The question asks about a specific ratio related to the heat absorbed by water in a boiler under different conditions. Let's break down the terms involved.
The ratio mentioned is:
$\frac{\text{Heat received by 1 kg of water under working conditions}}{\text{Heat received by 1 kg of water evaporated from and at 100 °C}}$
Let's define the terms in the ratio:
Now let's look at the given options:
Based on the definitions, the ratio of heat received by 1 kg of water under working conditions to that received by 1 kg of water evaporated from and at 100 °C is known as the factor of evaporation.
| Term | Definition | Formula/Description |
|---|---|---|
| Factor of Evaporation | Ratio of heat absorbed per kg of water under actual conditions to the heat of evaporation at 100 °C. | $\frac{(h_s - h_{f(T_{fw})})}{h_{fg(100 °C)}}$ |
| Boiler Efficiency | Ratio of heat output in steam to heat input from fuel. | $\frac{\text{Mass of steam} \times \text{Heat absorbed per kg}}{\text{Mass of fuel} \times \text{Calorific value of fuel}}$ |
| Equivalent Evaporation | The amount of water that would be evaporated from and at 100 °C by the same amount of heat absorbed under actual conditions. | $\text{Mass of water evaporated (actual)} \times \text{Factor of Evaporation}$ |
The factor of evaporation is an important parameter in evaluating boiler performance, especially when comparing boilers operating under different conditions. It essentially normalizes the heat absorption performance to a standard condition (evaporation from and at 100 °C). This standard condition corresponds to receiving the latent heat of vaporization at atmospheric pressure.
Related to the factor of evaporation is the concept of Equivalent Evaporation. Equivalent evaporation is defined as the mass of water that would be evaporated from and at 100 °C by the same amount of heat absorbed by the water under the actual boiler operating conditions. It is calculated by multiplying the actual mass of water evaporated by the factor of evaporation.
$\text{Equivalent Evaporation (kg/hr)} = \text{Mass of water evaporated (kg/hr)} \times \text{Factor of Evaporation}$
Using equivalent evaporation allows for a direct comparison of the steaming capacity of different boilers, regardless of their operating pressures, temperatures, or feedwater temperatures.
Boiler efficiency, on the other hand, is a measure of how well the energy from the fuel is utilized. While factor of evaporation relates to the quality of steam produced relative to a standard, boiler efficiency relates to the overall energy conversion process from fuel to steam.
When once a pocket of smoke, containing air pollutants, is released into the atmosphere from a source like an automobile or a factory chimney, it gets dispersed into the atmosphere into various directions depending upon the
1. prevailing winds
2. temperature
3. pressure conditions
Select the correct answer.
During the compaction test, the weight of compacted soil specimen along with mould is 38.2 N. The volume and weight of mould are 0.95×10-3 m³ and 20.5 N respectively and the water content is 12%. The dry unit weight of the compacted specimen will be nearly