To calculate the required thickness of the mild steel plate for the boiler, we can use the formula for the thickness of a pressure vessel. The thicker section will be determined by the circumferential (hoop) stress and longitudinal stress.
Step 1: Calculate the Effective Stresses
The allowable stress for the material can be calculated based on the given efficiency of the joints.
Longitudinal Stress (\( \sigma_L \))
The formula for the longitudinal stress is given by:
\[ \sigma_L = \frac{P D}{4 t} \]
Where:
- \( P \) = internal pressure (2 MPa = 2 N/mm²)
- \( D \) = diameter of the boiler (2.1 m = 2100 mm)
- \( t \) = thickness of the plate in mm
Circumferential (Hoop) Stress (\( \sigma_H \))
The formula for the circumferential stress is given by:
\[ \sigma_H = \frac{P D}{2 t} \]
Step 2: Determine the Allowable Stress
The allowable stress (\( \sigma_{all} \)) for the mild steel plate taking into account the joint efficiencies is calculated as follows:
\[ \sigma_{all} = \text{Tensile Stress} \times \text{Efficiency} \]
-
For longitudinal joints: \[ \sigma_{all,L} = 78 \text{ MPa} \times 0.72 = 56.16 \text{ MPa} \]
-
For circumferential joints: \[ \sigma_{all,H} = 78 \text{ MPa} \times 0.80 = 62.4 \text{ MPa} \]
Step 3: Set Up Equations
Using the allowable stresses, we need to set the stresses equal to the corresponding allowable stresses.
For Longitudinal Stress
\[ \frac{P D}{4 t} \leq \sigma_{all,L} \]
Substituting the values:
\[ \frac{2 \times 2100}{4 t} \leq 56.16 \]
\[ \frac{4200}{4 t} \leq 56.16 \]
\[ \frac{4200}{56.16} \leq 4 t \]
\[ t \geq \frac{4200}{4 \times 56.16} \]
Calculating \( t \):
\[ t \geq \frac{4200}{224.64} \approx 18.66 \text{ mm} \]
For Circumferential Stress
\[ \frac{P D}{2 t} \leq \sigma_{all,H} \]
Substituting the values:
\[ \frac{2 \times 2100}{2 t} \leq 62.4 \]
\[ \frac{4200}{2 t} \leq 62.4 \]
\[ \frac{4200}{62.4} \leq 2 t \]
\[ t \geq \frac{4200}{2 \times 62.4} \]
Calculating \( t \):
\[ t \geq \frac{4200}{124.8} \approx 33.61 \text{ mm} \]
Step 4: Determine the Thicker Requirement
From the calculations:
- The minimum thickness required for longitudinal stress is approximately 18.66 mm.
- The minimum thickness required for circumferential stress is approximately 33.61 mm.
Thus, the thickness required should ensure that both stresses are satisfied. Therefore, the required thickness of the mild steel plate will be:
\[ \text{Required Thickness} = 33.61 \text{ mm} \text{ (rounded up as necessary)} \]
Conclusion
The mild steel plate thickness of the boiler should be a minimum of approximately 34 mm to safely withstand the operating conditions specified.
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