2.6 Heat Inside Mines
59
• L H 2 O : Latent heat of water vaporization (kJ kg
−1 ).
Between 3 and 10 l of water per l of fuel consumed are produced.
4. Calculate the sensible heat (Eq. 2.20):
q s = q
T − q L
(2.20)
5. Determine the temperature decrease considering the airflow (Eq. 2.21):
T =
q s
˙
m air C air
(2.21)
where
• T: Increase in air temperature (K),
• ˙
m air : Mass flow rate of air (kg s
−1 ), and
• C air : Specific heat of air (kJ kg
−1 K
−1 ).
In a similar manner to the previous case, all the energy generated in the electrical
equipment is transformed into either potential energy (e.g. pumps, conveyor belts)
or heat. This can be shown by means of Eq. 2.22 (Calizaya and Marks 2011):
q em = P abs −
˙
mgh
1000
(2.22)
In which
• q em : Heat generated per unit time (kW),
• P abs : Electric power absorbed (kW),
• ˙
m: Mass flow rate (kg s
−1 ),
• g: Acceleration of gravity (N kg
−1 ), and
• h: Height difference (m).
With regard to the absorbed power, no efficiency is quoted because all the energy
that is not transformed into potential energy, useful or otherwise, becomes heat.
In the case of electric machines, whose efficiency is much higher than that of
internal combustion engines, the total efficiency can be approximated to 75–85%.
Regarding lighting, an incandescent bulb has a light emission efficiency close to
5%;
16 that of fluorescent bulbs is 20% and that of LED bulbs is 95% (potentially
100%). The remaining energy is immediately transformed into heat. Moreover, fans
can also increase air temperature and should also be considered during heat emission calculations. A good approximation here, is to consider that they increase air
temperature at a rate of 0.25 K per kPa.
16 Only 5% of the total electrical power is transformed into visible optical power and 95% is wasted
in the form of thermal power (mostly in the infrared).
59
• L H 2 O : Latent heat of water vaporization (kJ kg
−1 ).
Between 3 and 10 l of water per l of fuel consumed are produced.
4. Calculate the sensible heat (Eq. 2.20):
q s = q
T − q L
(2.20)
5. Determine the temperature decrease considering the airflow (Eq. 2.21):
T =
q s
˙
m air C air
(2.21)
where
• T: Increase in air temperature (K),
• ˙
m air : Mass flow rate of air (kg s
−1 ), and
• C air : Specific heat of air (kJ kg
−1 K
−1 ).
In a similar manner to the previous case, all the energy generated in the electrical
equipment is transformed into either potential energy (e.g. pumps, conveyor belts)
or heat. This can be shown by means of Eq. 2.22 (Calizaya and Marks 2011):
q em = P abs −
˙
mgh
1000
(2.22)
In which
• q em : Heat generated per unit time (kW),
• P abs : Electric power absorbed (kW),
• ˙
m: Mass flow rate (kg s
−1 ),
• g: Acceleration of gravity (N kg
−1 ), and
• h: Height difference (m).
With regard to the absorbed power, no efficiency is quoted because all the energy
that is not transformed into potential energy, useful or otherwise, becomes heat.
In the case of electric machines, whose efficiency is much higher than that of
internal combustion engines, the total efficiency can be approximated to 75–85%.
Regarding lighting, an incandescent bulb has a light emission efficiency close to
5%;
16 that of fluorescent bulbs is 20% and that of LED bulbs is 95% (potentially
100%). The remaining energy is immediately transformed into heat. Moreover, fans
can also increase air temperature and should also be considered during heat emission calculations. A good approximation here, is to consider that they increase air
temperature at a rate of 0.25 K per kPa.
16 Only 5% of the total electrical power is transformed into visible optical power and 95% is wasted
in the form of thermal power (mostly in the infrared).