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                                                                                      5.8 Materials of Construction  183


                     EXAMPLE 5.13  FREEZING INDEX AND FROST DEPTH REQUIREMENT IN NORTH AMERICA
                                         In the absence of daily readings, the value of freezing index F may be approximated by equa-
                                         tions 5.12a and 5.12b, using the U.S. customary units and the SI units, respectively. Determine
                                         the freezing index of a northern state in the U.S. assuming the coldest monthly temperatures
                                         recorded are:


                                                                            T m ,  F  T m ,  C
                                                                December      23       5
                                                                January       26       3.33
                                                                February      24       4.44
                                                                March         31       0.55


                                          Solution 1 (U.S. Customary System):
                                                          ©T m = 23 + 26 + 24 + 31 = 104
                                                             F = (32n -©T m ) * 30.2
                                                              = (32 * 4 - 104) * 30.2
                                                              = 724.8
                                                              d = 1.65F 0.468
                                                              = 1.65(724.8) 0.468
                                                              = 35.98 in. 7 12 in. Select 36 in. of fill material

                                          Solution 2 (SI System):
                                                         ©T m =-5 - 3.33 - 4.44 - 0.55 =-13.32
                                                            F = [0 -©T m ] * 30.2
                                                             = [0 - (-13.32)] * 30.2
                                                             = 402.26
                                                            d = 55.18F 0.468
                                                             = 55.18(402.26) 0.468
                                                             = 913.46 mm 7 305 mm. Select 1 m of fill material





                    5.8  MATERIALS OF CONSTRUCTION

                                         Selection of pipeline materials is based on carrying capacity, strength, life or durability,
                                         ease of transportation, handling and laying, safety, availability, cost in place, and cost of
                                         maintenance. Various types of iron, steel, reinforced concrete, and fiberglass are most used
                                         for water transmission pipes, but plastic pipes are now being made in the smaller sizes.
                                         Other materials may come into use in the future.


                    5.8.1  Carrying Capacity
                                         The initial value of the Hazen-Williams coefficient C hovers around 140 for all types of
                                         well-laid pipelines but tends to be somewhat higher for reinforced-concrete and fiberglass
                                         lines and to drop to a normal value of about 130 for unlined cast-iron pipe. Cast-iron and
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