The largest avoidable loss in most industrial furnaces is
Answer: C
Furnace exhausts leave far hotter than a boiler's, so recovering that heat into the combustion air or into a waste heat boiler is the first measure considered. Heat in the product is the useful output rather than a loss.
Please log in or register to report this question.
Heat stored in a furnace's refractory is lost mainly when the furnace is
Answer: D
The lining must be brought back to temperature after every shutdown, and on an intermittent furnace that is a large share of the fuel. Lightweight ceramic fibre linings store far less and are therefore preferred for batch duty.
Please log in or register to report this question.
Cogeneration is attractive where a site has
Answer: A
Correct answer: a substantial and steady demand for both heat and power
Cogeneration earns its return by using the heat rejected from generating power, so it needs a real and continuous use for that heat. A site wanting power alone must dump the heat and gains nothing, and a site wanting heat alone is better served by a boiler. The steadiness matters as much as the size, since a plant sized for a peak and run at part load rarely pays back.
Please log in or register to report this question.
A topping cycle cogeneration plant generates
Answer: B
The fuel is burnt at high temperature to drive a turbine or engine and the exhaust, still hot, supplies the process. A bottoming cycle reverses the order and is used where the process itself needs the highest temperature.
Please log in or register to report this question.
A boiler raises 5000 kg/h of steam with an enthalpy rise of 2700 kJ/kg, burning 400 kg/h of oil of net calorific value 42 MJ/kg. Its efficiency is about
Answer: C
Correct answer: 80 per cent
\[ \eta = \frac{5000 \times 2700}{400 \times 42{,}000} = \frac{13{,}500{,}000}{16{,}800{,}000} = 0.804 \approx 80\%. \] This is the direct method: useful heat in the steam over energy in the fuel. It is quick and needs only two measurements, but it says nothing about where the missing 20 per cent went - which is what the indirect method is for.
Please log in or register to report this question.
A flue gas mass of 14 kg per kg of fuel leaves 180 \( ^\circ \)C above ambient, with a specific heat of 1.0 kJ/kgK and a fuel of 42 MJ/kg. The dry flue gas loss is about
Answer: D
Correct answer: 6 per cent
\[ \frac{m c_p \Delta T}{\mathrm{NCV}} = \frac{14 \times 1.0 \times 180}{42{,}000} = \frac{2520}{42{,}000} = 0.06 = 6\%. \] The dry flue gas loss is usually the largest single loss in a well-run boiler, and both of its drivers - the gas mass and the stack temperature above ambient - are within the operator’s control.
Please log in or register to report this question.
Reducing that flue gas mass from 14 to 11 kg per kg of fuel, other things equal, changes the dry flue gas loss to about
Answer: A
Correct answer: 4.7 per cent
\[ \frac{11 \times 1.0 \times 180}{42{,}000} = \frac{1980}{42{,}000} = 0.0471 \approx 4.7\%. \] The loss is directly proportional to the gas mass, so cutting the mass by a fifth cuts the loss by a fifth - 1.3 percentage points of boiler efficiency recovered by air trimming alone, which costs nothing but attention.
Please log in or register to report this question.
Feedwater of 300 ppm total dissolved solids is used in a boiler whose limit is 3000 ppm. The blowdown required is about
Answer: B
The solids entering must leave, so the blowdown fraction is the feed concentration divided by the boiler concentration, that is a tenth. Halving the feed solids by better treatment would halve the blowdown and its heat loss.
Please log in or register to report this question.
A boiler at 10 bar blows down 500 kg/h of water whose enthalpy is 763 kJ/kg above the make-up temperature. The heat discarded is about
Answer: C
Correct answer: 106 kW
\[ \frac{500 \times 763}{3600} = \frac{381{,}500}{3600} = 106\ \mathrm{kW}. \] The division by 3600 converts kJ/h to kW and is where this calculation is usually got wrong. Note that the heat is recoverable: a blowdown heat exchanger can return most of it to the make-up water.
Please log in or register to report this question.
A bare surface at 60 \( ^\circ \)C in a 30 \( ^\circ \)C room loses, with a coefficient of 10 W/m\( ^2 \)K, about
Answer: D
Correct answer: 300 W/m\( ^2 \)
\[ q = h\,\Delta T = 10 \times (60-30) = 300\ \mathrm{W/m}^2. \] Only the temperature difference enters, not the absolute temperatures. Spread over the area of an uninsulated steam main this is a large and entirely continuous loss, which is why surface temperature surveys are a routine part of a thermal audit.
Please log in or register to report this question.
