An ideal gas of density rho = 0.2 kg m^-3 enters a chimney of height h at the rate of alpha = 0.8 kg

Q: Why do we use $P^{1-\gamma}T^\gamma = \text{constant}$ instead of $PV^\gamma = \text{constant}$?

Since we are given the temperatures and pressures, the $P-T$ relation is much more direct. You can easily derive it by substituting $V = \frac{nRT}{P}$ into the standard $PV^\gamma = \text{constant}$ equation.

Q: How did we get the expression for the change in specific enthalpy?

For an ideal gas, the change in specific enthalpy is $\Delta h = c_p \Delta T$. Using the relation $c_p = \frac{\gamma R}{M(\gamma - 1)}$ and the ideal gas equation $\frac{P}{\rho} = \frac{RT}{M}$, we can rewrite $c_p \Delta T$ purely in terms of pressure and density as $\frac{\gamma}{\gamma - 1} \left( \frac{P_2}{\rho_2} - \frac{P_1}{\rho_1} \right)$.

Q: Why did the velocity decrease even though the gas expanded?

Although the gas expanded (its density decreased from $0.2$ to $0.1$), the cross-sectional area of the chimney increased significantly (from $0.1$ to $0.4$). The area increased by a factor of 4, while the density only halved. To maintain a constant mass flow rate, the velocity had to decrease.

Q: Can we use standard Bernoulli's equation here?

No. The standard Bernoulli's equation assumes incompressible flow (constant density). Since the gas density changes significantly as it rises and expands, we must use the compressible form of Bernoulli's equation, which is essentially the Steady Flow Energy Equation.

Q: What does the term $\frac{P}{\rho}$ represent physically in the energy equation?

The term $\frac{P}{\rho}$ represents the 'flow work' or 'flow energy' per unit mass. It is the work required to push a unit mass of fluid into or out of the control volume against the surrounding pressure.

Question

An ideal gas of density $ho = 0.2 	ext{ kg m}^{-3}$ enters a chimney of height $h$ at the rate of $\alpha = 0.8 	ext{ kg s}^{-1}$ from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is $A_1 = 0.1 	ext{ m}^2$ and the upper end is $A_2 = 0.4 	ext{ m}^2$. The pressure and the temperature of the gas at the lower end are $600 	ext{ Pa}$ and $300 	ext{ K}$, respectively, while its temperature at the upper end is $150 	ext{ K}$. The chimney is heat insulated so that the gas undergoes adiabatic expansion. [Take: $g = 10 	ext{ ms}^{-2}$ and the ratio of specific heats of the gas $\gamma = 2$. Ignore atmospheric pressure.]
Which of the following statement(s) is(are) correct?

Options:

(A) The pressure of the gas at the upper end of the chimney is $300 	ext{ Pa}$.

(B) The velocity of the gas at the lower end of the chimney is $40 	ext{ ms}^{-1}$ and at the upper end is $20 	ext{ ms}^{-1}$.

(C) The height of the chimney is $590 	ext{ m}$.

(D) The density of the gas at the upper end is $0.05 	ext{ kg m}^{-3}$.

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of fluid dynamics and thermodynamics. We are tasked with analyzing an ideal gas flowing steadily through a heat-insulated chimney. To conquer this, we must weave together mass conservation, adiabatic processes, the ideal gas law, and the Steady Flow Energy Equation (SFEE).

### The Inlet Dynamics

Let's start at the bottom of the chimney. We are given a steady mass flow rate $\alpha = 0.8 	ext{ kg/s}$. Since the flow is steady, this mass flow rate must be conserved throughout the chimney. At the lower end, the mass flow rate is the product of density, cross-sectional area, and velocity:

$$\alpha = ho_1 A_1 v_1$$

By rearranging this, we can easily find the inlet velocity $v_1$:

$$v_1 = \frac{\alpha}{ho_1 A_1} = \frac{0.8}{0.2 	imes 0.1} = 40 	ext{ m/s}$$

### The Adiabatic Journey

As the gas rises, it expands. Because the chimney is perfectly heat-insulated, this expansion is strictly **adiabatic**. For an ideal gas undergoing an adiabatic process, the relationship between pressure and temperature is governed by:

$$P^{1-\gamma} T^\gamma = 	ext{constant}$$

We can set up a ratio between the upper and lower ends:

$$\frac{P_2}{P_1} = \left(\frac{T_1}{T_2}ight)^{\frac{\gamma}{1-\gamma}}$$

Given that the ratio of specific heats $\gamma = 2$, the exponent becomes $\frac{2}{1-2} = -2$. Plugging in our known temperatures ($T_1 = 300 	ext{ K}$, $T_2 = 150 	ext{ K}$) and the initial pressure ($P_1 = 600 	ext{ Pa}$):

$$P_2 = 600 	imes \left(\frac{300}{150}ight)^{-2} = 600 	imes (2)^{-2} = \frac{600}{4} = 150 	ext{ Pa}$$

### Finding the New Density

With the new pressure and temperature in hand, we can determine the density at the top using the **Ideal Gas Equation**, $P = \frac{ho R T}{M}$. This tells us that density is directly proportional to pressure and inversely proportional to temperature:

$$\frac{ho_2}{ho_1} = \left(\frac{P_2}{P_1}ight) \left(\frac{T_1}{T_2}ight)$$

Substituting our values:

$$ho_2 = 0.2 	imes \left(\frac{150}{600}ight) 	imes \left(\frac{300}{150}ight) = 0.2 	imes \frac{1}{4} 	imes 2 = 0.1 	ext{ kg/m}^3$$

Now, armed with the top density $ho_2$, we return to our trusty mass conservation principle to find the exit velocity $v_2$:

$$v_2 = \frac{\alpha}{ho_2 A_2} = \frac{0.8}{0.1 	imes 0.4} = 20 	ext{ m/s}$$

### The Master Equation: SFEE

Here is where the magic happens. To find the height of the chimney, we must apply the **Steady Flow Energy Equation (SFEE)**. For an adiabatic flow with no external shaft work, the total energy per unit mass is conserved. This means the sum of specific enthalpy ($h$), kinetic energy, and potential energy remains constant:

$$h_1 + \frac{v_1^2}{2} + g z_1 = h_2 + \frac{v_2^2}{2} + g z_2$$

Rearranging to solve for the height difference $h = z_2 - z_1$:

$$h_1 - h_2 = \frac{v_2^2 - v_1^2}{2} + g h$$

For an ideal gas, the change in specific enthalpy is $h_2 - h_1 = c_p (T_2 - T_1)$. Using the relation $c_p = \frac{\gamma R}{M(\gamma - 1)}$ and the ideal gas law $\frac{P}{ho} = \frac{RT}{M}$, we can express the enthalpy change purely in terms of pressure and density:

$$h_2 - h_1 = \frac{\gamma}{\gamma - 1} \left( \frac{P_2}{ho_2} - \frac{P_1}{ho_1} ight)$$

Since $\gamma = 2$, the coefficient $\frac{\gamma}{\gamma - 1}$ simplifies beautifully to $2$. Therefore:

$$h_1 - h_2 = 2 \left( \frac{P_1}{ho_1} - \frac{P_2}{ho_2} ight)$$

### The Final Calculation

Let's substitute this back into our energy equation:

$$2 \left( \frac{P_1}{ho_1} - \frac{P_2}{ho_2} ight) = \frac{v_2^2 - v_1^2}{2} + g h$$

Now, we carefully plug in every value we've painstakingly calculated:

$$2 \left( \frac{600}{0.2} ight) - 2 \left( \frac{150}{0.1} ight) = \frac{20^2 - 40^2}{2} + 10 h$$

$$2(3000) - 2(1500) = \frac{400 - 1600}{2} + 10 h$$

$$6000 - 3000 = -600 + 10 h$$

$$3000 = -600 + 10 h \implies 10 h = 3600 \implies h = 360 	ext{ m}$$

Reviewing the options, we see that only the velocities match our derived results. The correct statement is indeed **(B)**.

Select Answer:

Visualized Solution

$A na l y z in g t h e C himn ey F l o w$

$M a ss C o n ser v a t i o n a t I n l e t$

$A d iaba t i c E x p an s i o n$

$I d e a l G a s E q u a t i o n$

$M a ss C o n ser v a t i o n a t O u tl e t$

$S t e a d y F l o w E n er g y E q u a t i o n$

$E n t ha l p y o f I d e a l G a s$

$S o l v in g f or H e i g h t h$

$E v a l u a t in g t h e O pt i o n s$

The Sigma Insight: First Law of Thermodynamics

The Inlet Dynamics

The Adiabatic Journey

Finding the New Density

The Master Equation

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Chimney Flow

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Conservation at Inlet

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adiabatic Expansion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Conservation at Outlet

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steady Flow Energy Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Enthalpy of Ideal Gas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Height h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Options