Artist LOC AND FOC I Part II....

LOC AND FOC I Artist.....

My Pre-boss Girl friend

Calculating steam requirements – m cp ΔT.

A process needs heat at

• the correct temperature and
• the correct rate of heat transfer

Heat is being generated in the boiler in the form of steam. This heat is being distributed by steam lines to the process. Steam pressure determines the temperature at which heat is supplied, as saturated steam temperature is directly proportional to pressure. We need a ΔT of minimum 15-30°C to have efficient heat transfer (rate of heat transfer).

Consider a heat exchange process. The primary side is the steam space, and the secondary side is the process. Steam is condensing on the primary side into water. It is changing phase into liquid and giving off its latent heat to the process. This is Primary Heat (Q).

Primary Q = m x hfg

Where,
Primary Q = Quantity of heat energy released (in kcals)
m = Mass of steam releasing the heat (in kgs)
hfg = Specific enthalpy of evaporation of steam (in kcals/kg)

On the secondary side, this heat is being used for two things:

• 'heating up' heat - to increase the product temperature to the degree desired
• 'maintainance' heat - to maintain the product temperature as heat is lost by radiation, etc

Where,
Secondary Q = Quantity of heat energy absorbed (in kcals)
m = Mass of the substance absorbing the heat (in kgs)
cp = Specific heat capacity of the substance (in kcals / kg °C )
ΔT = Temperature rise of the substance (in °C)

This equation is also modified and used to establish the amount of heat required to raise the temperature of a substance, for a range of different heat transfer processes.
The above equations are very important. As Heat energy is being transferred from the primary to the secondary side, in an ideal condition,

Primary Q = Secondary Q

And this is the equation to calculate the theoretical heat balance of the entire system.

Example 1. Calculate steam flow rate for an autoclave which is heating 10,000 bottles of 1 litre each to a temperature of 120°C in 30 minutes. Steam supply is at 3 kg/cm2g.

Solution. What we are asking for is - what is the mass of steam that is supplied to the autoclave to heat these 10 bottles. This is 'm' on the primary side. First we will calculate the heat absorbed by the bottles (process), ie, secondary Q.

The formula

Secondary Q = m x cp x ΔT

Where,
Secondary Q = Quantity of heat absorbed by the bottles (in kcals)
m = Mass of water in the bottles which is absorbing the heat (in kgs)
= 10,000 bottles X 1lt = 10,000 lt = 10,000 kg
cp = Specific heat capacity of water (in kJ/kg °C ) = 1 kcal/kg °C
ΔT = Temperature rise of water (°C) assuming ambient is 30°C
= 120°C – 30°C = 90°C
Gives,
Secondary Q = 10,000 kg x 1 kcal/kg °C x 90°C = 9,00,000 kcal

So, 9,00,000 kcal is the heat energy absorbed by this autoclave on the secondary (process) side in 30 minutes. Steam at 3 kg/cm2g has 510 kcal/kg latent heat hfg (from steam tables).

As Sec Q = Pri Q,
9,00,000 kcal = m x 510 Kcal/kg
m = 9,00,000 / 510 = 1765 kgs

1765 kgs is the steam required in 30 mins. So, steam flowrate is 1765 X 60/30 = 3530 kgs/hr for this autoclave.

Suppose steam is supplied to a heat exchanger at 3 kg/cm2g - hg 630 kcal/kg. Condensate is coming out of the traps at 3 kg/cm2g hfg 130 kcal/kg. Ideally, the product should absorb 511 kcal/kg. But, it doesnt. Heat gets absorbed by the heat transfer barriers and is also lost via radiation. So, the actual heat absorbed is less than 511 kcal/kg.

Specific Volume

Specific volume vs. Pressure
We can see below, as the steam pressure increases from 1atm to 4 atm, the density of the steam molecules is increasing. As the specific volume is inversely related to the density, the specific volume will decrease with increasing pressure. We can see the reduced volume in the last jar.

This diagram clearly shows that the greatest change in specific volume occurs at lower pressures, whereas at the higher end of the pressure scale there is much less change in specific volume.

The extract from the steam tables below, shows specific volume, and other data related to saturated steam.

At 7 kg/cm2g, the saturation temperature of water is 170°C. More heat energy 'hf' is required to raise its temperature to saturation point at 7 bar g than would be needed if the water were at atmospheric pressure. The table gives a value of 171.96 kcals to raise 1 kg of water from 0°C to its saturation temperature of 170°C.

The heat energy (enthalpy of evaporation 'hfg') needed by the water at 7 bar g to change it into steam is actually less than the heat energy required at atmospheric pressure. This is because the specific enthalpy of evaporation decreases as the steam pressure increases.

However, as the specific volume also decreases with increasing pressure, the amount of heat energy transferred in the same volume actually increases with steam pressure.