Question
Determine the mild steel plate thickness of a 2,1 m diameter boiler subjected to a pressure of 2 MPa, if:
Longitudinal joint efficiency = 72%
Circumferential joint efficiency = 80%
Tensile stress in plate = 78 MPa
Longitudinal joint efficiency = 72%
Circumferential joint efficiency = 80%
Tensile stress in plate = 78 MPa
Answers
GPT-4o mini
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.
### 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.