1. (a) A patch is to be placed in the fire sheet of a horizontal return tubular boiler having a shell plate seven-sixteenths-inch (7/16″) thick, a longitudinal seam efficiency of seventy-four percent (74%), and a length of patch of thirty-six inches (36″). Find the width, W, of patch to be applied so that there will be no reduction in pressure, using a single-riveted seam of normal design.

   Referring to Table 1, it is found that seven-sixteenths-inch (7/16″) plate with fifteen-sixteenths-inch (15/16″) diameter rivet holes, pitched two and one-fourth (2 ¼ ) gives a seam efficiency of fifty-six percent (56%).

   Referring to Table 2, E = 0.74 and e = 0.56 gives a constant C = 1.75; then width W = L x C = 36 x 1.75 = sixty-three inches (63″).

2. (b) Pressure Allowed on an Existing Patch.

   A crescent-shaped patch has already been installed on a horizontal tubular boiler. It is thirty inches (30″) long and forty-eight inches (48″) wide. The seam is single riveted with thirteen-sixteenths-inch (13/16″) rivet holes pitched one and fifteen-sixteenths inches (1 15/16″). The boiler shell plate is three-eighths-inch (3/8″) thick. The longitudinal seam is of the double-riveted butt-strap type having an efficiency of eighty-two percent (82%). The safety valve is set for one hundred twenty-five pounds (125#) pressure. What maximum pressure should be allowed on the boiler? Reference to Table 1, shows that the normal efficiency of the patch seam is fifty-seven percent (57%). (If the efficiency is not found in the Table, refer to any other available table or determine it in the manner described in paragraphs A-1, A-2 and A-3 of the Appendix of the A.S.M.E. Boiler Code.)

   Divide the width of the patch, W = forty-eight inches (48″) by the length, L = thirty inches (30″) to find the constant, C = 48/30 = 1.60. Follow down column 0.57 of Table 2 until 1.60 is found. It will be noted that this is somewhere between 1.56 and 1.62 representing E somewhere between 0.72 and 0.73. As the difference between 1.56 and 1.62 is 6, and the difference between 1.56 and 1.60 is 4, E will be 0.72 plus ⁴/₆ of 0.01 which is 0.7266.

   The maximum allowable working pressure varies directly as the seam efficiency. Accordingly, P = (0.7266/0.82) x 125 = 110#. If this pressure allowance interferes with the operation of the plant, the patch shall be replaced by a new one with the proper dimensions giving a diagonal efficiency of eighty-two percent (82%).

3. (c) Design of Patch for Water-tube Boiler.

   Reference to Table 1 shows that a single-riveted lap seam with seven-sixteenths-inch (7/16″) plate, fifteen-sixteenths-inch (15/16″) diameter rivet holes, and two-and-one-half-inch (2 1/2″) pitch has a normal efficiency of fifty-six percent (56%).

   Reference to Table 3, shows E = 0.82 and e = 0.56, the constant C is 3.16.

   Then width W = C x 1 / 2.

   W = 3.16 x 36 / 2 = 56.88 or practically fifty-seven inches (57″).

   

   

   

   

History

1. Repealer and new subsections (a) and (b) filed 8-12-76; effective thirtieth day thereafter (Register 76, No. 33).

2. Change without regulatory effect repealing and adopting new Figures 1-6 within subsection (b) filed 5-19-2008 pursuant to section 100, title 1, California Code of Regulations (Register 2008, No. 21).