Water Hammer transient analysis (simplified) - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Software Libraries (/forum-10.html) +--- Forum: HP-41C Software Library (/forum-11.html) +--- Thread: Water Hammer transient analysis (simplified) (/thread-7166.html) Water Hammer transient analysis (simplified) - Ángel Martin - 11-04-2016 08:23 AM Water Hammer Transients. [ ARIETE ] From the author’s Engineering Collection, included in the ETSII4 module (ETI4 on the CL Library) This program calculates a simplified transient analysis of the pressure wave inside of a forced pipe, when the exit valve follows a given closing curve. The program does not take into account the friction so the pressure waves would continue to transient indefinitely. Let H0 be the stationary head pressure inside of the pipe, let “a” be the speed at which the pressure wave propagates, L the pipe length, and v(t) the valve closing speed equation. In ideal conditions (no friction dampening) the instant pressure change at the analysis point placed at a distance d from the valve (exit), is modeled by the following expression (eq. 1): H(t) = H0 + a/g SUM { (-1)^int(k/2) [ v0 – v(t”)] }; k=1,2..n where t“ is an auxiliary time defined as: t“ = t – t0 int(k/2) + (-1^k) td and: n = 2 int[(t+td)/t0] iif t< tc = td + t0 int[(t+td)/t0] , or: n = n-1 iif t>= tc t0 = 2L/a is the time it takes for the pressure wave to return to the valve for the first time; v0 = v(t0) is the initial speed before the pressure waves affect the analysis point. td = d/a is the time it takes for the pressure wave to reach the analysis point for the first time; The valve closing speed equation is supposed to be relatively slow – as referenced to the characteristic parameters t0 and td. The equation needs to be entered as a separate program using a global label. It can also have discontinuities, used to simulate sudden closing steps. This is handled by providing the left-value in the Y register, and the right-value in the X register. Because of this, two pressure results per time instant will be calculated, one before (H- in the Y register) and another after (H+ in the Z register) the instant t (in the X register). The user must provide a step size value to calculate the successive pressure values at the point of analysis. These can be plotted on the thermal printer using the PRPLOT program as well. The program uses the OS/X function PMTA to prompt for the closing equation label name. Note: Equation (1) above was deduced from a conceptual approach, extending the simple cases to a general-purpose scheme valid for any time instant. To the author’s knowledge there is no reference in the literature for an equivalent expression. The closer I’ve seen is the Bergeron Method, used to calculate reflection of electrical signals. Example. Study the pressure transients at the closing valve (thus d=0) during a time interval of 10 seconds, in a forced pipe with 1,200 m length. Assume that the pressure wave moves at 800 m/s (hence td = 2L/a = 2,400/800 = 3 s). For the gravity acceleration use g = 10 m/s^2 to simplify the numeric results. The closing valve equation is given as follows: v(t) = 1.2 m/s if 0 <= t < 1 v(t) = 0.8 m/s if 1 <= t < 3 v(t) = 0.4 m/s if 3 <= t < 4 v(t) = 0 if t >= 4 Which has been programmed under the global label “V” using the “from right to left” rule, as follows: (for your convenience, the ETSII4 module includes pre-programmed this valve closing equation, and another one that you can also use to test the operation - under the labels “V” and “V(T)”) Code: ``` 01    LBL “V” 02    4 03    X<>Y 04    X>Y? 05    GTO 06 06    X=Y? 07    GTO 05 08    3 09    X<>Y 10    X>Y? 11    GTO 04 12    X=Y? 13    GTO 03 14    1 15    X<>Y 16    X>Y? 17    GTO 02 18    X=Y? 19    GTO 01 20    1.2 21    RCL X 22    RTN 23    LBL 01 24    1.2 25    0,8 26    RTN 27    LBL 02 28    ,8 29    RCL X 30    RTN 31    LBL 03 32    0,8 33    0,4 34    RTN 35    LBL 04 36    0,4 37    RCL X 38    RTN 39    LBL 05 40    0,4 41    0 42    RTN 43    LBL 06 44    0 45    RCL X 46    END[/size]``` Using a step size of 0.25 s the results for the pressure changes are shown in the table below, with the critical instants t=1, 3, 4, 6, and 9; with the maximum pressure change occurring at t=6. Code: ``` t       H(-)  H(+)    t     H(-)  H(+)    t     H(-) H(+)      t     H(-)  H(+)    t     H(-)  H(+) ------------------------------------------------------------------------------------------------------- 0        0     0      2     32    32      4     64   32        6      32   -32     8     -32   -32 0.25     0     0      2.25  32    32      4.25  32   32        6.25  -32   -32     8.25  -32   -32 0.50     0     0      2.5   32    32      4.5   32   32        6.5   -32   -32     8.5   -32   -32 0.75     0     0      2.75  32    32      4.75  32   32        6.75  -32   -32     8.75  -32   -32 1        0    32      3     32    64      5     32   32        7     -32   -32     9     -32    32 1.25    32    32      3.25  64    64      5.25  32   32        7.25  -32   -32     9.25   32    32 1.5     32    32      3.5   64    64      5.5   32   32        7.5   -32   -32     9.5    32    32 1.75    32    32      3.75  64    64      5.75  32   32        7.75  -32   -32     9.75   32    32```