How to convert gas flow from M3/hr to Nm3/hr and Sm3/hr???

 

How to convert M3/hr to NM3/hr and Sm3/hr

How to convert Nm3/hr to SM3/hr and M3/hr

 

In order to understand the above, need to understand the basics of STP & NTP

STP - Standard Temperature and Pressure

STP is commonly used to define standard conditions for temperature and pressure of air or gases.

As per IUPAC STP is  air at 0 ^oC (273.15 K, 32 ^o F) and 10 ⁵ pascals (1.03 kg/cm2).

 As per  Imperial and USA system of units STP is air at 60 ^o F (520 ^o R, 15.6 ^oC ) and 14.696 psia (1 atm,  1.01325 bara)

Note! The earlier IUAPC definition of STP to 273.15 K and 1 atm (1.03kg/cm2) is discontinued. 

NTP - Normal Temperature and Pressure

NTP is commonly used  as a standard condition  for testing and documentation of fan capacities:

NTP - Normal Temperature and Pressure - is defined as air at 20 oC (293.15 K, 68 o F) and 1 atm ( 101.325 kN/m2, 101.325 kPa, 14.7 psia, 0 psig, 29.92 in Hg, 407 in H2O, 760 torr). Density 1.204 kg/m 3 (0.075 pounds per cubic foot)

 

In order to convert M3/Hr to NM3/hr, we use below formula

PaVa/Ta = PnVn/Tn

Vn = PaVaTn / TaPn

Where, Pa, Va & Ta are air parameters at actual condition

Pn,Vn & Tn are air parameters at Normal condition

In order to convert M3/Hr to SM3/hr, we use below formula

PaVa/Ta = PsVs/Ts

Vs = PaVaTs / TaPs

Where, Pa, Va & Ta are air parameters at actual condition

Ps,Vs & Ts are air parameters at Normal condition

In order to convert NM3/Hr to SM3/hr, we use below formula

PnVn/Tn = PsVs/Ts

Vs = PnVnTs / TnPs

Pn,Vn & Tn are air parameters at Normal condition

Ps,Vs & Ts are air parameters at Normal condition

Solved examples:

1-A 25 m3/sec capacity forced draft fan  discharges air at 250 mmwc static pressure and 30 deg C temperature, calculate the air flow at Nm3/hr and Sm3/hr

Pa = 250mmwc = 250/10000 = 0.025 kg/cm2

Va = 25 m3/sec

Convert it into M3/hr = 25 X 3600 = 90000 m3/hr

Ta = 30 + 273.15 = 303.15 K

Tn = 20 + 273.15 = 293.15 K

Ts = 15.6 + 273.15 =288.75 K

We have the formula

PaVa/Ta = PnVn/Tn

Vn = PaVaTn / TaPn

Vn = ((0.0025+1.033) X 90000 X 293.15) / (303.15 X 1.033)

Vn =89137.445 NM3/hr

Similarly

PaVa/Ta = PsVs/Ts

Vs = PaVaTs / TaPs

Vs =(0.025+1.033) X 90000 X 288.75) / (303.15 X 1.033)

Vs = 87,799.54 Sm3/hr

Now convert Nm3/hr to SM3/hr to cross check the flow

PnVn/Tn = PsVs/Ts

Vs = PnVnTs/TnPs

Vs =1.033 X 89137.445 X 288.75 / (293.15 X 1.033)

Vs = 87,799.54 Sm3/hr

2-A boiler of 100 TPH produces flue gas 27 MT/hr of flue gases at pressure 350 mmwc (ID fan inlet) and 145 deg C temperature, calculate the flue gas flow in Nm3/hr  and Sm3/hr

Given data,

Flue gas flow = 27MT/hr = 27 X 1000 = 27000 kg/hr

Convert to M3/hr

Density of flue gas at 145 deg C temperature = 1.293 X 273.15 / (273.15 + 145) = 0.84 kg/m3

Flue gas flow in M3/hr = 27000 kg/hr / 0.84 kg/m3 = 32142.85 M3/hr

Pa = 350 mmwc = 350 / 10000 =(1.033+ 0.035=1.068 kg/cm2)

Ta = 273.15 + 145 = 418.15 K

We have,

PaVa/Ta = PnVn/Tn

Vn = 1.068 X 32142.85 X (20 + 273.15) / (418.15 X 1.033)

Vn = 23,297.70 Nm3/hr

Similarly,

PaVa/Ta = PsVs/Ts

Vs =1.068 X 32142.85 X (15.6 + 273.15) / (418.15 X 1.033)

Vs = 22,948.01 Sm3/hr

Now convert Nm3/hr to SM3/hr to cross check the flow

PnVn/Tn = PsVs/Ts

Vs = PnVnTs/TnPs

Vs =1.033 X 23297.70 X 288.75 / (293.15 X 1.033)

Vs = 22,948.015 Sm3/hr

 

For more>>>>> read Powerplant and calculations

 

 

How to calculate the quantity of oxygen required for gas cutting operation?

 We know that, for gas cutting operation generally we use combination of oxygen and LPG or Oxygen and acetylene.But now a days for some industry LPG is banned for safety point of view.

 

Following table gives the difference between Oxy-acetyle and Oxy-LPG gas cutting

COMPARISON BETWEEN ACETYLENE AND LPG FUELS FOR GAS CUTTING OPERATION
SL No.	Acetylene	LPG (Propane)
1	Highest flame temperature up to 3100 Deg C	Flame temperature up to 2800 Deg C
2	Flame speed up to 7.5 m/sec	Flame speed up to 3.3 m/sec
3	Most of the heat released is in inner cone	Most of the heat released is in outer cone
4	Higher flame GCV of inner cone (4500 kcal/M3)	Lower flame GCV of inner cone (2500 kcal/M3) as compared to acetylene
5	Stoichiometric air fuel ratio1.2:1 (Requires 2.5 to 3 Oxygen cylinders for burning one Acetylene cylinder)	Stoichiometric  air fuel ratio 4.3:1 (Requires 7 to 8 Oxygen cylinders for burning one LPG cylinder)
6	Can be used in gas welding, as acetylene when burning with air creates reducing zone that cleans the steel surface	Cannot be used in gas welding as it does not create reducing zone
7	Acetylene has Specific gravity 0.9 kg/m3, so if it leaks it will raise in air without harming much	Propane  has Specific gravity 1.6 kg/m3,which is heavier than air.So if it leaks it will concentrate in deck level or any other closed/corner area
8	Acetylene requires less air for complete combustion	Propane requires more air for complete combustion, so there may be chances of formation of carbon monoxide (CO) in case of incomplete combustion. This incomplete combustion may result into poisoning of working area, as CO is poisonous gas
9	Can be used for cutting & welding applications in industry	Used only for domestic applications

 

Calculate the number of Oxygen cylinders required to consume 1 no.of industrial LPG cylinder for gas cutting operation

Commercial LPG (C3H8) has 19 kg weight that is 19 kg of propane

Combustion equation of propane

C3H8 + 5O2 = 3CO2 + 4 H2O

44 + 160 = 132 + 72 (Molecular weight of C = 12, O = 16, H = 1)

 Divide equation by 44

1 + 3.63 = 3 +1.63

 From above result it is clear that 3.63 kg of Oxygen is required to burn 1 kg of Propane to achieve 100% combustion.

So for burning 19 kg of commercial LPG, need 19 X 3.63 = 68.97 Kg of oxygen

 Volume of oxygen cylinder in cylinder = 6.9 M3 compressed at 140-150 kg/cm2

Convert 6.9 to kg by dividing oxygen density, we get weight of O2 in cylinder = 9.1 kg

 So total O2 cylinders required = 68.97 / 9.1 =7.58 Nos for consuming 1 LPG cylinder

Calculate the number of Oxygen cylinders required to consume 1 no.of dilute acetylene cylinder for gas cutting operation

DA (C2H2) cylinder has 8 m3 of acetylene

 Convert volume to kg by multiplying the density of the gas

8 X 0.899 = 7.192 kg

Combustion equation of propane

2C2H2 + 5O2 = 4CO2 + 2H2O

52 + 160 = 176 + 36 (Molecular weight of C = 12, O = 16, H = 1)

 Divide equation by 52

1 + 3.07 = 3.38 +0.69

So for burning 7.192 kg of DA, need 7.192 X 3.07 = 22.07 Kg of oxygen

So total O2 cylinders required = 22.07 / 9.1 =2.42 Nos

 Read Powerplant maintenance calculations

Note:

DA (Dilute Acetylene): OD 265 mm X Height 1 meter (Appx) and thickness 4.0 mm.Volume of acetylene in cylinder is 8.5 m3

 Oxygen cylinder size : OD 235 mm X Height 1.34 meter (Appx) and thickness 4.0 mm.Volume of O2 in cylinder is 6.9 m3

 

For more read>>>power plant and calculations

 

How to calculate the quantity of lime stone required to reduce SO2 emission ??

We know that, Lime stone is nothing but Calcium carbonate (CacO3)

 

Lime stone (CaCo3) on heating gets converted into slaked lime and carbon dioxide

 

i.e,

CaCo3 + Heat = Cao + Co2

Shall calculate, quantity of CacO3 required for converting into Cao.

Molecular weight of Ca = 40 g/mol

Molecular weight of Oxygen = 16 g/mol

Molecular weight of Carbon = 12 g/mol

Molecular weight of Sulphur = 32 g/mol

Sulphur on heating gets converted into Sulphur di-oxide (SO2)

Therefore we have,

CaCo3 + Heat = Cao + CO2

(40+12+3X16) + Heat = (40+16) + (12+2X16)

100 + Heat           = 56 + 44

1 + Heat = 0.56 + 0.44-----------I

This implies, 1 kg of calcium carbonate (CaCO3) on heating gets converted into 0.56 kg of slaked lime & 0.44 kg of CO2

Further,Sulphur present in coal, on combustion gets converted into sulphur dioxide (SO2)

Shall calculate, quantity of SO2 generated on combustion of 1 kg of sulphur

 

S + O2 = SO2

32 + 2X16 = 32+2X16

32 + 32 = 64

1 + 1 = 2-----II

This means, 2 kg of SO2 will produce on combustion of 1 kg of sulphur.

Further,Slaked lime (Cao) reacts with Sulphur dioxide (SO2) & converts into Calcium sulphate (CaSO4)

i.e

Cao + SO2 + O2 = CaSO4

(40+16) + (32 + 2X16) + 16 = (40+32+4X16)

56 +(32 + 2X16) + 16 = 136

56 + 64 + 16 = 136

0.875 + 1 + 0.25 = 2.12------III

This means, 1 kg of So2 needs 0.875 kg of Cao to produce  2.42 kg of Calcium sulphate

i.e, On burning 1 kg of sulphur there produces 2 kg of SO2 (refer equation-II), hence Cao required to reduce sulphur from 1 kg So2 = 2 X 0.875 = 1.75 kg

Similarly referring equation-I, 1 kg of calcium carbonate (CaCO3) on heating gets converted into 0.56 kg of slaked lime(Cao)

Therefore  lime stone required for 1 kg sulphur to convert it into CaSO4 is;

 

1.75/0.56 = 3.125 kg

Considering 95% efficiency for above combustion, total lime stone required for 1 kg sulphur to get converted it into CaSo4 is;

3.125 X 105% = 3.28 kgs-------IV

 

Demonstration with example.

A 50 MW coal based power plant has specific fuel consumption 0.75 kg/kwh.The sulphur content in the coal is 0.75%, the maximum permissible limit of sulphur in the coal is 0.5%.Calculate the amount of lime stone required to reduce SO2 emission.

 

Assuming SO2 emission at 0.5% sulphur in coal is normal.

Extra % of sulphur in coal is = 0.75-0.5 = 0.25%

Coal consumption per day

Specific fuel consumption = 0.75 kg/kwh

Total power generation in a day = 50 X 1000 X 24 = 1200000 kw

Total coal consumption = 1200000 X 0.75 / 1000 = 900 MT/day

Extra Sulphur burned  = 900 X 0.25% = 2.25 MT or 2250 kg/day

Lime stone required to reduce SO2 emission is 2250 X 3.28 = 7380 kg (refer equation-IV)

 For more read>>>>>powerplantandcalculations