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)

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.