFERENT TTB ALARMS – HOW YOU KNOW THERE IS ANY ALARM. 25. WHAT IS CIRCULATING CURRENT AND HOW IT WILL COME. 26. IF ONE TR. TRIPPED ON DIFF. RELAY WHAT YOU WILL DO. 27. IF SUB-STATION TRIP ON BUSBAR PROTECTION, HOW YOU WILL RESTORE. 28. WHAT IS CT SATURATION. HOW IT IS CARRIED OUT. 29. D.C EARTH FAULT – HOW YOU WILL FIND OUT. Relay Page 4 of5 30. PILOT WIRE SUPERVISION – HOW IT WORKS 31. IN 11KV SWITCHGEAR DBB SCHEME, ONE E/F RELAY IS FAULTY IN ONE FEEDER AND THERE IS NO SPARE FEEDER – HOW YOU WILL NORMALISE THE FEEDER. 32. WHICH PROTECTION IS USED FOR LONG CABLE FEEDER 33. DIFFERENCE BETWEEN TRAFO. DIFF. AND CABLE DIFF. PROTECTION. 34. WHAT TYPE CHARACTERISTIC AVAILABLE FOR O/C AND E/F RELAY. 35. WHAT IS THE DIFFERENCE BETWEEN STATIC AND ELECTRO- MECHANICAL O/C RELAYS. 36. WHAT IS DEAD ZONE 37. HOW TRIPPING OF OTHER ZONE AVOIDED WHEN BUS SECTION CB IN OPEN CONDITION FOR DEAD ZONE FAULT. 38. IF OUT OF STEP ALARM COMES WHILE ALL TR. IN SAME TAP POSITION, WHAT YOU WILL DO. 39. WHAT YOU WILL DO IF AN 11KV FEEDER FAILED TO TRIP ON: O/C or E/F or SOLKOR. HOW YOU RESTORE THE SUPPLY. Relay Page 5 of5

2 Answers   KSEB, Siemens,

---