You are designing a horizontal pathway system where you will provide 100 mm (4 in) conduit and pull boxes. The building architectural elements will only allow for a single layer of conduits (not stacked). The main horizontal pathway system consists of five 100 mm (4 in) conduits. Large single sheath copper cables are not expected to be placed. Which of the following is the MINIMUM width pull box that is to be provided?
The minimum width of a pull box required for a main horizontal pathway system consisting of five 100 mm (4 in) conduits can be determined based on the conduit size and configuration. For a single layer configuration, the width of the pull box is calculated as follows: The initial width for a single conduit is generally around 15 inches (380 mm), and one needs to add 8 inches (200 mm) for each additional conduit. Thus, for four additional conduits (5 - 1), the calculation would be 15 inches + (4 * 8 inches) = 15 inches + 32 inches = 47 inches, which converts approximately to 1200 mm. Therefore, the closest option provided that exceeds this requirement is 1220 mm (48 in), making it the appropriate choice.
TDMM page 5-68 (page 464) table 5.8 Pull Box would be 15in width for (1) 4in conduit. The box width increases 8in for each additional conduit. * 15in + 8in x 4 = 15in + 32in = 47in
"C" is still correct answer, but the math is a bit off. Remember, the length of the pull box is already 15in for one conduit. So you only need to multiply 8in three times to account for the three other conduits. 15in + (8in x 3) = 39in 39in is greater than 24in and 36in, so "A" and "B" are incorrect. 39in is smaller than 60, so "C" is the answer.
confused
which publication are you using for the answers?
why c? The minimum length of a pull box shall be at least 16 times the diameter of the largest conduit entering the pull box, I believe the answer is 4x16=64 C: is only 60 D: 72 - greater than 64
Because they asking for the width not the length. Also because the on the table on pg 5-83 is says the box width increased 8 in for each additional conduit. So: =Box width 15 in + (4 conduits * 8in) =15in +32in =47in