A 10-hp split-phase motor running at 1750 rev/min is used to drive a rotary pump, which operates 24 hours per day. An engineer has specified a 7.4-in small sheave, an 11-in large sheave, and three B112 belts. The service factor of 1.2 was augmented by 0.1 because of the continuous-duty requirement.

Question

A 10-hp split-phase motor running at 1750 rev/min is used to drive a rotary pump, which operates 24 hours per day. An engineer has specified a 7.4-in small sheave, an 11-in large sheave, and three B112 belts. The service factor of 1.2 was augmented by 0.1 because of the continuous-duty requirement. Analyze the drive and estimate the belt life in passes and hours.

Accepted Answer

The peripheral speed V of the belt is

[latex]V=\pi d n / 12=\pi(7.4) 1750 / 12=3390 ft / min[/latex]

Table 17–11: [latex]L_{p}=L+L_{c}=112+1.8=113.8 ext { in }[/latex]

Eq. (17–16b): [latex]\begin{aligned}C=& 0.25\left\{\left[113.8-\frac{\pi}{2}(11+7.4) ight] ight.\&\left.+\sqrt{\left[113.8-\frac{\pi}{2}(11+7.4) ight]^{2}-2(11-7.4)^{2}} ight\}\end{aligned}[/latex]

= 42.4 in

[latex]C=0.25\left\{\left[L_{p}-\frac{\pi}{2}(D+d) ight]+\sqrt{\left[L_{p}-\frac{\pi}{2}(D+d) ight]^{2}-2(D-d)^{2}} ight\}[/latex] (17–16b)

Eq. (17–1):[latex]\begin{array}{c}\phi= heta_{d}=\pi-2 \sin ^{-1}(11-7.4) /[2(42.4)]=3.057 rad \\exp[0.5123(3.057)]=4.788\end{array}[/latex]

[latex] heta_{D}=\pi+2 \sin ^{-1} \frac{D-d}{2 C}[/latex] (17–1)

Interpolating in Table 17–12 for V = 3390 ft/min gives[latex]H_{ tab }=4.693 hp[/latex]. The wrap angle in degrees is [latex]3.057(180) / \pi=175^{\circ} .[/latex]. From Table 17–13, K1 5 0.99. From Table 17–14, [latex]K_{2}=1.05 .[/latex] Thus, from Eq. (17–17),

[latex]H_{a}=K_{1} K_{2} H_{ tab }[/latex] (17–17)

[latex]H_{a}=K_{1} K_{2} H_{ ext {tab }}=0.99(1.05) 4.693=4.878 hp[/latex]

Eq. (17–19): [latex]H_{d}=H_{ nom } K_{s} n_{d}=10(1.2+0.1)(1)=13 hp[/latex]

[latex]H_{d}=H_{ nom } K_{s} n_{d}[/latex] (17–19)

Eq. (17–20): [latex]N\geq H_{d}/H_{a}=13/4.878=2.67 ightarrow 3[/latex]

[latex]N_{b} \geq \frac{H_{d}}{H_{a}} \quad N_{b}=1,2,3, \ldots[/latex] (17–20)

From Table 17–16, [latex]K_{c}=0.965[/latex]. Thus, from Eq. (17–21),

[latex]F_{c}=K_{c}\left(\frac{V}{1000} ight)^{2}[/latex] (17–21)

[latex]F_{c}=0.965(3390 / 1000)^{2}=11.1 lbf[/latex]

Eq. (17–22): [latex]\Delta F=\frac{63025(13) / 3}{1750(7.4 / 2)}=42.2 lbf[/latex]

[latex]\Delta F=\frac{63025 H_{d} / N_{b}}{n(d / 2)}[/latex] (17–22)

Eq. (17–23): [latex]F_{1}=11.1+\frac{42.2(4.788)}{4.788-1}=64.4 lbf[/latex]

[latex]F_{1}=F_{c}+\frac{\Delta F \exp (f \phi)}{\exp (f \phi)-1}[/latex] (17–23)

Eq. (17–24): [latex]F_{2}=F_{1}-\Delta F=64.4-42.2=22.2 lbf[/latex]

[latex]F_{2}=F_{1}-\Delta F[/latex] (17–24)

Eq. (17–25): [latex]F_{i}=\frac{64.4+22.2}{2}-11.1=32.2 lbf[/latex]

[latex]F_{i}=\frac{F_{1}+F_{2}}{2}-F_{c}[/latex] (17–25)

Eq. (17–26): [latex]n_{f s}=\frac{H_{a} N_{b}}{H_{ nom } K_{s}}=\frac{4.878(3)}{10(1.3)}=1.13[/latex]

[latex]n_{f s}=\frac{H_{a} N_{b}}{H_{ nom } K_{s}}[/latex] (17–26)

Life: From Table 17–16, [latex]K_{b}=576 .[/latex]

[latex]\left(F_{b} ight)_{1}=\frac{K_{b}}{d}=\frac{576}{7.4}=77.8 lbf[/latex]

[latex]\left(F_{b} ight)_{2}=\frac{576}{11}=52.4 lbf[/latex]

[latex]T_{1}=F_{1}+\left(F_{b} ight)_{1}=64.4+77.8=142.2 lbf[/latex]

[latex]T_{2}=F_{1}+\left(F_{b} ight)_{2}=64.4+52.4=116.8 lbf[/latex]

From Table 17–17, K = 1193 and b = 10.926.

[latex] ext { Eq. (17-27): } \quad N_{P}=\left[\left(\frac{1193}{142.2} ight)^{-10.926}+\left(\frac{1193}{116.8} ight)^{-10.926} ight]^{-1}=11\left(10^{9} ight) ext { passes }[/latex]

[latex]N_{P}=\left[\left(\frac{K}{T_{1}} ight)^{-b}+\left(\frac{K}{T_{2}} ight)^{-b} ight]^{-1}[/latex] (17–27)

Since [latex]N_{P}[/latex] is out of the validity range of Eq. (17–27), life is reported as greater than [latex]10^{9}[/latex] passes. Then

Eq. (17–28): [latex]t>\frac{10^{9}(113.8)}{720(3390)}=46600 h[/latex]

[latex]t=\frac{N_{P} L_{p}}{720 V }[/latex] (17–28)

Question 17–4: A 10-hp split-phase motor running at 1750 rev/min is used to...

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