Question 10.13: A welding-neck flange with the same geometry as that mention...
A welding-neck flange with the same geometry as that mentioned 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?
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 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 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) ×π(17.317)(1.75)(320) = 178,986
H_{ p }^{\prime}=\left(\frac{h_{ G }}{h_{ G }^{\prime}}\right) H_{ p }=\left(\frac{2.5915}{1.0272}\right)(178,986)=451,559
W_{ m 1}=H+H_{ p }+H_{ p }^{\prime}=75,368+178,986 +451,559 = 705,913
H_{ G y} = bπGy = (2.9375)π(17.317)(1100)
= 175,790
H_{ G y}^{\prime}=\left(\frac{h_{ g }}{h_{ G }^{\prime}}\right) H_{ Gy }=\frac{(2.5915)(175,790)}{(1.0272)}= 443,497
W_{ m 2}=H_{ G y}+H_{ G y}^{\prime} =175,790+443,496=619,286.
4) Determine bolting requirements. A_{m} is the greater of W_{ m 1} / S_{ a } \text { or } W_{ m 2} / S_{ a } :
A_{ m }=36.76 in. ^{2} based on W_{m1}
A_{ b }=36.8 in ^{2} based on 16-2in. diameter bolts
W_{ a }=0.5\left(A_{ m }+A_{ b }\right) S_{ a } = 0.5(36.76 + 36.8) ×(19,200) = 706,176.
5) Determine the flange moments at operating condition. Flange loads:
H_{ D }=\frac{\pi}{4} B^{2} P=\frac{\pi}{4}(10.75)^{2}(320)=29,044
H_{ T }=H-H_{ D } = 75,368 − 29,044 = 46,324.
Lever arms:
h_{ D }=R+0.5 g_{1} = 2.5 + 0.5(3.375) = 4.1875in.
h_{ T }=0.5\left(R+g_{1}+h_{ G }\right)= 0.5(2.5 + 3.375 + 2.5915)
= 4.2333in.
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
M_{0}=M_{ D }+M_{ T } = 121,622 + 196,104
= 317,726.
6) Determine the flange moment at gasket seating condition. Flange load:
H_{ G }=W_{ a }-H=706,176-75,368=630,808Lever 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)}= 0.7356in.
Flange moment:
M_{ g }=H_{ G } h_{ G }^{\prime \prime} = (630,808)(0.7356) = 464,022.
All flange geometry constants are the same as those mentioned in Example 10.9.
7) Calculate the 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 }^{ bc }=\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}(10.72)}
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).