Tuesday, March 30, 2021

SCREW COMPRESSOR CALCULATION

 SCREW COMPRESSOR  CALCULATION:


A small machine shop has installed 220 cfm screw compressor to meet air requirement for various operation. The operating details are given below:

Shift reference

(8 hrs/ Shift)

Load time in sec

Un-Load time in sec

I

60

10

II

45

25

III

25

45

Load Power = 37 KW

Un-load power = 11 KW

Calculate the following:

1. Energy loss per day 

2. Shift wise average air requirement in cfm 

3. The plant has proposed to install a VFD for the compressor. Calculate the energy savings after installing the VFD operated compressor, if the VFD loss is 3 % of load power.


CALCULATION:

Ist shift consumption = ((60 / 70) x 37) + (10 / 70) x 11) x 8)

= (31.71+1.57) x 8

= 266.24 kWh

IInd shift consumption = ((0.64 x 37 + 0.36 x 11) x 8 )

= (23.68 + 3.96) x 8

= 221.12 kWh

IIIrd shift consumption = ((0.36 x 37 + 0.64 x 11) x 8)

= (13.32 +7.04) x 8)

= 162.88 kWh

Daily Total Energy consumption = (266.24 + 221.12 + 162.88)

= 650.24 kWh

Daily Energy loss due to unloading = (1.57 +3.96 +7.04) x 8

= 100.56 kWh

Daily load cycle Energy consumption = (650.24 – 100.56)

= 549.68 kWh

Daily energy consumption with VFD = (549.68 / 0.97)

= 566.68 kWh

Paper 3 Code : Pink

_______________________

Bureau of Energy Efficiency

15

Daily Energy loss due to VFD = (566.68 – 549.68)

= 17 kWh

Daily Net Energy savings with VFD compressor = (100.56 – 17)

= 83.56 kWh

Ist shift air requirement = (0.86 x 220)

= 189.2 cfm

IInd shift air requirement = (0.64 x 220)

= 140.8 cfm

IIIrd shift air requirement = (0.36 x 220)

= 79.2 cfm

ELECTRICAL CALCULATION OF 10 MW COGENERATION PLANT

ELECTRICAL CALCULATION OF 10 MW COGENERATION PLANT:


 A 10 MW co-generation plant is operating at a daily load factor of 85 %. Power is generated at 11 KV.

➢ 35 % of the power generated, is exported to grid, through a 7.5 MVA Transformer with 99 % efficiency.

➢ 32 % power generated, is supplied to mill motors, at 600 Volts, through a 5 MVA step down transformer, with 98 % efficiency.

➢ The balance power generated is supplied to other LT Loads and auxiliaries, at 415 Volts, through a 2 MVA transformer, with 98 % efficiency.

Calculate the following:

1) Daily energy exported to grid at 33 KV.

2) Daily mill motors consumption at 600 V.

3) Daily LT loads and auxiliary consumption at 415 V.

4) Daily transformers losses in kWh and % transformers losses


CALCULATION:

1.

Daily generation = (10,000 x 0.85 x 24)

= 2,04,000 kWh

Daily energy generation for export purpose = (2,04,000 x 0.35)

= 71,400 KWh

7.5 MVA transformer loss = [ 71,400 - (71,400 x 0.99) ]

= (71,400 – 70,686)

= 714 kWh

Net energy export to the Grid at 33 KV level = (71,400 kWh - 714 kWh)

=70,686 KWh

2.

Daily energy generation for mill motor consumption = (2,04,000 x 0.32)

= 65,280 kWh

5 MVA Transformer loss = [ 65,280 - (65,280 x 0.98 ) ]

= (65,280 - 63,974.4)

= 1,306 kWh

Net mill Consumption = 63,974 KWh

3.

Daily generation for LT loads & Auxiliary consumption = (2,04,000 x 0.33)

= 67,320 kWh

2MVA Transformer loss = [ 67320 - (67320 x 0.98) ]

= 67,320 - 65,974

= 1,346 kWh

Net LT loads & Auxiliary Consumption = 65,974 kWh

4.

Transformers losses = (714 +1306 +1346)

= 3,366 kWh day

% transformers losses = (3,366 / 2,04,000) x 100

= 1.65 %

(Or)

To meet the plant LT loads and co-gen auxiliary load, the transformer capacity should be more than 2 MVA.


CALCULATION OF COMPRESSOR DAILY ENERGY CONSUMPTION:

 CALCULATION OF COMPRESSOR DAILY ENERGY CONSUMPTION:


One of the Machining centres has installed 2 No’s of 270 cfm compressors for pneumatic operation and also for cleaning operation of components after machining. The compressors are operated at 7 kg/cm2(g) and are on-load for 80 % of the time. The load Power and the un-load Power of each 270 cfm compressor is, 40 kW and 15 kW respectively. The energy audit estimated that cleaning air requirement is 60% of the air generated.

Calculate the daily energy consumption for cleaning air alone, assuming continuous operation of the compressor.

CALCULATION:

Compressor capacity = 270 cfm

% Loading = 80 %

Air Delivered by 2 compressors = (270 X 0.80 x 2)

= 432 cfm

Loading Power drawn by the compressors = (40 + 40)

= 80 kW

Un-Loading power drawn by the compressors = (15 + 15)

= 30 kW

Average kW drawn by the compressors = [(80 x (0.8 x24))+ (30x (0.2 x 24))]/(24)

= 70 kW

SEC of compressor = (70/432)

= 0.162 kW/cfm

Cleaning air consumption at 7 Kg/cm2 = (60 % of generation)

= (0.60 x 432)

= 259 cfm

Energy requirement for Cleaning air per day = (259 x 0.162 x 24)

= 1007 kWh/day

(or) Alternate Solution

= (Load Power x load time) + (Unload Power x Unload time)

= (40 x 0.8) + (15 x 0.2)

= 32+3

= 35 KW

Average KW drawn by the compressors = 35 x 2 = 70 KW

Energy requirement for Cleaning air per day = (70 kW x 0.6) x 24

=1008 kWh/day


CALCULATION OF DRY FLUE GAS LOSSES & COAL REQUIREMENT

         

Quantity of annual coal requirement in tonnes/year

                                                

                         Calculate % dry flue gas losses 


A process industry consuming 10 TPH of saturated steam at 10 kg/sq.cm(g) pressure has been using coal as fuel in boiler.

Typical ultimate analysis of the coal:

Carbon : 41.11%

Hydrogen : 2.76 %

Nitrogen : 1.22 %

Oxygen : 9.89 %

Sulphur : 0.41%

Ash : 38.63

Water : 5.89

Flue gas temperature = 200°C

Ambient temperature = 30°C

Enthalpy of steam = 668 kcal/kg

Feed water temperature = 80°C

Specific heat of flue gases = 0.23 kcal/kgoC

Boiler efficiency with Indian coal =72 %

GCV of coal = 4,000 kCal/kg

Oxygen content in dry flue gases = 10%

Annual Hours of operation = 8000 hrs.

Determine:

(i) Quantity of annual coal requirement in tonnes/year 

(ii) Calculate % dry flue gas losses 


Solution:

a)

Coal requirement Q = Steam (q) x (hg – hf)/( Efficiency x GCV)

= 10 x (668-80) / (0.72 x 4000)

= 2.042 T/Hr

= 2.042 x 8000 hrs

= 16336 Tonnes/year

Theoretical air requirement for coal

= [(11.6 x C%) + {34.8 x(H2% - O2%/8)} + (4.35 x S%)] kg / kg of coal

100

= [(11.6 x 41.11) + {34.8 x(2.76 – 9.89/8)} + (4.35 x 0.41)]

100

= 5.31 kg / kg of coal

(Or)

C+ O2 = CO2

12+32= 44

(C%*32)/12

2H2+ O2 = 2H2O

4+32=36

(H%*32)/4

S+ O2 = SO2

32+32=64

(S%*32)/32

Total oxygen required = (41.11* 32/12) + (2.76*32/4)+ (0.41*32/32)

= (109.63) + (22.08) + (0.41)

= 132.1 kg/ 100 kg fuel

Oxygen already present in 100 kg fuel = 9.89 kg/ 100 kg fuel

Additional oxygen required = 132.1 – 9.89 kg/ 100 kg fuel

= 122.21 kg/ 100 kg fuel

Quantity of dry air required

(Air contains 23% O2 by weight ) = 122.1/ 0.23

= 531.35 kg/ 100 kg fuel


Theoretical air required = 531.35/100

= 5.31 kg air/ kg fuel

Excess air = O2x100/(21-O2)

Excess air = 10 x 100/ (21 – 10) = 90.9%

Actual air= 5.31 * (100+90.9)/100

=10.137 kg air/kg coal

Heat loss in dry flue gas = m x CP (Tf – Ta) x 100 /GCV

= (10.137+ 1) x 0.23 x (200 – 30) x 100

4000

= 10.89 %

PIPE HEAT EXCHANGER CALCULATION & OPERATION

PIPE HEAT EXCHANGER CALCULATION & OPERATION

A) 

In a double pipe heat exchanger, flow rates of the hot and the cold water streams flowing through a heat exchanger are 10 and 25 kg/min, respectively. Hot and cold water stream inlet temperatures are 70 °C and 27 °C, respectively. The exit temperature of the hot stream is required to be 50°C. The specific heat of water is 4.179 kJ/ kg K. The overall heat transfer coefficient is 900 W/m2 K. Neglecting the effect of fouling, calculate the heat transfer area for a) Parallel-flow b) Counter-flow.


B) Write a brief note on the operation and application of plate heat exchangers in process industries.


Ans:

A)

Rate of heat transfer, Q (watts)

Q= m x Cp x 1000 x (T2-T1)

=(10/60)× 4.179 x 1000 ×(70 – 50) = 13930 W

Cold water exit temperature, T2

T2 = [Q/(mx Cp x 1000)]+ T1

= (13930/((25/60)* (4.179*1000)))+27

= 35°

Terminal temperature differences for parallel flow heat exchangers

= (70-27) & (50 – 35); i.e.,

43°C and 15°C respectively.

LMTD

(43 – 15)/ln(43/15) = 26.59°C

Overall heat transfer coefficient U

900 W/m2 °K

Heat transfer area required for parallel flow

A = Q/ (U*LMTD)

= [13930/ (900 × 26.59)]

= 0.582 m2

Terminal temperature differences for counter flow heat exchangers

(70–35) and (50–27) 0C i.e.,

35 °C and 23 °C respectively.

LMTD

(35–23)/ln(35/23) = 28.58

Overall heat transfer coefficient U

900 W/m2 °K

Heat transfer area required for counter flow

A = Q/ (U*LMTD)

=[13930/ (900 × 28.58)]

= 0.542 m2

(B)

Plate heat exchangers consist of a stack of parallel thin plates that lie between heavy end plates. Each fluid streampasses alternately between adjoining plates in the stack, exchanging heat through the plates. The plates are corrugated for strength and to enhance heat transfer by directing the flow and increasing turbulence. These exchangers have high heat-transfer coefficients and area, the pressure drop is also typically low, and they often provide very high effectiveness.

However, they have relatively low pressure capability. The biggest advantage of the plate and frame heat exchanger, and a situation where it is most often used, is when the heat transfer application calls for the cold side fluid to exit the exchanger at a temperature significantly higher than the hot side fluid exit temperature i.e. “temperature cross”. This would require several shell and tube exchangers in series due to the lack of purely counter-current flow.

The overall heat transfer coefficient of plate heat exchangers under favorable circumstances can be as high as 8,000 W/m2°C. With traditional shell and tube heat exchangers, the U-value will be below 2,500 W/m2°C.


CALCULATION OF DIAMETER OF STEAM PIPE LINE

 CALCULATION OF DIAMETER OF STEAM PIPE LINE:


In an industry the process equipment needs 5000 kg/hr of saturated steam at 10 kg/ cm2(g). For a steam velocity of 25 m/sec, what will be the diameter of the steam pipe in ‘mm’, given that the specific volume of steam at 10 kg/ cm2(g) is 0.1802 m3/kg.


CALCULATION:

Specific volume of steam at 10 kg/cm2(g) = 0.1802 m3/kg

Flow rate = 25m/sec

Mass flow rate = 5000 kg/hr

= 1.389 kg/sec

Volume flow rate = 1.389 x 0.1802

= 0.25 m3/sec

Volume flow rate is also = (π/4 x D2) x25

Therefore, (π/4 x D2) x 25 = 0.25

Hence, diameter of steam pipe line ‘D’ = [(0.25/((π/4) x 25)]0.5

= 0.1128m

or 112.8 mm

FLY ASH & UN-BURNT CALCULATION IN BOILER

 FLY ASH & UN - BURNT  CALCULATION IN BOILER

a) A sample of fuel being used in a boiler is found to contain 40% carbon and 23% ash. The refuse obtained after combustion is analyzed and found to contain 7% carbon & the rest is ash. Compute the percentage of the original carbon in fuel which remains as unburnt in the refuse.

b) During an ESP performance evaluation study, the inlet gas stream to ESP is 2,89,920 Nm3/hr and the dust loading is 5,500 mg/ Nm3. The outlet gas stream from ESP is 3,01,100 Nm3/hr and the dust loading is 110 mg/Nm3. How much fly ash is collected in the system in kg/hr?


CALCULATION:-

a)

Let the quantity of Refuse sample = 100 kg

Amount of unburnt Carbon in Refuse = 7 kg

Amount of Ash in the Refuse = 93 kg

Total ash in the fuel that has come into the Refuse = 23% of fuel

93 kg of Ash corresponds to 23% ash in the fuel

Therefore, quantity of total raw fuel = 93 / 0.23

= 404.35 kg

Quantity of original Carbon in the fuel = 0.40 x 404.35

= 161.74 kg

Quantity of unburnt fuel in Refuse = 7 kg

%age of the original carbon unburnt in the refuse = (7 / 161.74) x 100

= 4.32%

b)

Based on Mass balance,

Inlet gas stream dust = Outlet Gas stream dust + Fly ash

i) Inlet gas stream flow = 289920 NM3/hr

Dust Concentration = 5500 mg/NM3

Inlet dust quantity = 289920 x 5500

                                            -----------

                                            1000000

= 1594.56 kg/hr


ii) Outlet dust quantity = 301100 (NM3/hr) x 110 (mg/NM3) x 1

                                                                    ----------

                                                                    1000000

= 33.12 kg/hr


iii) Fly Ash = Inlet gas stream dust – Outlet gas stream dust

= 1594.56 – 33.12 = 1561.44 kg/hr

DEMAND CALCULATION OF PROCESS PLANT

 DEMAND CALCULATION OF PROCESS PLANT


A food processing plant has a contract demand of 2500 kVA with the power supply

company. The average maximum demand of the plant is 2000 kVA at a power factor of

0.95.

The maximum demand is billed at the rate of Rs.300/kVA. The minimum billable

maximum demand is 75 % of the contract demand. An incentive of 0.5 % reduction in

energy charges component of electricity bill are provided for every 0.01 increase in power

factor over and above 0.95. The average energy charge component of the electricity bill per

month for the company is Rs.10 lakhs.

The plant decides to improve the power factor to unity. Determine the power factor

capacitor kVAr required, annual reduction in maximum demand charges and energy

charge component. What will be the simple payback period if the cost of power factor

capacitors is Rs.800/kVAr ?


CALCULATION:


kW drawn 2000 x 0.95 = 1900 kW

Kvar required to improve power factor from

0.95 to 1

kW ( tan 1 – tan 2)

kW ( tan (cos-1) – tan (cos-2)

1900 ( tan (cos-0.95) – tan (cos-1)

1900 (0.329 - 0)

625 kVAr

Cost of capacitors @Rs.800/kVAr Rs.5,00,000

Maximum demand at unity power factor 1900/1 = 1900 kVA

75 % of contract demand 1875 kVA

Reduction in Demand charges 100 kVA x Rs.300

Rs.30000 x 12

Rs.3,60,000

Percentage reduction in energy charge from

0.95 to 1 @ 0.5 % for every 0.01 increase

2.5 %

Monthly energy cost component of the bill Rs.10,00,000

Reduction in energy cost component 10,00,000 x (2.5/100)

Rs.25,000/month

Annual reduction Rs.25,000 x 12

Rs.3,00,000

Savings in electricity bill Rs.6,60,000

Investment Rs.5,00,000

Payback period 5,00,000/6,60,000

0.76 years or 9 months

INDUSTRIAL FAN SHAFT POWER CALCULATION

INDUSTRIAL FAN SHAFT POWER CALCULATION:


 In a boiler, the forced draught fan develops a total static pressure of 300 mmWC.

Determine the shaft power (in kW) required to drive the fan if 10,000 kg of coal is burnt

per hour with 13 kg of air per kg of coal burnt. The boiler house temperature is 20 0C and

static efficiency of the fan is 80 %.

The operating air density may be calculated from the following:

R = 847.84 mmWC m3/kg mole K and Molecular weight of air, M = 28.92 kg/kg mole.


ANSWER:

Total Pressure = 300 mm of WC

Mass of air handled, m = 10000 × 13/ 3600 = 36.11 kg/s

Atmospheric pressure, P = 1 kg/ cm2 = 10 mtr of WC = 10,000 mm of WC.

Temperature T = 20 + 273 = 293 K

Gas Constant for air, R = 847.84 mm WC m3/kg mole K

Molecular weight of air, M = 28.92 kg/kg mole

Density, kg/m3 = (P x M) / (R X T) = (10000 x 28.92) / (847.84 x 293)

= 1.164 kg/m3

Volume in m3/s = mass (kg/s) / density (kg/m3)

= 36.11 / 1.164

= 31.02 m3/s

Power to fan shaft, kW

= [Volume (m3/s) x Total pressure (mm of WC)] / [102 x fan efficiency]

= [31.02 x 300] / [102 x 0.8]

= 114 kW


AIR CONDITIONER ANNUAL COST SAVING CALCULATION

 A 7.5 TR package air conditioner is provided for a UPS room for removing the heat

generated from the UPS of rated capacity 40 kVA. The following parameters were noticed

while performing the assessment of the total system.

UPS Parameters:

Rating Input Power (kW) Output Power (kW)

40 kVA

On Load (16 hrs) 11.94 8.61

No Load (8 hrs) 1.16 0.00

Air conditioner parameters:

Installed capacity of Air conditioner 7.5 TR

Outdoor unit (condenser) air velocity 6.1 m/s

Radius of the fan opening at the point of velocity measurement

in outdoor unit

0.30 m

Air Density 1.174 kg/m3

Ambient temperature 305 0K

Temperature of hot air (condenser outlet) 313.5 0K

Specific heat of air 1.009 kJ/kg K

Power drawn by the compressor 5.40 kW

Efficiency of the compressor motor 90 %

Calculate

a) Present delivery capacity of air conditioner (TR) (3 Marks)

b) Power drawn per TR of refrigeration (3 Marks)

c) Calculate the annual energy savings for 7200 hrs, if the UPS is relocated to a non

air-conditioned ventilated area. Assume energy cost Rs.8/kWh.




ANSWER: 


Capacity Installed 7.5 TR

Outdoor unit air velocity 6.1 m/s

Radius of the opening 0.30 m

Area of cross section (3.14x0.3^2) 0.283 m2

Total Air flow (0.283x6.1) 1.72 m3/s

Density of the air 1.174 kg/m3

Mass of air, m (1.72x1.174) 2.02 kg/s

Ambient temperature, T1 305 oK

Air temperature, T2 313.5 oK

Difference in Temperature (T2-T1), (dT) 8.5 oK

Specific Heat at Constant pressure, cp 1.009 kJ/kgK

Heat Transfer (mxCpx(T2-T1)) 17.32 kJ/s

Heat transfer per hour 62352 kJ/hr

14917 kcal/Hr

Heat input from the compressor (5.4x0.9x860) 4180 kcal/Hr

Evaporator heat load (14949-4180) 10737 kcal/Hr

1 Tonne of refrigeration 3024 kCal/Hr


Effective TR 3.55 TR

Power drawn by the compressor 5.40 kW

power taken per TR of refrigeration 1.52 kW/ TR


Heat Load generated by UPS in Conditioned Space

Rating/ Location

Input

Power

(kW)

Output

Power

(kW)

Heat Load

(kW) kCal/Sec kCal/Hr TR/hr

Total

TR/day

40 kVA

On

Load

(16hrs)

11.94 8.61 3.33 0.80 2880 0.95 15.2

No Load

(8hrs)

1.16 0 1.16 0.28 1008 0.33 2.64

Total 17.84

The savings that can be achieved by providing clean, cool and dust free environment for UPS

operation is given below.

AC Load generated by UPS/ day = 17.84 TR

Power taken by AC to generate 17.82 TR at 1.52 kW/ TR = 27.12 kW

Annual energy savings at 300 days of operation = 8136 kWh

Cost of power = Rs.8/ kWh

Annual Cost Savings = Rs.65,088/-

CALCULATION Q & A

TECHNICAL SOLUTION:-

QUESTION 1


In a Thermal Power Station, the steam input to a turbine operating on a fully condensing

mode is 100 TPH. The heat rejection requirement of the steam turbine condenser is 555

kcal/kg of steam condensed. The temperature of cooling water at the inlet and outlet of the

turbine condenser is 27 0C and 37 0C respectively. Find out the circulating cooling water

flow.


ANSWER 1

The quantum of heat rejected in the turbine condenser

= Quantum of steam condensed (kg) x heat rejection (kcal/kg)

= 100,000 x 555 = 55.5 Million kcal/h.

Heat gained by circulating cooling water = Heat rejected in the condenser

Circulating cooling water flow

= 100,000 x 555 / (37-27) x specific heat (1)

= 5550 m3/hr


QUESTION 2


A medium sized engineering industry has installed two 480 CFM screw compressors, A &

B. Compressor-A is operating at full load and Compressor-B is running in load – unload

condition. The load power of both the compressor is 74 kW and the unload power of the

Compressor-B is 26 kW. Both the compressors are operated during working day.The

percentage loading of the Compressor-B during working day is 64 %. After arresting the

leakage in the system, the loading of the compressor was found to be 35 %. Estimate the

energy savings per day.

 

ANSWER 2

Existing Case:

Energy consumed per hour by Compressor -A= 74 kW

Energy consumed per hour by Compressor -B= 0.64 x 74 + 0.36 X 26 = 56.72 kW

Total energy consumed (Compressor A& B) = 74 + 56.72 = 130.72 kW/hr

Energy consumed per day= 130.72 X 24 hrs = 3137.3 kWh/day

Leakage Calculation:

Energy consumed per hour by Compressor -B= 0.64 x 74 + 0.36 X 26 = 56.72 kW

Energy consumed per hour by Compressor -B= 0.35 x 74 + 0.65 X 26 = 42.8 kW

Difference in power consumption = 56.72 - 42.8 = 13.92 kW/hr

Savings by arresting leakage per day= 13.92 X 24 = 334 kWh/day


QUESTION 3


A plant is operating a chilled water system always at full load. The chilled water inlet and

outlet temperatures are 12 0C and 7 0C respectively. The chilled water pump discharge

pressure is 3.6 kg/cm2g and the suction is 5 meters above the pump centreline. The power

drawn by the chilled water pump’s motor is 70 kW and an efficiency of 90 %. The chilled

water pump efficiency at the operating point from pump characteristic curve is 60 %. Find

out the operating refrigeration load in TR.


ANSWER 3

Total head 36 – 5 = 31 m

Pump shaft power 70 x 0.9

63 kW

Flow rate (63 x 1000) X 0.6 / 31 x 1000 x 9.81

0.124297 m3/s

447.5 m3/hr

Refrigeration load (447500 x 5) / 3024

740 TR


QUESTION 4


The operating data of an induced draft-cooling tower is as follows:

Observed range : 8 0C.

Cooling water flow rate : 12,500 m3/hr

Drift loss : 0.1 % of circulation rate

Wet Bulb Temperature : 27 0C

Ambient Dry Bulb Temperature : 35 0C

Effectiveness : 67 %

Cycle of Concentration : 3

Estimate the evaporation loss; make up water requirement and TR load of cooling tower.


ANSWER 4

Evaporation loss = 0.00085 X 1.8 X 12500 X 8 = 153 m3/hr

Blow Down =153 /(3-1)= 76.5 m3/hr

Make up =153 + 76.5 + (12500*0.001)= 242 m3/hr

Heat load =12500*1000*8/3024 = 33069 TR


QUESTION 5


 In an air washer of a textile humidification system with an airflow of 3000 m3/h at 25 0C

and 10 % relative humidity is humidified to 60 % relative humidity by adding water

through spray nozzles. The specific humidity of air at inlet and outlet are 0.002 kg/kg of

dry air and 0.0062 kg/kg of dry air respectively. The density of air at 25 0C is 1.184

kg/m3. Calculate the amount of water required in kg/hr.


ANSWER 5

The amount of water required:

mw =v ρ ( Wout - Win)

= 3000 X 1.184 X (0.0062- 0.002)

= 14.9 kg/h


Monday, March 29, 2021

BAGASSE % CANE CALCULATION

 BAGASSE % CANE

 • Bagasse % cane , can be calculated by two methods. 

• Cane + Added water = Mixed juice + Bagasse 

•OR 100 + A.W. % cane = M.J. % cane + Bag. % cane 

• Bag. % cane = 100 + A.W. % cane – M.J. % cane 

•Here cane, Added water & mixed juice are directly weighed & the quantity of bagasse & bag % cane is calculated as, 

• By determining fibre % cane & fibre % bagasse •Bagasse % cane can be determined as, 

• Bagasse % cane = Fibre % cane x 100 

• Fibre % bagasse

 • Here fibre % cane is determined by direct method and bagasse is analyzed for moisture % and Pol % & knowing the purity of L.M.J. fibre % bag. is calculated.

FIBRE % CANE CALCULATION

FIBRE % CANE CALCULATION 

• There are two methods for finding fibre % cane 

• (A) Direct Method 

• (B) Indirect Method •

 • Direct Method :-

(1) Laboratory Method :- 

• In this method a few cane sticks are taken & some portion of cane is shredded or cut into fine pieces by means of a laboratory shredder.

 • A sample of shredded cane is weighed and dried in air over to determine moisture % cane. • The remaining cane sticks are crushed into laboratory mill or crusher and brix of the juice so extracted is determined. • From these figs ,fibre ,cane is determined as follows. •

 • (2) By Rapi – pol Extractor – Method :-

 • The sugar cane sticks are cut into fine pieces. 

• 250 gms of this sample along with 2000 ml water is transferred to rapi-pol vessel. 

• Run the Rapi-pol extractor for 15 mints . 

• Decant the whole liquid from vessel. 

• Repeat the same above procedure for 2 or 3 times. Then take out the fibrous mass from vessel .

 • after well squeezing with hand & dry the fibrous mass completely in drier and weigh it. The fibre % cane will be calculated as, 

• Fibre%cane= Weight of the fibrous massx100/250 • 

•Indirect Method:- 

• Laboratory Method:- 

Crush the weighed cane in a laboratory mill. Determine the wt. of juice extracted and analyse the juice for Brix %, Pol % & Purity. Bagasse is also analyzed for Pol % & moisture % as usual.

 • From there analytical results fibre % cane is calculated as under - • 

•Bagasse % cane = wt.of cane –wt.of juice extracted x 100 

• Wt.of the cane •Brix % Bag. = Pol % bagx 100 

• Pty of extracted juice •Fibre % Bag = 100 – Moist. % Bag – Brix % Bag.

 •Fibre % cane = Fibre in bag. % cane

 •Fibre % cane = fibre % Bag. X Bag. % cane • 100 •

 • But ,in the factory ,fibre % cane is determined as follows 

• Cane + A.W. = Mixed juice + Bagasse • 100 + A.W. % cane = M.J. % cane + Bag. % cane

 • Bag. % cane = 100+A.W.%cane – M.J.% cane

 • Fibre % cane = 100 – moist. % Bag – Brix % Bag

 • Brix % Bag = Pol % Bag.x100/Purity LMJ 

• Fibre % cane = fibre % Bag. x Bag. % cane 100

MILL EXTRACTION IN SUGAR MILL CALCULATION

Mill Extraction :-

 The quantity of sugar extracted in the mixed juice per 100 sugar in cane i.e. percentage of sugar extracted in the mixed juice out of total sugar in cane

 • Mill Extraction = Sugar (Pol) in M.J. % cane x 100 • Sugar (Pol) % cane
 
• Exam.(1) Cane = 9500 M.T. ;Mixed juice =9120 M.T.
 
• Added water = 2327.5
 
• Analysis of M.J.: – Bx. %juice - 15.5 , Pol %juice -13.02 

• Pol % bagasse -2.5 

• Calculate A.W. % cane, Pol % cane & Mill Extraction

 • Solution Cane + A.W. = M.J. + Bagasse 

• 9500 + 2327.5 = 9120 + Bagasse 

• Bagasse = 9500 + 2327.50 – 9120 

• = 2707.50 
 • Bagasse % cane= Total bagasse x 100/ cane crushed 

• = 2707.5 x10/ 9500

 • = 28.50 % 

 • M.J. % cane = Total M.J. x 100/ cane crushed 
• = 9120 x 100/ 9500 •
 • = 96.00 % 
• A.W. % cane = Total Added water x 100/ cane crushed 
 = 2327.50 x 100/ 9500 = 24.50 % Pol in M.J. % cane = Pol % M.J. % cane Mill Extraction =Pol in M.J. % cane x 100/ Pol % cane = 12.50 x 100/ 13.21 = 94.625 = 94.63

Monday, March 15, 2021

CENTRIFUGAL PUMP TROUBLESHOOTING

INSULATION OF PIPING

 

Insulation of Piping

Piping plays a central role in many industrial processes in chemical or petrochemical installations such as power plants, as it connects core components such as appliances, columns, vessels, boilers, turbines etc. with one another and facilitates the flow of materials and energy.

To guarantee a correct process cycle, the condition of the media within the pipes must remain within the set limitations (e.g. temperature, viscosity, pressure, etc.).

In addition to the correct isometric construction and fastening of the piping, the piping insulation also has an important function. It must ensure that heat loss are effectively reduced and that the installation continues to operate economically and functionally on a permanent basis. This is the only way to guarantee the maximum efficiency of the process cycle throughout the design service life without losses as a result of faults.


Requirements for industrial piping

The basic efficiency and productivity factors of piping for the processing industry include: energy efficiency, dependability and reliability under different conditions, functionality of the process control, appropriate support structure suitable for the operating environment, as well as mechanical durability. The thermal insulation of piping plays a significant role in fulfilling these requirements.

THERMAL INSULATION

The functions of proper thermal insulation for piping include:

  • Reduction of heat losses (cost savings)
  • Reduction of CO2 emissions
  • Frost protection
  • Process control: ensuring the stability of the process temperature
  • Noise reduction
  • Condensation prevention
  • Personnel protection against high temperatures

APPLICABLE STANDARDS - A FEW EXAMPLES:

  • NACE SP0198 (Control of corrosion under thermal insulation and fireproofing materials - a systems approach)
  • MICA (National Commercial and Industrial Insulation Standards)
  • DIN 4140 (Insulation works on technical industrial plants and in technical facility equipment)
  • AGI Q101 (Insulation works on power plant components)
  • CINI-Manual "Insulation for industries"
  • BS 5970 (Code of practice for the thermal insulation of pipework, ductwork, associated equipment and other industrial installations)

Minimum pipe insulation thickness

Fluid Operating Temperature Range and Usage (°F)Insulation Conductivity
Conductivity
Btu · in. /(h · ft2 · °F)b
Mean
Rating
Temperature, °F
> 3500.32 - 0.34250
251 - 3500.29 - 0.32200
201 - 2500.27 - 0.30150
141 - 2000.25 - 0.29125
105 - 1400.21 - 0.28100
40 - 600.21 - 0.2775
< 400.20 - 0.2675
Nominal Pipe or Tube Size (inches)
< 11 to < 1-1/21-1/2 to < 44 to < 8≥ 8
4.55.05.05.05.0
3.04.04.54.54.5
2.52.52.53.03.0
1.51.52.02.02.0
1.01.01.51.51.5
0.50.51.01.01.0
0.51.01.01.01.5

A For piping smaller than 1-1/2 inch (38 mm) and located in partitions within conditioned spaces, reduction of these thicknesses by 1 inch (25 mm) shall be permitted (before thickness adjustment required in footnote b) but not to a thickness less than 1 inch (25 mm).

B For insulation outside the stated conductivity range, the minimum thickness (T) shall be determined as follows:

T = r{(1+t/r) K/k-1}

Where:

T = Minimum insulation thickness
r = Actual outside radius of pipe
T = Insulation thickness listed in the table for applicable fluid temperature and pipe size
K = Conductivity of alternate material at mean rating temperature indicated for the applicable fluid temperature (Btu x in/h x ft2 x °F) and
k = The upper value of the conductivity range listed in the table for the applicable fluid temperature

C For direct-buried heating and hot water system piping, reduction of these thicknesses by 1-1/2 inches (38 mm) shall be permitted (before thickness adjustment required in footnote b but not to thicknesses less than 1 inch (25 mm).


1. Pipe   2. Insulation   3. Clamp or binding wire   4. Sheet cladding
5. Sheet-metal screw or rivet

Cladding

Suitable cladding should be applied to protect the insulation from weather influences, mechanical loads and (potentially corrosive) pollution. Selecting the appropriate cladding depens on various factors, such as working loads, wind loads, ambient temperatures and conditions.

When selecting the appropriate cladding, take the following points into account:

  • As a general rule, galvanised steel more than aluminium is used indoors due to its mechanical strength, fire resistance and low surface temperature (in comparison to an aluminium cladding).
  • In corrosive environments like outdoors on deck where salty water leads to corrision, aluminised steel, stainless steel or glass reinforced polyester is used as cladding. Stainless steel is recommended for use in environments with a fire risk.
  • The surface temperature of the cladding is influenced by the material type. The following applies as a general rule: the shinier the surface, the higher the surface temperature.
  • To exclude the risk of galvanic corrosion, only use combinations of metals that do not tend to corrode due to their electrochemical potentials.
  • For acoustic insulation, a noise absorbent material (Lead layer, polyethyten foil) is installed on the insulation or inside the cladding. To reduce the risk of fire, limit the surface temperatures of the cladding to the maximum operating temperature of the noise absorbent material.

Types and Materials of Insulation

Any surface which is hotter than its surroundings will lose heat. The heat loss depends on many factors, but the surface temperature and its size are dominant.

Putting the insulation on a hot surface will reduce the external surface temperature. By insulation, the surface will increase on objects, but the relative effect of temperature reduction will be much greater and heat loss will be reduced.

A similar situation occurs when the surface temperature is lower than its surroundings. In both cases some energy is lost. These energy losses can be reduced by laying the practical and economical insulation on surfaces whose temperatures are quite different than the surrounding one.


Categories of Insulation Materials

INSULATION MATERIALS OR SYSTEMS MAY ALSO BE CATEGORIZED BY SERVICE TEMPERATURE RANGE.

There are varying opinions as to the classification of mechanical insulation by the service temperature range for which insulation is used. As an example, the word cryogenics means "the production of freezing cold"; however the term is used widely as a synonym for many low temperature applications. It is not well-defined at what point on the temperature scale refrigeration ends and cryogenics begins.

The National Institute of Standards and Technology in Boulder, Colorado considers the field of cryogenics as those involving temperatures below -180°C. They based their determination on the understanding that the normal boiling points of the so-called permanent gases, such as helium, hydrogen, nitrogen, oxygen and normal air, lie below -180°C while the Freon refrigerants, hydrogen sulfide and other common refrigerants have boiling points above -180°C.

Understanding that some may have a different range of service temperature by which to classify mechanical insulation, the mechanical insulation industry has generally adopted the following category definitions:

CategoryDefinition
Cryogenic Applications-50°F and Below
Thermal Applications:
Refrigeration, chill water and below ambient applications-49°F to +75°F
Medium to high temp. applications+76°F to +1200°F
Refractory Applications+1200°F and Above

CELLULAR INSULATIONS are composed of small individual cells either interconnecting or sealed from each other to form a cellular structure. Glass, plastics, and rubber may comprise the base material and a variety of foaming agents are used.

CELLULAR INSULATIONS are often further classified as either open cell (i.e. cells are interconnecting) or closed cell (cells sealed from each other). Generally, materials that have greater than 90% closed cell content are considered to be closed cell materials.

FIBROUS INSULATIONS are composed of small diameter fibers that finely divide the air space. The fibers may be organic or inorganic and they are normally (but not always) held together by a binder. Typical inorganic fibers include glass, rock wool, slag wool, and alumina silica.

Fibrous insulations are further classified as either wool or textile-based insulations. Textile-based insulations are composed of woven and non-woven fibers and yarns. The fibers and yarns may be organic or inorganic. These materials are sometimes supplied with coatings or as composites for specific properties, e.g. weather and chemical resistance, reflectivity, etc.

FLAKE INSULATIONS are composed of small particles or flakes which finely divide the air space. These flakes may or may not be bonded together. Vermiculite, or expanded mica, is flake insulation.

GRANULAR INSULATIONS are composed of small nodules that contain voids or hollow spaces. These materials are sometimes considered open cell materials since gases can be transferred between the individual spaces. Calcium silicate and molded perlite insulations are considered granular insulation.

REFLECTIVE INSULATIONS & treatments are added to surfaces to lower the long-wave emittance thereby reducing the radiant heat transfer to or from the surface. Some reflective insulation systems consist of multiple parallel thin sheets or foil spaced to minimize convective heat transfer. Low emittance jackets and facings are often used in combination with other insulation materials.

Some Insulation Type Examples

Cellular Insulations

ELASTOMERIC

Elastomeric insulations are defined by ASTM C 534, Type I (preformed tubes) and Type II (sheets). There are three grades in the ASTM standard which are widely available.


Elastomeric insulations
GradeBasic descriptionTemp. LimitsFlame Spread Index / Smoke Developed Index
1Widely used on typical commercial systems-297°F to 220°F25/50 through 1½in thickness.
2High temp. uses-297°F to 350°FNot 25/50 Rated
3Use on stainless steel applications above 125 °F-297°F to 250°FNot 25/50 Rated

All three grades are flexible and resilient closed-cell expanded foam insulation. The maximum water vapor permeability is 0.10 perm-inch and the maximum thermal conductivity at 75°F temperature is 0.28 BTU in/(h ft2 F) for grades 1 and 3 and grade 2 is 0.30 BTU in/(h ft2 F). Grade 3 formulation does not contain any leachable chlorides, fluorides or polyvinyl chloride or any halogens.

The preformed tubular insulation is available in ID sizes from 3/8" to 6 IPS and in wall thickness from 3/8" to 1½" and in typical length of 6 feet. The tubular product is available with and without pre-applied adhesive. The sheet insulation is available in continuous lengths of 4 feet widths or 3' x 4' and in wall thicknesses from 1/8" to 2". The sheet product is available with and without pre-applied adhesive.

These materials are normally installed without additional vapor retarders. Additional vapor-retarder protection may be necessary when installed on very-low-temperature piping or where exposed to continually high humidity conditions. All seams and termination points must be sealed with manufacturer recommended contact adhesive. For outdoor applications a weatherable jacket or manufacturer recommended coating must be applied to protect against UV and ozone.

CELLULAR GLASS

Cellular Glass is defined by ASTM as insulation composed of glass processed to form a rigid foam having a predominantly closed-cell structure. Cellular glass is covered by ASTM C552, "Standard Specification for Cellular Glass Thermal Insulation" and is intended for use on surfaces operating at temperatures between -450 and 800°F. The Standard defines two grades and four types, as follows:


Cellular Glass insulations
TypeForm and Grades Available
IFlat Block, Grades 1 and 2
IIPipe and Tubing, Fabricated, Grades 1 and 2
IIISpecial Fabricated Shapes, Grades 1 and 2
IVBoard, Fabricated, Grade 2

Cellular glass is produced in block form (Type I). Blocks of Type I product are typically shipped to fabricators who produce fabricated shapes (Types II, III, and IV) that are supplied to distributors and/or insulation contractors.

The maximum thermal conductivity is specified, by grade, as follows (for selected temperatures):

Temperature,°FGrade 1Grade 2
Type I, Block
-150°F0.200.26
-50°F0.240.29
50°F0.300.34
75°F0.310.35
100°F0.330.37
200°F0.400.44
400°F0.580.63
Type II, Pipe
100°F0.370.41
400°F0.690.69

The standard also contains requirements for density, compressive strength, flexural strength, water absorption, water-vapor permeability, combustibility, and surface burning characteristics.

Cellular glass insulation is a rigid inorganic non-combustible, impermeable, chemically resistant form of glass. It is available faced or un-faced (jacketed or un-jacketed). Because of the wide temperature range, different fabrication techniques are sometimes used at various operating temperature ranges.

Typically, fabrication of cellular glass insulation involves gluing multiple blocks together to form a "billet" which is then used to produce pipe insulation or special shapes. The glue or adhesives used vary with the intended end use and design operating temperatures. For below-ambient applications, hot melt adhesives such as ASTM D 312 Type III asphalt are usually used.

On above-ambient systems, or where organic adhesives could pose a problem (i.e., LOX service) an inorganic product such as gypsum cement is often used as fabricating adhesive. Other adhesives may be recommended for specific applications. When specifying cellular glass insulation, include system operating conditions to ensure proper fabrication.

Fibrous Insulations

Fibrous insulations are composed of small diameter fibers that finely divide the air space. The fibers may be organic or inorganic and they are normally (but not always) held together by a binder. Typical inorganic fibers include glass, rock wool, slag wool, and alumina silica.


Fibrous Insulations

MINERAL FIBER PIPE

Mineral Fiber Pipe insulation is covered in ASTM C 547. The standard contains five types classified primarily by maximum use temperature.

TypeFormMaximum Use
Temp,°F
IMolded850°F
IIMolded1200°F
IIIPrecision V-groove1200°F
IVMolded1000°F
VMolded1400°F

The standard further classifies products by grade. Grade A products may be "slapped-on" at the maximum use temperature indicated, while Grade B products are designed to be used with a heat-up schedule.

The specified maximum thermal conductivity for all types is 0.25 Btu in/(hr ft2 °F) at a mean temperature of 100°F.

The standard also contains requirements for sag resistance, linear shrinkage, water-vapor sorption, surface-burning characteristics, hot surface performance, and non-fibrous (shot) content. Further, there is an optional requirement in ASTM C 547 for stress corrosion performance if the product is to be used in contact with austenitic stainless steel piping.

Fiberglass pipe insulation products will generally fall into either Type I or Type IV. Mineral wool products will comply with the higher temperature requirements for Types II, III, and V.

These pipe insulation products may be specified with various factory-applied facings, or they may be jacketed in the field. Mineral fiber pipe insulations systems are also available with "self-drying" wicking material that wraps continuously around pipes, valves, and fittings. These products are intended to keep the insulation material dry for chilled water piping in high-humidity locations.

Mineral fiber pipe insulation sections are typically supplied in lengths of 36 inch, and are available for most standard pipe and tubing sizes. Available thicknesses range from 1/2in to 6in.

Granular Insulations

CALCIUM SILICATE

Calcium Silicate thermal insulation is defined by ASTM as insulation composed principally of hydrous calcium silicate, and which usually contains reinforcing fibers.

Calcium Silicate Pipe and Block Insulation are covered in ASTM C 533. The standard contains three types classified primarily by maximum use temperature and density.


Calcium Silicate thermal insulation
TypeMaximum Use Temp (°F) and Density
IMax Temp 1200°F, Max Density 15 pcf
IAMax Temp 1200°F, Max Density 22 pcf
IIMax Use Temp 1700°F

The standard limits the operating temperature between 80°F to 1700°F.

Calcium Silicate pipe insulation is supplied as hollow cylinder shapes split in half lengthwise or as curved segments. Pipe insulation sections are typically supplied in lengths of 36 inch, and are available in sizes to fit most standard pipe sizes. Available thicknesses range from 1" to 3" in one layer. Thicker insulation is supplied as nested sections.

Calcium Silicate block insulation is supplied as flat sections in lengths of 36", widths of 6", 12", and 18" and thickness from 1" to 4". Grooved block is available for fitting block to large diameter curved surfaces.

Special shapes such as valve or fitting insulation can be fabricated from standard sections.

Calcium Silicate is normally finished with a metal or fabric jacket for appearance and weather protection.

The specified maximum thermal conductivity for Type 1 is 0.41 Btu-in/(h ft2 °F) at a mean temperature of 100°F. The specified maximum thermal conductivity for Types 1A and Type 2 is 0.50 Btu-in/(h ft2 °F) at a mean temperature of 100°F.

The standard also contains requirements for flexural (bending) strength, compressive strength, linear shrinkage, surface-burning characteristics, and maximum moisture content as shipped.

Typical applications include piping and equipment operating at temperatures above 250°F, tanks, vessels, heat exchangers, steam piping, valve and fitting insulation, boilers, vents and exhaust ducts.

Space Requirements of Insulation

The space requirements of the insulation must be taken into account when the installation is being designed and planned. Therefore, the insulation thicknesses should be determined in the early planning stages and the distances between the individual objects should be taken into account in the piping isometrics. To guarantee systematic installation of the insulation materials and the cladding without increased expense, observe the minimum distances between the objects as specified in the following illustrations.

MINIMUM DISTANCES BETWEEN VESSELS AND COLUMNS; DIMENSIONS IN INCHES (MM)

MINIMUM DISTANCE

MINIMUM DISTANCES BETWEEN VESSELS AND COLUMNS; DIMENSIONS IN INCHES (MM)

MINIMUM DISTANCES BETWEEN INSULATED PIPES; DIMENSIONS IN INCHES (MM)

MINIMUM DISTANCES WITHIN RANGE OF PIPE FLANGES; DIMENSIONS IN INCHES (MM)

  • a = distance flange to normal insulation
  • a ≥ 2in (50 mm)
  • x = bolt length + 1.2in (30 mm)
  • s = insulation thickness
  • CES BETWEEN INSULATED PIPES; DIMENSIONS IN INCHES (MM)
  • MINIMUM DISTANCES WITHIN RANGE OF PIPE FLANGES; DIMENSIONS IN INCHES (MM)
  • a = distance flange to normal insulation
  • a ≥ 2in (50 mm)
  • x = bolt length + 1.2in (30 mm)
  • s = insulation thickness


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