PLEASE HELP WITH THE CALCLULATIONS NOT THE CONCEPT!

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Assigned ACSR Conductor: DOVE

DOVE Weight $=765\mathrm{.}2$ lb$/$Kft

DOVE CDR $=0\mathrm{.}927$

Conditions HEAVY $=$ Ice Thickness $=0\mathrm{.}5$ inch

Wind Pressure $=4$ lb$/$ft

Temp $=0$ degrees

K $=0\mathrm{.}3$ lb$/$ft

Span $($S$)=700$ ft

Ice loading $=0\mathrm{.}99$ lb$/$foot

Wind loading $=0\mathrm{.}7$ lb$/$ft

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Find the maximum sag, tension, and length of the conductor of ACSR conductor $($your assigned conductor$)$ to be used in heavy loading districts.

Note $1$: You will use the following specifications except the conductor$-$specific data that need to be obtained from the data sheet provided for your assigned conductor.

Specs:

Span $($S$)=700$ ft

Ice loading $=0\mathrm{.}99$ lb$/$foot

Wind loading $=0\mathrm{.}7$ lb$/$ft

K $=0\mathrm{.}3$ lb$/$ft

Note $2$: Find the conductor weight from the datasheet and compute the total weight with ice and wind loading $($see example in slide$-$set #$11).$

Weight, with ice and wind loading $($wT$)=-----$ lb$/$ft

Horizontal tension $($rated$,$ initial, Hi$)=6300$ lbs

Cross section $($total$,$ A$)=--------$ in$2$

$($Calculate from overall diameter of the conductor given in the datasheet$).$

Breaking strength $=-------$ lb

Elastic modulus of the conductor $($Aluminum & Steel, combined$)=11\mathrm{.}2$ x $106$ psi

Coefficient of thermal expansion $($Aluminum & Steel, combined$)=10\mathrm{.}6$ x $10-6$ per oF

Note: Once you have all the data for your conductor follow the step$-$by$-$step process outlined in this sample to complete the problem.

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Task: String the conductor applying the tension $($H$)$ and compute resulting sag $($D$)$ and the length $($L$),$ which will be used to determine the tower height to maintain the minimum ground clearance needed.

The sag $($D$)$ is computed for the following three scenarios:

$1.$ Snow and wind loading

$2.$ Maximum temperature

$3.10-$year creep

The worst sag $($longest$)$ is adapted. In this assignment you will do snow and wind loading only.

Problem: A conductor under tension would elongate, increasing the length of the conductor. Increase in L results in increase in D$,$ which in turn reduces the tension H$.$

Fix: Find a combination of L$,$ D $,$ and H$,$ which would result in minimum change in tension H before and after stringing the conductor using an iterative technique.

Initialization:

$1.$ Compute D using the given H and S

$2.$ Compute L using D in $(1)$ and the given S

$3.$ Compute ZTL at the normal temperature $(60$ degrees$)$

$4.$ Compute ZTL at the maximum temperature $(167$ degrees$),$ call it LHREF at

$5.$ Compute D using ZTL in $(4)$ first and then use that D to compute Hi $..$

Iterations:

Given the initial Hi compute the Li

$1.$ Compute Di for Li in $(1)$

$2.$ Compute Hf for Li and Di from $(1)$ and $(2)$

$3.$ If the difference between Hi and Hf is minimal $(0\mathrm{.}001\%$ or less than $1$ lb$),$ stop. Else, repeat with a new initial Hi$.$

$4.$ New Hi could be the average of Hi and Hf$.$

We need to do sag$-$tension calculations for the conditions given below.

$1.$ Snow and wind loading

$($This sample uses w $=2\mathrm{.}507$ lb$/$ft$,$ you need to calculate that for your assigned conductor. See the example of doing calculation in slides set $11.)$

ZTL calculation

$1.$ Compute D for a specific tension $(6300$ lb$)$ HREF.

$2.$ Compute LREF for D in $(1)$

$3.$ Use LREF and HREF in ZTL calculation.

$4.$ D $=($w$*$S$^2)/(8$HREF$)=24\mathrm{.}3736$

$5.$ LREF $=$ S $+((8$D$^2)/(3$S$))=702\mathrm{.}263$

$6.$ ZTL$(60$ degree$)=$ LH $=$ LHREF$(1+(($H$-$Href$)/($EcA$))=701\mathrm{.}719$

$7.$ ZTL$(0$ degree$)=$ LT $=$ LTREF $(1+$ alpha$($AS$)$ x $($T $$ TREF$))=701\mathrm{.}719(1+10\mathrm{.}6$ x $10^-6(0-60))=701\mathrm{.}273$ ft

initial tension calculation

$1.$ Compute D using ZTL $(0$ degree$)$

$2.$ Compute Hi using D in $(1)$

$3.$ D $=$ SQRT$((3$S$($ZTL $$ S$))/8)=$ SQRT $((3$x$700(701\mathrm{.}273700)/8)=18\mathrm{.}28$

$4.$ Hi $=($w$*$S$^2)/8$D $=(2\mathrm{.}507$ x $(700)^2)/(8$ x $18\mathrm{.}28)=8400$

Iterative Technique:

$1.$ Using the tension, Hi$,$ compute L$,$ where HREF $=0$ and LREF$=$ ZTL $(0$ degree$)$

$2.$ Compute D for L in $(1)$

$3.$ L $=$ LREF $(1+(($Hi $$ HREF$)/($EcA$)))$

$4.$ D $=$ SQRT$((3$S$($L$-$S$))/8)$

Iterative Technique $($continued$)$:

$5.$ Compute Hf for L and D from $(1)$ and $(2)$

$6.$ If the difference between Hi and Hf is minimal, stop. Else, repeat with a new initial Hi$.$

$7.$ New Hi could be the average of Hi and Hf$.$

$8.$ Hf $=($w $*$ S$^2)/(8$D$)$