How to Calculate Steam Consumption
or
How to calculate steam
requirements for flow and non-flow applications. Including warm-up, heat losses
and running loads.
The optimum design for a steam system will largely depend on whether the
steam consumption rate has been accurately established. This will enable pipe
sizes to be calculated, while ancillaries such as control valves and steam
traps can be sized to give the best possible results. The steam demand of the
plant can be determined using a number of different methods:
- change a product or fluid
temperature
- maintain a product or fluid
temperature
“A benefit with steam is the
large amount of heat energy that can be transferred. The energy released when
steam condenses to water is in the range 2000 - 2250 kJ/kg (depending
on the pressure) - compared to water with 80 - 120 kJ/kg (with
temperature difference 20 - 30 oC). “
Calculation :- By analysing the heat output on an item of
plant using heat transfer equations, it may be possible to obtain an estimate
for the steam consumption. Although heat transfer is not an exact science and
there may be many unknown variables, it is possible to utilise previous
experimental data from similar applications. The results acquired using this
method are usually accurate enough for most purposes.
Measurement
Steam consumption may be determined by direct
measurement, using flowmetering equipment. This will provide relatively
accurate data on the steam consumption for an existing plant. However, for a
plant which is still at the design stage, or is not up and running, this method
is of little use.
Thermal rating
The thermal rating (or design rating) is often displayed on the name-plate of an individual item of plant, as provided by the manufacturers. These ratings usually express the anticipated heat output in kW, but the steam consumption required in kg/h will depend on the recommended steam pressure.
Solution or Calculation
·
Changing Product Temperature - Heating up the Product
with Steam
·
To maintain the product temperature as heat is lost by
natural causes or by design, that is providing a ‘heat loss’ component.
In any heating process, the ‘heating up’
component will decrease as the product temperature rises, and the differential
temperature between the heating coil and the product reduces. However, the heat
loss component will increase as the product temperature rises and more heat is
lost to the environment from the vessel or pipe work.
The total heat demand at any time is the sum of these
two components.
Q = m cp dT…..(1)
where
Q = quantity of energy or heat (kJ)
m = mass of substance (kg)
cp = specific heat of substance (kJ/kg oC ) - Material
properties and heat capacities common materials
dT = temperature rise of substance (oC)
In its original form this equation can be
used to determine a total amount of heat energy over the whole process.
However, in its current form, it does not take into account the rate of heat
transfer. To establish the rates of heat transfer, the various types of heat
exchange application can be divided into two broad categories:
Non-flow type applications :- an
non-flow type applications the process fluid is kept as a single batch within a
tank or vessel. A steam coil or a steam jacket heats the fluid from a low to a
high temperature.
The mean rate of heat
transfer for such applications can be expressed as
q = m cp dT / t …..(2)
where
q = mean heat transfer rate (kW (kJ/s))
m = mass of the product (kg)
cp = specific heat of the product (kJ/kg.oC)
- Material properties and heat capacities common materials
dT = Change in temperature of the fluid (oC)
t = total time over which the heating process occurs
(seconds)
Calculating the mean heat transfer rate in a non-flow
application.
A quantity of oil is heated from a temperature of 30
°C to 120 °C over a period of 10 minutes (600 seconds). The volume of the oil
is 30 litres, its specific gravity is 0.8 and it specific heat capacity is 1.8
kJ/kg °C over that temperature range.
Determine the rate of heat transfer required:
As the density of water at Standard Temperature and
Pressure (STP) is 1 000 kg/m³
Solution :
The Density
of the oil Po = 0.8*1000
= 800
kg/m3
As 1000Ltr 3
m3 , Po= 800 kg/m3
Po= 0.8 kg/l
Therefore the Mass of the Oil = 0.8 x 30
m
= 24 kg
q = m cp dT / t
q= 24 x
1.8 x (120-30)/600
q = 6.48
kj/s (6.48 kw)
Equation
2 can be
applied whether the substance being heated is a solid, a liquid or a gas. However,
it does not take into account the transfer of heat involved when there is a
change of phase.
The quantity
of heat provided by the condensing of steam can be determined by Equation
|
Where
Q = Quantity
of Heat (KJ)
Ms = Mass of
Steam (kg)
Hfg =
Specific enthalpy of evaporation of Steam (kg)
It therefore
follows that the steam consumption can be determined from the heat transfer
rate and vice-versa
|
Where
Q* = Heat
Transfer Rate (kw)
Ms* = Steam Consumption
(kg/s)
Hfg = enthalpy of evaporation of Steam (kg)
Equation(4)
to Equation (2)
Where
Ms* = Steam Consumption
Rate (kg/s)
Hfg = enthalpy of evaporation of Steam (kg)
Q* = Heat Transfer Rate (kw)
M = Mass of
Secondary Fluid (kg)
Cp =
Specific Heat Capacity of Secondary fluid (kj/kg)
t =
Temperature Rise for Secondary Fluid ( c)
t = Time of
Heating Process (Second)
Example
-2
A tank
containing 400 kg of kerosene is to be heated from 10 °C to 40 °C in 20 minutes
(1 200 seconds), using 4 bar g steam. The kerosene has a specific heat capacity
of 2.0 kJ/kg °C over that temperature range. hfg at 4.0 bar g is 2 108.1 kJ/kg.
The tank is well insulated and heat losses are negligible.
q = m cp dT / t
= 400 x 2.0 x(40-10)/1200
= 20kj/s
Ms= Q/Hfg
= 20/2108.1
kg/h
Ms = 34.2
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