PPI Six minute HVAC - #30

[sayed]

i'm not understanding the use of these equations:

41.26 - 41.30

since the answer doesn't reference an equation, i am guessing they are using 41.27 with the assumption that dynamic head, h, is p/gamma

my problem is, why aren't these equations equivalent? using 41.30 seems like a good choice too, but get different answer. the book doesn't explain these equations much

 

[MikeGlass1969]

Is this must be a fluids problem..  and can you state or post them problem?

 

[Slay the P.E.]

i'm not understanding the use of these equations:

41.26 - 41.30

since the answer doesn't reference an equation, i am guessing they are using 41.27 with the assumption that dynamic head, h, is p/gamma

my problem is, why aren't these equations equivalent? using 41.30 seems like a good choice too, but get different answer. the book doesn't explain these equations much

HVAC SMS Problem #30 (I have 2nd edition, 7th printing) is a hydraulics problem solved by using the Bernoulli equation, but the equations you are referring to (41.26 - 41.30) in MERM 13th edition are fan affinity laws. Therefore, I am most confused by your post.

 

[sayed]

HVAC SMS Problem #30 (I have 2nd edition, 7th printing) is a hydraulics problem solved by using the Bernoulli equation, but the equations you are referring to (41.26 - 41.30) in MERM 13th edition are fan affinity laws. Therefore, I am most confused by your post.

View attachment 10220View attachment 10221

not the right problem.

the problem you posted seems to be a good problem. can you post the answer to see if i got it right (mostly number crunching i think)View attachment 10222

 

[sayed]

View attachment 10223f

I just want confirmation. The solution has the frequency wrong by 2*pi, correct?

 

[Slay the P.E.]

The problem you posted asks for an impeller diameter and the solution you posted obtains a force transmitted to a structure. This is not the solution to the problem you posted.

 

[sayed]

The problem you posted asks for an impeller diameter and the solution you posted obtains a force transmitted to a structure. This is not the solution to the problem you posted.

i should have mentioned that the solution i posted is a different problem. I just wanted confirmation that that frequency, f, was wrong.

for question 30, i want to know why using equation 41.27 gives a different result than 41.30 (QA = QB and nA = nB). Or better yet, why choose one over the other?

MERM doesn't explain anything about the usage for each of these equations.

 

[sayed]

View attachment 10223f

I just want confirmation. The solution has the frequency wrong by 2*pi, correct?

oh jeez. i shouldn't be working problems so late at night. simple goof on my end

 

[MikeGlass1969]

Frequency is in Hz..  Cycles per sec or revolutions per second.   Angular frequency is in radians per sec.  So angular frequency converts to 2*pi radians = 1 revolution.

 

[Audi Driver P.E.]

By far the most confusing post I have seen.

 

[sayed]

Frequency is in Hz..  Cycles per sec or revolutions per second.   Angular frequency is in radians per sec.  So angular frequency converts to 2*pi radians = 1 revolution.

yeah, i must have been too sleepy to get that one confused.

The issue that was never resolved was the original question. When do we obtain different answers when picking different equations 41.26-41.30 ?

they are all similarity equations. when do we choose one over the other?

 

[MikeGlass1969]

not the right problem.

the problem you posted seems to be a good problem. can you post the answer to see if i got it right (mostly number crunching i think)View attachment 10222

This is an Affinity Law Problem...  See Equations 18.46 in MERM

 

[Slay the P.E.]

The issue that was never resolved was the original question. When do we obtain different answers when picking different equations 41.26-41.30 ?

they are all similarity equations. when do we choose one over the other?

This is a bad problem and you should just ignore it. The way they have written it makes it impossible to solve, and their solution is wrong. Here's why:

Pump affinity laws essentially take one point (Q1, h1) on a pump curve (for a given diameter, D1) and assign to it one and only one point (Q2, h2) on a different pump curve (which corresponds to a different diameter D2). If you do this for enough points on the original curve, you can get enough points on the new curve that you can draw it. Note that since Q2 =( D2 /D1 )Q1 and h2 =(( D2 /D1 )^2)xh1 the flow rate and head between corresponding points are never the same. In graphic form (see below), pump affinity laws take each blue point in the original curve, and assign to it a red point on the new curve. Applying pump affinity laws is like using the green arrows in this graph:

​

The issue with the problem you've posted is that you are not given the whole pump curve for the original diameter. You are only given a single point of the original curve. The problem is not solvable, and as you've noted, there are issues with it. In simpler (and easier to understand) terms, the problem you posted can be re-written as:

“A pump curve for a known diameter passes through the blue point in the graph below. Find the impeller diameter that will make the curve pass through the red point”. Hint: Can't be done.

​

You can use h2 =(( D2 /D1 )^2)xh1 to solve for D2 with h2 = 115 ft and h1 = 140 ft. This gives you D2 = 11.3 in. However, this does not correspond to the red point. The flow rate would have to change. It will be Q2 =( 11.3 / 12.5 ) x 380 = 344 gpm. In graphical form (see below), applying affinity laws takes the blue point, along the green arrow and gives you the green point.

.​

From the graph above, its clear that trimming down the impeller to 11.3 inches will not make the pump add a head of 115 ft for a flow rate of 380 gpm. Their solution is wrong. The problem as presented cannot be solved.

As always, I could be mistaken and welcome any critique of this write-up.