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As everyone who has used a fireplace knows, when a fire burns in a furnace, a draft, or slight vacuum, is induced that causes the hot combustion gases and entrained particulate matter to flow up and out of the stack. The reason is that the hot gas in the stack is less dense than air at ambient temperature, leading to a lower hydrostatic head inside the stack than at the furnace inlet. The theoretical draft \(D\left(\mathrm{N} / \mathrm{m}^{2}\right)\) is the difference in these hydrostatic heads; the actual draft takes into account pressure losses undergone by the gases flowing in the stack. Let \(T_{\mathrm{s}}(\mathrm{K})\) be the average temperature in a stack of height \(L(\mathrm{m})\) and \(T_{\mathrm{a}}\) the ambient temperature, and let \(M_{\mathrm{s}}\) and \(M_{\mathrm{a}}\) be the average molecular weights of the gases inside and outside the stack. Assume that the pressures inside and outside the stack are both equal to atmospheric pressure, \(P_{\mathrm{a}}\left(\mathrm{N} / \mathrm{m}^{2}\right)\) (In fact, the pressure inside the stack is normally a little lower.) (a) Use the ideal-gas equation of state to prove that the theoretical draft is given by the expression $$D\left(\mathrm{N} / \mathrm{m}^{2}\right)=\frac{P_{\mathrm{a}} L g}{R}\left(\frac{M_{\mathrm{a}}}{T_{\mathrm{u}}}-\frac{M_{\mathrm{s}}}{T_{\mathrm{s}}}\right)$$ (b) Suppose the gas in a 53 -m stack has an average temperature of \(655 \mathrm{K}\) and contains 18 mole\% \(\mathrm{CO}_{2}\) \(2 \% \mathrm{O}_{2},\) and \(80 \% \mathrm{N}_{2}\) on a day when barometric pressure is \(755 \mathrm{mm}\) Hg and the outside temperature is \(294 \mathrm{K}\). Calculate the theoretical draft \(\left(\mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) induced in the furnace.

Step by step solution

01

Derive Theoretical Draft Expression

Using the ideal-gas equation of state \( P = 蟻RT \), we can write an expression for the densities of the stack gas (\(蟻_s\)) and ambient air (\(蟻_a\)) as follows: \( 蟻_s = \frac{P_a M_s} {RT_s} \) and \( 蟻_a = \frac{P_a M_a} {RT_a} \). The pressure difference between the bottom and the top of the stack is due to the difference in hydrostatic head of the inside and outside gases. Therefore, 'D' is given by \( D = 蟻_a gL - 蟻_s gL \) which simplifies to \( D = gL (蟻_a - 蟻_s) \). Substituting the values of \( 蟻_s \) and \( 蟻_a \) from above, we get \( D = \frac{P_a L g}{R} \left(\frac{M_a}{T_a} - \frac{M_s}{T_s}\right) \).

02

Calculate Average Molecular Weight

Given the composition of the gas in the furnace stack, the average molecular weight \(M_s\) is the sum of the weight fractions times their molecular weights. The molecular weights of \( CO_2 \), \( O_2 \), and \( N_2 \) are \(44 gm/mol\), \(32 gm/mol\), \(28 gm/mol\) respectively. So, \( M_s = 0.18*44 + 0.02*32 + 0.80*28 = 30.8 g/mol \).

03

Convert Pressure to appropriate units

The given pressure is in mm of Hg. We need to convert it to \(N/m^2\) (Pa). Using the conversion factor \(1 mm Hg = 133.322 Pa\), we get \( P_a = 755*133.322 = 100705.51 Pa \).

04

Calculate Theoretical Draft

Substituting \(P_a = 100705.51 Pa\), \(T_s = 655 K\), \(T_a = 294 K\), \(M_s = 30.8 \), \(M_a = 28.97 g/mol \) (for air), \(L = 53 m\) and \(g = 9.81 m/s^2\) in the draft equation from step 1, we calculate the theoretical draft in pascals. Further convert it to pressure in cm of water using the conversion factor 1 Pa = 0.00001 cm H2O.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Hydrostatic Head

The term hydrostatic head refers to the height of fluid above a particular point that contributes to the pressure at that point due to gravity. In the context of a furnace stack, the concept becomes particularly interesting. Imagine you're standing outside and you can see the chimney with smoke rising from it. The hot gas inside that chimney is less dense than the colder air outside, creating a pressure difference between them. This pressure difference powers the draft which helps the smoke rise up and out of the chimney.

When discussing hydrostatic head in this scenario, we're dealing with the weight of the gas column within the stack compared to the weight of an equivalent column of outside air. It's this weight difference that causes a pressure differential, or draft, that draws the combustion gases upwards. Essentially, the greater the hydrostatic head of the ambient air compared to that of the hot gases, the stronger the draft.

In our theoretical draft calculation, the hydrostatic head contributes to the difference in pressures at the bottom of the stack, which can be translated into an appealing and simple formula for draft. Applying this to real-life situations can help in designing efficient chimney systems for households and industrial setups alike.

The Ideal-Gas Equation of State

Ideal-gas equation of state is a cornerstone of understanding how gases behave under different temperatures, pressures, and volumes. This equation is beautifully simple: \(PV = nRT\), where

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles of gas,
• \(R\) is the universal gas constant,
• and \(T\) is the temperature in kelvins.

For our draft calculation, we modify it slightly to relate the density \(\rho\) of the gas to its molecular weight \(M\), thus: \(\rho = \frac{PM}{RT}\).

This version of the equation helps us understand the relationship between the molecular weight of the gas, its temperature, and the resulting density, all of which are critical when calculating the draft in a chimney. By using it, we can derive a theoretical draft expression considering the molecular weights and temperatures of the gases inside and outside the stack. It showcases the power of theoretical physics: with some simplifying assumptions, we can model and predict the behaviour of gases in practical applications such as chimney design.

The Role of Molecular Weights in Draft Calculation

The average molecular weight of a gas plays a pivotal role in determining its density, and thus its behaviour within the context of hydrostatic head. Molecular weight refers to the mass of one mole of molecules for a particular substance, measured in grams per mole (g/mol).

When dealing with a mixture of gases, such as the hot combustion gases in our furnace stack, we need to find the weighted average of the molecular weights of each individual component. This is obtained by multiplying the fraction of each gas (by moles) by its respective molecular weight and summing them up. In our problem, the stack gas consists of \(CO_2\), \(O_2\), and \(N_2\), each with different molecular weights.

Average Molecular Weight Calculation

Following the composition percentages, you'd sum the products of each gas's mole percentage and its molecular weight to get the overall average molecular weight. This figure is crucial because, together with the temperature and gas constant, it determines the density of the gas. The density then feeds into the hydrostatic head calculation, directly affecting the theoretical draft calculation - a perfect example of how a seemingly simple characteristic like molecular weight plays into the broader physics of everyday phenomena.