A vertical circular stack 100 ft high converges uniformly from a diameter of 20 ft at the bottom to 16 ft at the top. Coal gas with a unit weight of 0.030 pcf enters the bottom of the stack with a velocity of 10 fps. The unit weight of the gas increases uniformly to 0.042 pcf at the top. What is the mean velocity 25 ft above the bottom of the stack?
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What have you tried? What is the diameter 25 ft up? What is the density 25 feet up? – Ross Millikan Oct 03 '14 at 03:59
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i tried using continuity of flow, the diameter of the stack 25 ft up is 19 ft. but i'm getting the wrong answer – Edison Ulanday Oct 03 '14 at 04:07
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will the increasing density affect the velocity? – Edison Ulanday Oct 03 '14 at 04:07
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It is hard to help starting from "I'm getting the wrong answer" as we don't see your answer or how you found it. If you show your work, we can probably find the problem. Yes, the diameter is 19 feet. – Ross Millikan Oct 03 '14 at 04:09
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i'm trying to equate the discharge and getting the equation (d1/d2)^2*v1 = v2, the v2 i'm getting is 11.08 fps – Edison Ulanday Oct 03 '14 at 04:14
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the choices given to this question were 10.07, 10.17, 10.12, 10.22 fps – Edison Ulanday Oct 03 '14 at 04:17
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Yes, the increasing density will affect the velocity. What is conserved is mass flow. Your equation ignores the density variation-it assumes that volume flow is constant. You need a factor of the density ratio. – Ross Millikan Oct 03 '14 at 04:20
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how should my equation be? – Edison Ulanday Oct 03 '14 at 04:24
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You need to convert volumes to masses, so need a factor of the ratio of the densities at 0 feet and 25 feet. I'll leave it to you to figure out which is on top. – Ross Millikan Oct 03 '14 at 04:29
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still don't know what to do, sorry – Edison Ulanday Oct 03 '14 at 05:05
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@EdisonUlanday, I have provided the full solution and is in the lines of what Ross was suggesting. Hopefully this answers your question. Good Luck. – Satish Ramanathan Oct 03 '14 at 12:20
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@RossMillikan, Out of curiosity, are you by any chance trained on engineering or sciences besides mathematics, how could you solve problems like this with such clarity? – Satish Ramanathan Oct 03 '14 at 13:36
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Yes, my college education was in physics. I took a lot of math as well. – Ross Millikan Oct 03 '14 at 13:55
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No wonder!!. hoping to learn from you as well. Thanks – Satish Ramanathan Oct 03 '14 at 14:48
1 Answers
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Converting velocity in to volumetric flow rate
At the bottom of stack:$$v_1 = \frac{Q_1}{A_1}$$
From this find $Q_1$
Thus $$Q_1 = 10\times \pi\times 10^2 = 1000\pi$$
Now since the unit weight changes, equate the mass flow rate such as below:
$$\rho_1\times Q_1 = \rho_2\times Q_2$$
Unit weight changes uniformly down the stack at the rate of $\frac{(.042-.030)}{100} = 0.00012$
At the stack height of 25 ft , the unit weight $= .03+0.00012*25 = 0.033$
Thus $\rho_2 = 0.033$. Apply this in the equation above to get $Q_2$
$$Q_2 = \frac{0.03*1000\pi}{0.033} = 909.09\pi$$
Now $$V_2 = \frac{Q_2}{A_2} = \frac{909.09\pi}{9.5^2\pi}$$ If you solve this you get $v_2 = 10.07$ fps
Satish Ramanathan
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