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Chapter 10

Q. 10.13

A welding-neck flange with the same geometry as that in Example 10.9 except for the thickness is used with a full-face gasket. The design pressure is 320 psi and the “soft” gasket is vegetable fiber with m = 1.75 and y = 1100. What is the minimum required thickness?


Verified Solution

1 . Determine the lever arms of the inner and outer parts of the gasket:

h_{G}=\frac{(C-B)(2 B+C)}{6(B+C)}=\frac{(22.5-10.75)(2 \times 10.75+22.5)}{6(10.75+22.5)}

h_{G}=2.5915 \ in.

h_{G}^{\prime}=\frac{(A-C)(2 A+C)}{6(C+A)}=\frac{(26.5-22.5)(2 \times 26.5+22.5)}{6(22.5+26.5)}

h_{G}^{\prime}=1.0272 \text { in. }

2 . Determine the gasket dimensions:

G=C-2 h_{G}=22.5-2 \times 2.5915=17.317 \text { in. }

b=\frac{(C-B)}{4}=\frac{(22.5-10.75)}{4}=2.9375 \text { in. }

y = 1100     and      m = 1.75

3 . Determine the loads:

H=\frac{\pi}{4} G^{2} p=\frac{\pi}{4}(17.317)^{2}(320)=75,368

H_{p}=2 b \pi G m p=2(2.9375) \pi(17.317)(1.75)(320)=178,986

H_{p}^{\prime}=\left(\frac{h_{G}}{h_{G}^{\prime}}\right) H_{p}=\frac{(2.5915)}{(1.0272)(178,986)}=451,559


H_{G y}=b \pi G y=(2.9375) \pi(17.317)(1100)=175,790

H_{G y}^{\prime}=\left(\frac{h_{G}}{h_{G}^{\prime}}\right) H_{G y}=\frac{(2.5915)}{(1.0272)(175,790)}=443,497


4 . Determine bolting requirements: A_{m} is the greater of W_{m 1} / S_{a} or W_{m 2} / S_{a}

A_{m}=36.76 \text { in. }^{2} based on W_{m 1}

A_{b}=36.8 in.^{2}  based on 16-2-in. diameter bolts


5 . Determine flange moments at operating condition.

Flange Loads

H_{D}=\frac{\pi}{4} B^{2} p=\frac{\pi}{4}(10.75)^{2}(320)=29,044


Lever Arms

h_{D}=R+0.5 g_{1}=2.5+0.5(3.375)=4.1875 \text { in. }

h_{T}=0.5\left(R+g_{1}+h_{G}\right)=0.5(2.5+3.375+2.5915)=4.2333 \text { in. }

Flange Moments

M_{D}=H_{D} h_{D}=(29,044)(4.1875)=121,622

M_{T}=H_{T} h_{T}=(46,324)(4.2333)=196,104


6 . Determine flange moment at gasket seating condition.

Flange Load


Lever arm

h_{G}^{\prime \prime}=\frac{h_{G} h_{G}^{\prime}}{h_{G}+h_{G}^{\prime}}=\frac{(2.5915)(1.0272)}{(2.5915)+(1.0272)}=0.7356 \text { in. }

Flange moment

M_{g}=H_{G} h_{G}^{\prime \prime}=(630,808)(0.7356)=464,022

All flange geometry constants are the same as in Example 10.9.

7 . Calculate flange stresses. Assume flange thickness t = 2.03 in. This is set directly from the radial flange stress at the bolt circle which is

S_{R}^{b c}=\frac{6 M_{g}}{t^{2}(\pi C-N d)}=\frac{6(464,022)}{(2.03)^{2}(22.5 \pi-16 \times 2)}

S_{R}^{b c}=17,464 \ psi <17,500 \ psi     allowable stress

Longitudinal Hub Stress

L=\frac{t e+1}{T}+\frac{t^{3}}{d}=1.0021+0.0407=1.0428

S_{H}=\frac{f M_{g}}{{L_g}_{ 1}^{2} B}=\frac{1(464,022)}{(1.0428)(3.375)^{2}(10.75)}

S_{H}=3634 \ psi <26,250 \ psi    allowable stress

Radial Flange Stress

S_{R}=\frac{\left(\frac{4}{3} t e+1\right) M_{g}}{L t^{2} B}=\frac{(1.4704)(464,022)}{(1.0428)(2.03)^{2}(10.75)}

S_{R}=14,770 \ psi <17,500 \ psi    allowable stress

Tangential Flange Stress

S_{T}=\frac{Y M_{g}}{t^{2} B}-Z S_{R}=\frac{2.29(464,022)}{(2.03)^{2}(10.75)-(1.39)(14,770)}

S_{T}=3457 \ psi <17,500 \ psi    allowable stress