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Chapter 11

Q. 11.6

Determine the reinforcement requirements of an 8 in. ID nozzle that is centrally located in a 2:1 ellipsoidal head. The inside diameter of the head skirt is 41.75 in. The allowable stress of both the head and nozzle material is 17.5 ksi. The design pressure is 700 psi and the design temperature is 500°F. There is no corrosion and the weld joint efficiency is E = 1.0. See Fig. 11.13.1 for details of a nozzle.

Determine the reinforcement requirements of an 8 in. ID nozzle that is centrally located in a 2:1 ellipsoidal head. The inside diameter of the head skirt is 41.75 in. The allowable stress of both the head and nozzle material is 17.5 ksi. The design pressure is 700 psi and the design temperature is

Step-by-Step

Verified Solution

1 . The minimum required thickness of a 2 : 1 ellipsoidal head without an opening is determined from UG-32(d) of the ASME Code, VIII-1 as

t_{r}=\frac{P D}{2 S E-0.2 P}=\frac{700(41.75)}{2(17,500 \times 1.0)-0.2(700)}=0.838 \text { in. };

use 1.0 in.

2 . As noted in the definition of t_{r} to use with Eq. 11.20 and shown in Fig. 11.13.1, when an opening and its reinforcement are located in an ellipsoidal head and within a circle equal to 80% of the shell diameter, t_{r} to be used in reinforcement calculations is the thickness required for a seamless sphere of radius K_{1}D, where D is the shell ID and K_{1} for a 2 : 1 ellipsoidal head is 0.9 from Table 11.3. For this head, the opening and reinforcement are within 0.8 D = 0.8(41.75) = 33.4 in.

A_{2}=2\left(T_{n}-t_{n}\right)\left(2.5 T_{n}+T_{e}\right)                      (11.20)

\text { Table } 11.3^{a}
D/2h 3.0 2.8 2.6 2.4 2.2 2.0
K_{1} 1.36 1.27 1.18 1.08 0.99 0.90
D/2h 1.8 1.6 1.4 1.2 1.0
K_{1} 0.81 0.73 0.65 0.57 0.50

3 . Using the spherical shell radius of R = K_{1}D = 0.9(41.75) = 37.575 in. in the hemispherical head formula gives

t_{r}=\frac{P R}{2 S E-0.2 P}=\frac{700(37.575)}{2(17,500 \times 1.0)-0.2(700)}=0.755 \ in.

4 . The minimum required thickness of the nozzle is

t_{rn}=\frac{P R_{n}}{S E-0.6 P}=\frac{700(4)}{(17,500 \times 1.0)-0.6(700)}=0.164 \ in.

5 . Limits parallel to head surface = X =d or \left(T_{s}+T_{n}+r\right), whichever is larger. X = 8 in. or (4 + 1 + 1.125 = 6.125 in. ); use 8 in.

6 . Limits perpendicular to head surface is Y = 2.5T_{s} or 2.5T_{n}, whichever is smaller. Y = 2.5(1) = 2.5 in. or 2.5(1.125) = 2.813 in.; use 2.5 in.

7 . Because the limit of 2X = 2(8) = 16 in. is less than 33.4 in. of item 2 above, the provision of the spherical head may be used.

8 . Reinforcement area required following Eq. 11.20 is

A_{r}=d t_{r} F=8(0.755)(1.0)=6.040 \ in.^{2}

9 . Reinforcement area available in head according to Eq. 11.21 is

A=d t_{r} F                         (11.21)

A_{1}=\left(E T_{s}-F t_{r}\right)(2 d-d)=(1.0-0.755)(16-8)=1.960 \text { in. }^{2}

10 . Reinforcement area available in nozzle according to Eq. 11.23 is

A_{1}=2\left(T_{s}+T_{n}+0.5 d\right)-d\left(E T_{s}-F t_{r}\right)                            (11.23)

A_{2}=5 T_{s}\left(T_{n}-t_{rn}\right)=5(1.0)(1.125-0.164)=4.805 \text { in. }^{2}

11 . Total reinforcement available from head and nozzle is

A_{t}=A_{1}+A_{2}=1.960+4.805=6.765 \text { in. }^{2}

Area provided = 6.765 in² >  area required = 6.040 in.² If additional area is needed, use fillet weld area.