Welcome to a thrilling journey through one of the most debated fluid dynamics problems from JEE Advanced! This comprehension passage presents a fascinating scenario involving a cylindrical furnace and a tall chimney. We are tasked with finding the mass flow rate of the exiting air and the pressure difference if the chimney is capped.
Let's dive deep into the physics, decode the examiner's intended logic, and uncover the hidden paradox that ultimately caused this question to be dropped from the official evaluation!
Analyzing the Setup
Imagine a robust cylindrical furnace with a height H=1 m and a diameter D=1 m
Sitting right on top of it is a vertical chimney with a height h=9 m and a much narrower diameter d=0.1 m.
Cold atmospheric air, with a density ρa=1.2 kg/m3 and temperature Ta=300 K, is drawn into the furnace. Inside, it gets heated to a scorching T=360 K. Because hot air is less dense, it becomes buoyant and rushes up the chimney, creating a steady streamline flow.
The Density of Hot Air
Before we can analyze the flow, we must determine the density of the hot air inside the furnace
The problem explicitly states that the air is heated at a constant pressure Pa.
Using the Ideal Gas Law for a constant pressure process, we know that density is inversely proportional to temperature:
ρaTa=ρT
Substituting our known values:
1.2×300=ρ×360
ρ=1.0 kg/m3
This density difference between the cold outside air (1.2 kg/m3) and the hot inside air (1.0 kg/m3) is the engine driving the entire flow!
The "Constant Pressure" Assumption
Here is where we must carefully interpret the examiner's intent
The problem states that the air is heated "at constant pressure Pa". The intended mathematical translation of this phrase is that the pressure throughout the entire volume of the furnace is exactly Pa.
Therefore, at the very top of the furnace (which is the entrance to the chimney), the pressure is Pa.
Meanwhile, what is the pressure at the top of the chimney, outside in the atmosphere? The atmospheric pressure drops as we go higher. At a height of
H+h from the ground, the outside pressure is:
PC=Pa−ρag(H+h)
Bernoulli's Equation in Action
Now, let's apply Bernoulli's equation along a streamline starting from the top of the furnace and ending at the top of the chimney.
Pfurnace_top+21ρV02+ρgH=Pchimney_top+21ρV2+ρg(H+h)
Substituting our pressure values:
Pa+21ρV02+ρgH=(Pa−ρag(H+h))+21ρV2+ρg(H+h)
By the equation of continuity, the velocity V0 at the top of the wide furnace is related to the velocity V in the narrow chimney by the ratio of their areas. Since the diameter ratio is 10:1, the area ratio is 100:1. Thus, V0=0.01V. Squaring this makes V02 incredibly small, allowing us to safely approximate it as zero.
Simplifying our Bernoulli equation:
21ρV2=ρag(H+h)−ρgh
Let's plug in the numbers:
21(1)V2=1.2×10×10−1×10×9
21V2=120−90=30
Calculating the Mass Flow Rate
With the velocity in hand, calculating the steady mass flow rate (m˙) is a breeze.
m˙=ρAV=ρ(4πd2)V
m˙=0.00785×7.746≈0.0608 kg/s
Converting this to grams per second, we get 60.8 g/s. This is the intended answer for the first question!
The Closed Chimney Scenario
For the second question, imagine we place a tight cap at the top of the chimney
The air flow completely halts, meaning V=0. We are now dealing with pure fluid statics.
We need to find the pressure difference ΔP across the cap.
Inside the cap: The pressure drops hydrostatically from the top of the furnace (
Pa) up through the chimney (height
h) filled with hot air (
ρ).
Pin=Pa−ρgh
Outside the cap: The pressure drops hydrostatically from the ground (
Pa) up through the atmosphere (height
H+h) filled with cold air (
ρa).
Pout=Pa−ρag(H+h)
The pressure difference is simply:
ΔP=Pin−Pout
ΔP=(Pa−ρgh)−(Pa−ρag(H+h))
ΔP=ρag(H+h)−ρgh
Plugging in the numbers:
ΔP=1.2×10×10−1×10×9=120−90=30 N/m2
This gives us the intended answer of 30 N/m2 for the second question.
The Hydrostatic Paradox
Why JEE Dropped This Question
If you followed the logic above, you arrived at the exact answers the examiner intended. However, this question was officially dropped by JEE Advanced! Why?
The fatal flaw lies in the assumption that the pressure inside the 1-meter tall furnace is uniformly Pa. In reality, gravity acts on the hot air inside the furnace just as it does everywhere else. According to strict fluid statics, the pressure at the top of the furnace should actually be Pa−ρgH.
If we use this physically rigorous hydrostatic pressure drop inside the furnace, the calculations change drastically. The velocity squared becomes 40 instead of 60, and the static pressure difference becomes 20 instead of 30. Because the problem's wording forced an unphysical assumption, it created an unsolvable paradox for top-tier physics students, leading to its removal.
Always trust your core physics principles—even when the question itself bends the rules!
