Let's break down the problem step-by-step: --- ## **Given Data** - **Applied Pressure, \(P\)** = 1.2 MPa - **Young's Modulus, \(E\)** = 24 MPa - **Roller Radii, \(R_1\)** = 200 mm, \(R_2\)** = 300 mm - **Roller Length, \(L\)** = 2000 mm - **Web speed, \(V\)** = 200 m/min = \( \frac{200}{60} = 3.33 \) m/s - **Thickness in, \(T_{in}\)** = 200 mm - **Thickness out, \(T_{out}\)** = 60 mm - **Porosity** = 50% - **Thermal Diffusivity**: - Polypropylene (\(\alpha_{PP}\)) = \(0.1 \times 10^{-6}\) m\(^2\)/s - Air (\(\alpha_{air}\)) = \(2 \times 10^{-5}\) m\(^2\)/s --- ## **1. Contact Time (\(t_c\))** **Contact time** is the time for which a point on the web remains in contact with the roller. \[ \text{Contact time, } t_c = \frac{\text{Contact width}}{\text{Web speed}} \] The **contact width (w)** can be estimated using the formula for the width of the flattened contact between two cylinders under pressure: \[ w = \sqrt{\frac{4 P R_{eq} L}{\pi E}} \] Where, - \(P\) = Applied Pressure - \(E\) = Young's Modulus (of web) - \(R_{eq}\) = Equivalent radius of rollers For two cylinders: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \] ### **Step 1: Calculate \(R_{eq}\)** \[ \frac{1}{R_{eq}} = \frac{1}{200} + \frac{1}{300} = \frac{3 + 2}{600} = \frac{5}{600} \] \[ R_{eq} = \frac{600}{5} = 120 \text{ mm} = 0.12 \text{ m} \] ### **Step 2: Convert Units** - \(P = 1.2 \) MPa = \(1.2 \times 10^6\) Pa - \(E = 24 \) MPa = \(24 \times 10^6\) Pa - \(L = 2000 \) mm = 2 m - \(R_{eq} = 0.12\) m ### **Step 3: Calculate Contact Width (w)** \[ w = \sqrt{ \frac{4 \times 1.2 \times 10^6 \times 0.12 \times 2 }{ \pi \times 24 \times 10^6 }} \] \[ = \sqrt{ \frac{4 \times 1.2 \times 0.12 \times 2 \times 10^6}{\pi \times 24 \times 10^6}} \] \[ = \sqrt{ \frac{1.152 \times 10^6}{75.398 \times 10^6} } \] \[ = \sqrt{ \frac{1.152}{75.398} } \] \[ = \sqrt{0.01528} \] \[ = 0.1236 \text{ m} = 123.6 \text{ mm} \] ### **Step 4: Calculate Contact Time (\(t_c\))** \[ t_c = \frac{w}{\text{Web speed}} = \frac{0.1236}{3.33} = 0.037 \text{ s} = 37 \text{ ms} \] --- ## **2. Gap Between Rollers (\(g\))** The gap is the thickness of the web after compression: \[ g = T_{out} \] Given: - \(T_{out} = 60 \) mm --- ## **3. Will the Fabric Be Uniformly Heated?** Polypropylene web (50% porosity) is a composite of PP and air. To be uniformly heated, heat must penetrate the web during the contact time. ### **Thermal Diffusivity of the Web:** \[ \alpha_{web} = (\text{fraction of PP}) \times \alpha_{PP} + (\text{fraction of air}) \times \alpha_{air} \] \[ = 0.5 \times 0.1 \times 10^{-6} + 0.5 \times 2 \times 10^{-5} \] \[ = 0.05 \times 10^{-6} + 1 \times 10^{-5} \] \[ = 0.05 \times 10^{-6} + 10 \times 10^{-6} \] \[ = 10.05 \times 10^{-6} \text{ m}^2/\text{s} \] ### **Thermal Penetration Depth (\(\delta\)):** \[ \delta = \sqrt{\alpha_{web} \cdot t_c} \] \[ \delta = \sqrt{10.05 \times 10^{-6} \times 0.037} \] \[ = \sqrt{0.37185 \times 10^{-6}} \] \[ = \sqrt{3.72 \times 10^{-7}} \] \[ = 6.1 \times 10^{-4} \text{ m} = 0.61 \text{ mm} \] **Compressed thickness = 60 mm** **Penetration depth = 0.61 mm** So, **the fabric will NOT be uniformly heated** during the contact time, as the penetration depth is much smaller than the compressed thickness. --- ## **### Final Answers** **1. Contact time:** \[ \boxed{0.037\, \text{s} \text{ or } 37\, \text{ms}} \] **2. Gap between rollers:** \[ \boxed{60\, \text{mm}} \] **3. Will the fabric be uniformly heated?** \[ \boxed{\text{No, because the thermal penetration depth (0.61 mm) is much less than the compressed web thickness (60 mm).}} \]