Products

Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

Holooly Ads. Manager

Advertise your business, and reach millions of students around the world.

Holooly Tables

All the data tables that you may search for.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Help Desk

Need Help? We got you covered.

Chapter 10

Q. 10.9

What is the minimum required thickness of a welding neck flange as shown in Fig. 10.14a with the following design data? (Note: These data are the same as those used for the blind flange in Example 10.8. In Fig. 10.15 is a sample calculation of a welding neck flange.

Design pressure, p, = 2,500 psi.

Design temperature = 250°F.

Bolt-up and gasket seating temperature = 70°F.

Flange material is SA-105.

Bolting material is SA-325 Grade 1 .

Gasket details are spiral-wound metal, fiber filled, stainless steel, inside diameter is 13.75 in. and width is 1.0 in.

What is the minimum required thickness of a welding neck flange as shown in Fig. 10.14a with the following design data? (Note: These data are the same as those used for the blind flange in Example 10.8. In Fig. 10.15 is a sample calculation of a welding neck flange. Design pressure, p, = 2,500 psi.
What is the minimum required thickness of a welding neck flange as shown in Fig. 10.14a with the following design data? (Note: These data are the same as those used for the blind flange in Example 10.8. In Fig. 10.15 is a sample calculation of a welding neck flange. Design pressure, p, = 2,500 psi.

Step-by-Step

Verified Solution

1 . Allowable bolt stress at design and seating temperatures = S_{b} = 19,200 psi.

2 . Allowable flange stress at design and seating temperatures = S_{f} = 17,500 psi.

3 . Gasket dimensions are

b_{o}=N / 2=0.5 \ in.              and              b=0.5 \sqrt{b_{0}}=0.3535

G = 13.75 + (2 × 1) – (2 × 0.3535) = 15.043 in.

4 . Determine bolt loadings and sizing of bolts with N = 1 ; b = 0.3535; y = 10,000 ; m = 3.0.

H=\frac{\pi}{4} G^{2} p=\frac{\pi}{4}(15.043)^{2}(2500)=444,323

H_{p}=2 b \pi G m p=2(0.3535) \pi(15.043)(3.0)(2500)=250,591

W_{m 1}=H+H_{p}=(444,323)+(250,591)=694,914

W_{m 2}=\pi b G y=\pi(0.3535)(15.043)(10,000)=167,060

A_{m}=\text { the greater of } W_{m 1} / S_{b h}=(694,914) /(19,200)=36.2 \text { in.}^{2}

\text { or } W_{m 2} / S_{b c}=(167,060) /(19,200)=8.7 \text { in. }^{2}

A_{b} = actual bolt area = 36.8 in.²                    16 bolts at 2-in. diameter

W_{a}=0.5\left(A_{m}+A_{b}\right) S_{b c}=0.5(36.2+36.8)(19,200)=700,800

W_{0}=W_{m 1}=694,914

5 . Calculate total flange moment for the design condition.

Flange Loads

H_{D}=\frac{\pi}{4} B^{2} p=\frac{\pi}{4}(10.75)^{2}(2500)=226,906

H_{G}=H_{p}=250,591

H_{T}=H-H_{D}=(444,323)-(226,906)=217,417

Lever Arms

h_{D}=R+0.5 g_{1}=(2.5)+0.5(3.375)=4.1875

h_{G}=0.5(C-G)=0.5(22.5-15.043)=3.7285

h_{T}=0.5\left(R+g_{1}+h_{G}\right)=0.5(2.5+3.375+3.7285)=4.8018

Flange Moments

M_{D}=H_{D} \times h_{D}=(226,910)(4.1875)=950,170

M_{G}=H_{G} \times h_{G}=(250,590)(3.7285)=934,330

M_{T}=H_{T} \times h_{T}=(217,420)(4.8018)=1,043,990

M_{d e}=M_{D}+M_{G}+M_{T}=2,928,490

6 . Calculate total flange moment for bolt-up condition.

Flange Load

H_{G}=W_{a}=700,800

Lever Arm

h_{G}=0.5(C-G)=3.7285

Flange Moment

M_{b u}=H_{G} \times h_{G}=(700,800)(3.7285)=2,612,930

7 . Use the greater of M_{d e} or M_{bu}\left(S_{h} / S_{c}\right) ; M_{0}=2,928,490.

8 . Shape constants from the ASME Code, VIII-1, Appendix 2: K = A/B = (26.5)/(10.75) = 2.465. From Fig. 2-7.1, Section VIII-1;

T = 1.35                  Z = 1.39                Y = 2.29              U = 2.51.

g_{1} / g_{0}=3.375 / 1.0=3.375

h_{0}=\sqrt{B g_{0}}=\sqrt{(10.75)(1.0)}=3.279

From Fig. 2-7.2, Section VIII-1, F = 0.57.

From Fig. 2-7.3, Section VIII-1, V = 0.04.

From Fig. 2-7.6, Section VIII-1; f = 1.0.

e=F / h_{0}=(.57) /(3.279)=0.1738

d=(U / V) h_{0} g_{0}^{2}=(2.51 / .04)(3.279)(1)^{2}=205.76

9 . Calculate stresses. Assume a flange thickness t = 4.5 in.

L = (te + 1) / T + t³ / d = (1.320) + (0.443) = 1.763

Longitudinal Hub Stress

S_{H}=f M_{0} / \operatorname{Lg}_{1}^{2} B=(1)(2,928,490) /(1.763)(3.375)^{2}(10.75)

S_{H}=13,570  \ psi

Radial Flange Stress

S_{R}=\left(\frac{4}{3} t e+1\right) M_{0} / L t^{2} B=(2.0428)(2,928,490) /(1.763)(4.5)^{2}(10.75)

S_{R}=15,590 \ psi

Tangential Flange Stress

\begin{aligned}S_{T} &=\left(Y M_{0} / t^{2} B\right)-Z S_{R} \\&=(2.29)(2,928,490) /(4.5)^{2}(10.75)-(1.39)(15,590)\end{aligned}

S_{T}=9140 \ psi

10 . Allowable stresses

S_{H} \leq 1.5 S_{f}:(1.5)(17,500)=26,250>13,570 \ psi

S_{R} \leq S_{f}: 17,500>15,590 \ psi

S_{T} \leq S_{f}: 17,500>9140 \ psi