Answer / jugal pandya

Pressure and temp at 1. Superheater Outlet : 67 Kg/cm2 & 485 C
2. Steam Drum : 73 Kg/cm2 & Saturated
3. Economizer inlet : Water inlet at 105 C

From Steam tables,
Enthalpy of Superheated steam , Hsh = 809 Kcal/ kg = 1456 BTU/lb
Enthalpy of Drum water , Hdwat = 305 Kcal/kg = 549 BTU/lb
Enthalpy of inlet water , Hwat = 105 Kcal/kg = 189 BTU/lb

Assume 3% Blowdown from Boiler.

Total Heat Load of the Boiler = Total heat absorbed by water to convert to steam + heat absorbed to get superheated + Blow down losses
= 50000(809-305) + 50000 x 1.03 x (305-105)
= 35.5e06 Kcal/hr = 140.87e06 BTU/hr

Fuel consumption = Heat Load/ (HHV x Efficiency)
= 35.5e06/ (7278 x 0.8649)
= 5639 Kg/hr = 12428 Lb/hr of coal

From previous article on Combustion and efficiency,
Wet gases = 14.05 and Air = 13.12 kg / kg of coal

Therefore, Exhaust gases produced = Fuel consumption x UnitWetGas
= 5639 x 14.05
= 79,228 Kg/hr of wet gases
Combustion Air required = 5639 x 13.12
= 73,984 Kg/hr of combustion air

Feed Water required = 50,000 x 1.03 : 3% Blowdown
= 51,00 Kg/hr

Sizing Calculations :
a) Boiler feed Water Pumps :

Two pumps of 100 % capacity are required one for working and one for standby.

Each pump discharge capacity minimum= 51500 Kg/hr
= 51500/950 : 950 kg/m3 water density
= 53.8 m3/hr
Margin on discharge capacity : 15- 25 %.
Take 20% margin in this case.

So discharge capacity of each pump : 53.8 x 1.2
= 64.6 m3/hr =say 65 m3/hr

If Recirculation valves are not provided, you need to add min recirculation flow to the above figure, which may be about 6-10 m3/hr depending up on pump type and make.

Pump head required = Drum Pressure + Drum elevation + Piping Losses + Control Valve Loss + Other valve losses

= 75 Kg/cm2 + 2.0 + 2.0 +5.0 +2.0
= 86 Kg/cm2
= 86 x 10/0.95 mts of water head at 105C
= 905 mts of WC

Provide up to 5% margin on head. So final Pump head is 905 x 1.05 = 950 m of WC
So BFW pumps (2 nos) rating is 65 m3/hr at 950 m of WC with feed water at 105 C.

b) Sizing calculations of FD Fan :
Forced Draft Fan is required to pump in primary combustion Air into the Boiler furnace. Air from FD fan passes through Air Heater before entering furnace through Grate. Secondary Air Fan (SA fan) supplies secondary combustion air in to the furnace. Usually primary air is 70 -80 % of the total air and balance is supplied as secondary air through SA fan. Secondary air is supplied at a higher pressure to help fuel spreading on the grate called as pneumatic spreading.

Total combustion Air, Kg/hr = 73,984
= 73994/(1.17 x 3600) m3/s :Air density-1.17kg/m3
= 17.56 m3/s

Primary Air , 70% of total , m3/s = 0.7 x 17.56
= 12.3 m3/s

Take 20% margin on discharge capacity. So FD Fan flow is 1.2 x 12.3 = 14.76 m3/s

Head required = Draft loss across Air Heater + Grate + Ducting & others
= 75 mmWC + 75 + 50 mm : Approximate
= 200 mm WC (approximate)
Take 15-20 % margin on head. So FD fan head should be about 230 mm of WC.

Therefore, FD fan rating is 15 m3/s of air at 230 mm WC static head.

Power requirements of FD Fan :
Let us assume Fan efficiency as 75% and Motor Efficiency as 90%.
Power required for FD Fan, BHP = Flow x Head / ( Efficiency x 75.8 )
= 15 x 230 / ( 0.75 x 75.8 )
= 60.7 HP

Motor HP required = 60.7 / 0.9 = 68 HP
Annual cost of operation assuming 7 cents per KWH and 7200 hrs of operation per annum. 0.74 is factor for converting HP to KW. Pl note that unit Electricity charges vary widely across different countries.

= 68 x 0.74 x 0.07 x 7200
= $ 25, 362 /-

c) Sizing calculations of SA Fan :
Secondary Air Fan (SA fan) supplies secondary combustion air in to the furnace.
Secondary Air , 30% of total , m3/s = 0.3 x 17.56
= 5.27 m3/s

Take 20% margin on discharge capacity. So SA Fan flow is 1.2 x 5.27 = 6.3 m3/s
SA fan static head is about 630 mm WC.
Therefore, SA fan rating is 6.3 m3/s of air at 650 mm WC static head.

Power requirements of SA Fan :
Let us assume Fan efficiency as 70% and Motor Efficiency as 90%.
Power required for FD Fan, BHP = Flow x Head / ( Efficiency x 75.8 )
= 6.3 x 650 / ( 0.7 x 75.8 )
= 77.1 HP

Motor HP required = 77.1 / 0.9 = 86 HP
Annual cost of operation assuming 7 cents per KWH and 7200 hrs of operation per annum. 0.74 is factor for converting HP to KW. Pl note that unit Electricity charges vary widely across different countries.

= 86 x 0.74 x 0.07 x 7200
= $ 32,075 /-

d) Sizing calculations of ID Fan :
Induced draft fan or ID Fan is required to evacuate the exhaust gases from Boiler to atmosphere through Duct collectors and chimney. Usually ID should take care of draft loss across the Boiler from furnace to Air heater and then draft loss across Duct Collectors like ESP, Wet Scrubber or mechanical type Cyclone dust collectors .etc.
Total wet gases, Kg/hr = 79,228

Gas Density = 1.3265 Kg/Nm3
Therefore, gas flow in Nm3/hr = 79,228 / 1.3265
= 59227 Nm3/hr
= 16.6 Nm3/s

Gas flow at 150C in m3/s = 16.6 x (273+150)/273 = 25.7
ID Fan capacity taking 20% margin on flow = 25.7 x 1.2
= 30 m3/s

ID Fan static Head = Draft Loss in (Boiler + Duct + Dust collector)
= 150 + 50 + 50 mm WC : Approximate
= 250 mmWC

Taking 20% margin on head, ID Fan head = 250 * 1.2 = 300 mm WC

Power requirements of ID Fan :
Let us assume Fan efficiency as 75% and Motor Efficiency as 90%.
Power required for ID Fan, BHP = Flow x Head / ( Efficiency x 75.8 )
= 30 x 300 / ( 0.75 x 75.8 )
= 158 HP

Motor HP required = 158 / 0.9 = 175 HP
Annual cost of operation assuming 7 cents per KWH and 7200 hrs of operation per annum. 0.74 is factor for converting HP to KW. Pl note that unit Electricity charges vary widely across different countries.
= 175 x 0.74 x 0.07 x 7200 = $ 65,268 /-

As a foreign applicant, what do you understand about our company before sending your application CV for consideration?

0 Answers   Nexel,

---

What are the minerals needed to make or build a car?

0 Answers  

---

What is polytropic process ? Under what conditions it approaches isobaric, isothermal, and isometric process ? In which reversible process no work is done ?

0 Answers  

---

Explain the difference between pearlite and cementile ?

0 Answers  

---

What is the purpose of a turning gear?

3 Answers  

---

---

I need the complete syllabus and model Question paper for Written exam of JINDAL STEEL AND POWER LTD.,for MECHNAICAL ENGINEERING branch

0 Answers   Jindal, TAL Manufacturing Solutions,

---

how we calculate the induced air capacity of id fan of a boiler?

0 Answers   Delight Dairy,

---

Maximum combustion temperature in gas turbines is of the order of 1100 to 10°c whereas same is around 00°c in i.c. Engine ? Why ?

0 Answers  

---

Explain how the cylinder lubrication on main engine is done?

0 Answers  

---

1: When mesuring the flow of liquid what are the two things to measure? 2: In which industrial process you will use centrifugal separator? 3: If any machinery or transmission needs periodic and regular lubrication what is the best method to give it? 4: Oil melts, grease melts. Is there any other lubrication which is effect at 200 deg c? 5: What is a torque converter? 6: Instead of having individual transmission what is the benefit of unified transmission? 7: What are the different types of mechanism to engage primary moving engine and transmission (in other words how many different types of clutches are there)? 8: In order to ensure constant pressure and constant flow rate of compressed air what should we put between the compressor and end user? 9: What can be the reasons for a metal gasket to leak? 10: The old welding machines used electromagnetic winding to step up the current for welding rods. What is the new principle inside the welding machines?

0 Answers   Petrotech,

---

How to protect the pipe line from crrossion.?

9 Answers   MAHINDRA,

---

What are the modes of heat transfer?

0 Answers  

---