To calculate the tension in the slack side of the driving belt, we will use the formula related to belt tension in open belt drives:
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First, we need to find the usable torque, which can be translated into tension on the belt. The relationship between power (P), torque (T), and angular velocity (ω) is given by:
\[ P = T \cdot ω \]
where:
- \( P = 25 , \text{kW} = 25000 , \text{W} \)
- \( ω = \frac{2 \pi \cdot N}{60} \) (N is in rpm, so we convert it to rad/s)
Given that \( N = 200 , \text{rpm} \):
\[ ω = \frac{2 \pi \cdot 200}{60} \approx 20.94 , \text{rad/s} \]
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Now, we can calculate the torque (T):
\[ T = \frac{P}{ω} = \frac{25000}{20.94} \approx 1195.5 , \text{Nm} \]
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Now, let's determine the effective radius of the driving pulley. The diameter of the driving pulley is 200 mm, so:
\[ r_1 = \frac{200 , \text{mm}}{2} = 100 , \text{mm} = 0.1 , \text{m} \]
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We can find the effective tension in the belt (T1) using the formula:
\[ T_1 = \frac{T}{r_1} \]
Rearranging gives us:
\[ T_1 = \frac{1195.5 , \text{Nm}}{0.1 , \text{m}} = 11955 , \text{N} \]
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Next, we will calculate the tension in the slack side of the belt (T2) using the relationship which incorporates the angle of contact and the coefficient of friction (μ):
The general formula relating the tension in the tight side (T1) and slack side (T2) is given by:
\[ \frac{T_1}{T_2} = e^{\mu \theta} \]
where:
- \( \mu = 0.37 \) (coefficient of friction)
- \( \theta \) is the angle of contact in radians. We must convert \( 176° \) to radians:
\[ \theta = 176° \times \frac{\pi}{180} \approx 3.07 , \text{rad} \]
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Now substituting into the equation:
\[ \frac{T_1}{T_2} = e^{0.37 \cdot 3.07} = e^{1.1369} \approx 3.11 \]
Rearranging gives:
\[ T_2 = \frac{T_1}{3.11} = \frac{11955 , \text{N}}{3.11} \approx 3843.14 , \text{N} \]
Thus, the tension in the slack side of the driving belt is approximately 3843.14 N.
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