Slay the P.E. 52 Report post Posted October 10 On 10/8/2018 at 6:12 PM, MikeGlass1969 said: Slay, I tried this... However I converted the straight pipe, for each branch, into a K and summed all the K's for each branch. Plugged into the Bernoulli equation. And Came up with something in the area of (B). But, in the TFS the answer was reported as (D). In the solution posted in the other forum, added the head losses from friction in the energy equation. Why? Shouldn't they be subtracted? I found a mistake, got 93ft which is close to (D) Mike, the "extended" Bernoulli equation is actually a statement of conservation of energy. as such, it is essentially this (for steady state): The rate at which energy enters a control volume = The rate at which energy leaves the control volume In this problem a control volume is defined with boundaries just below the water level in the reservoirs. It is at these locations where mass enters/leaves the control volume. The left hand side of the energy balance will have the enthalpy (pressure head, which is zero), kinetic energy (zero), and potential energy (height) entering the control volume at the reservoir in the right side of the drawing. That is all we have on the left side of the equation. (If we had a pump, the pump head would be on this side, because it is energy entering the control volume) The right hand side of the energy balance will have the enthalpy (pressure head, which is zero), kinetic energy (zero), and potential energy (height) leaving the control volume at the reservoir in the left side of the drawing. The right hand side of the equation will also contain the energy that is "lost" (stray heat is an enthalpy, which in hydraulics ends up being pressure loss) so the friction and minor losses are forms of energy leaving the control volume. These terms therefore, belong on the right side of the equation. A much shorter way to say all this is that in the extended Bernoulli equation, the left side of the equation (subscript 1) is the inlet reservoir, and the right side (subscript 2) is the discharge reservoir. Pump head (if non-zero) goes in the left side of the equation. Turbine head (if non-zero) goes in the right side of the equation. Friction and minor losses go in the right side of the equation. Share this post Link to post Share on other sites

Slay the P.E. 52 Report post Posted October 10 (edited) @mcc515 @MikeGlass1969 Thanks for working on the double pane window problem. Here is my solution. I get -4F which is different from what both of you are getting. Let me know if you see anything wrong Edited October 11 by Slay the P.E. Share this post Link to post Share on other sites

mcc515 0 Report post Posted October 11 @Slay the P.E. I used the same method, but with a Tdp of 70F instead of 70.2F to get -7. If I sub 70.2 in for Tsurface, I also end up with -3.82F Rtotal = 0.2+3+(1/12) = 3.28 Rair= 0.2 Ts=70F (Tin-Toa)/Rtotal = (Tin-Ts)/Rair Toa = 75F - (3.28/.2)*(75F-70F) Toa = -7F Share this post Link to post Share on other sites

Slay the P.E. 52 Report post Posted October 11 2 hours ago, mcc515 said: @Slay the P.E. I used the same method, but with a Tdp of 70F instead of 70.2F to get -7. If I sub 70.2 in for Tsurface, I also end up with -3.82F Rtotal = 0.2+3+(1/12) = 3.28 Rair= 0.2 Ts=70F (Tin-Toa)/Rtotal = (Tin-Ts)/Rair Toa = 75F - (3.28/.2)*(75F-70F) Toa = -7F Thanks. I’m going to have to re-write some of the given data in that problem so the final answer isn’t so sensitive to small variations of Tdp. I’ll make sure that using 70F or 70.2F yields roughly the same answer. Share this post Link to post Share on other sites

MikeGlass1969 30 Report post Posted October 11 2 hours ago, Slay the P.E. said: Thanks. I’m going to have to re-write some of the given data in that problem so the final answer isn’t so sensitive to small variations of Tdp. I’ll make sure that using 70F or 70.2F yields roughly the same answer. I used 71F for the Tdp... I got 9.3F. But when I substitute in 70.2F I get -3.8F Pretty large swing. All answers were possible in this problem Share this post Link to post Share on other sites

Slay the P.E. 52 Report post Posted October 11 1 minute ago, MikeGlass1969 said: I used 71F for the Tdp... I got 9.3F. But when I substitute in 70.2F I get -3.8F Pretty large swing. All answers were possible in this problem Yep. Its badly crafted as is. The factor (3.28/.2) or (.1368/.-0083) is too big and makes the final answer too sensitive to Tdp. I need to change the relative magnitude of the given R-values to improve the problem. I think the problem itself is still pretty cool... just needs better numerical data. We shouldn't get a -7F to 9.3F swing by changing Tdp from 70F to 71F Share this post Link to post Share on other sites