**Step 1: Given Data and Conceptual Introduction** - Length of each arm (l) = 240 mm = 0.24 m - Mass of each ball (m) = 3 kg - Central load on sleeve (M) = 30 kg - Distance of arm pivots from axis (e) = 30 mm = 0.03 m - Radius of rotation in mid-position (r) = 195 mm = 0.195 m - Vertical height of governor (h) = 190 mm = 0.19 m - Governor speed (N) = 180 rpm *Explanation Block:* A Proell governor is similar to a Porter governor but with the balls attached to extensions of the lower arms, which allows a greater radius of rotation for the same height. The problem requires us to determine the length of the extension links and the tension in the upper arms when the governor is in the mid-position. --- **Step 2: Step-by-Step Calculation Part (Length of Extension Link)** 1. **Draw the Geometry:** The upper arm is hinged at 30 mm from the axis. The ball rotates at a radius of 195 mm from the axis. 2. **Find Horizontal Distance of Ball from Pivot (A):** - Let O = axis of rotation, A = upper arm pivot (30 mm from O), B = Ball center - OA = 30 mm, AB (upper arm) = 240 mm, OB = 195 mm (from axis) The horizontal distance from pivot A to ball B: \[ AB_{horizontal} = r - e = 195\,mm - 30\,mm = 165\,mm \] 3. **Calculate Vertical Distance (h₁) from A to B:** \[ AB^2 = (AB_{horizontal})^2 + (h_1)^2 \] \[ (240)^2 = (165)^2 + (h_1)^2 \] \[ 57600 = 27225 + (h_1)^2 \] \[ (h_1)^2 = 57600 - 27225 = 30375 \] \[ h_1 = \sqrt{30375} \approx 174.3\,mm \] 4. **Find Length of Extension Link (BC):** - The vertical height from A to C is 174.3 mm (above), but the total height from O to C is 190 mm. - Therefore, the vertical portion from B to C (extension link) is: \[ BC = h - h_1 = 190\,mm - 174.3\,mm = 15.7\,mm \] (Since the extension link is vertical in mid-position.) **Explanation Block:** The geometry of the governor arms is analyzed using the given dimensions. By representing the arm as a right-angled triangle, we determine the vertical segment between the ball and the sleeve, giving the extension link length. --- **Step 3: Step-by-Step Calculation Part (Tension in Upper Arms)** 1. **Calculate the Centrifugal Force (F):** \[ F = m \omega^2 r \] Where: \[ \omega = 2\pi N/60 = 2\pi \times 180/60 = 18.85\,\text{rad/s} \] \[ F = 3 \times (18.85)^2 \times 0.195 = 3 \times 355.3 \times 0.195 = 208.2\,N \] 2. **Find Angle of Upper Arm (θ):** \[ \sin\theta = \frac{r-e}{l} = \frac{165}{240} = 0.6875 \implies \theta = 43.3^\circ \] 3. **Tension in Upper Arm (T):** - The upper arm is subjected to two forces: the vertical component balances the weight of the ball and the sleeve, the horizontal component balances the centrifugal force. - For upper arm: \[ T = \frac{F}{\sin\theta} \] \[ T = \frac{208.2}{0.6875} = 302.8\,N \] **Explanation Block:** The tension in the upper arms is found by resolving the centrifugal force along the direction of the arm. The angle of the arm from the vertical is used to resolve the force correctly. --- **Final Answer:** **Length of extension link = 15.7 mm, Tension in upper arm = 302.8 N.**

**Step 1: Given Data and Conceptual Introduction** - Mass flow rate of steam, \( \dot{m}_s = 3.0\,kg/s \) - Saturation temperature of steam, \( T_{steam} = 330\,K \) - Water inlet temperature, \( T_{w,in} = 20^\circ C = 293\,K \) - Water outlet temperature, \( T_{w,out} = 30^\circ C = 303\,K \) - Tube inner diameter, \( D = 20\,mm = 0.02\,m \) - Convection coefficient (initial), \( h_{water} = 10,000\,W/m^2\cdot K \) - Water flow velocity, \( v = 0.5\,m/s \) - Number of shell passes: 1, with 2 tube passes (shell-and-tube configuration) *Explanation Block:* The problem involves designing a shell-and-tube heat exchanger to condense saturated steam using water flowing inside tubes. It requires calculating the minimum number of tubes and their length to achieve the desired heat transfer, considering an improved convection coefficient. --- **Step 2: Step-by-Step Calculation of Heat Duty and Overall Heat Transfer Coefficient** 1. **Determine the heat transfer rate \( Q \):** Since the steam condenses at 330 K, the heat removed from steam is: \[ Q = \dot{m}_s \times h_{fg} \] Where \( h_{fg} \) is the latent heat of vaporization at 330 K. From saturation tables, approximately: \[ h_{fg} \approx 2,260\,kJ/kg = 2,260,000\,J/kg \] Thus: \[ Q = 3.0\,kg/s \times 2,260,000\,J/kg = 6,780,000\,W \] 2. **Determine the temperature difference:** \[ \Delta T_{lm} \approx T_{steam} - T_{water\,avg} \] Average water temperature: \[ T_{water\,avg} = \frac{20 + 30}{2} = 25^\circ C = 298\,K \] Temperature difference: \[ \Delta T_{lm} = 330\,K - 298\,K = 32\,K \] (For precise calculations, logarithmic mean temperature difference (LMTD) should be used, but for an initial estimate, this is sufficient.) 3. **Calculate overall heat transfer coefficient \( U \):** Assuming the dominant resistance is the water-side convection, and ignoring tube wall resistance: \[ Q = U \times A_{total} \times \Delta T_{lm} \] Rearranged: \[ A_{total} = \frac{Q}{U \times \Delta T_{lm}} \] 4. **Calculate the heat transfer area per tube:** - Outer surface area per unit length: \[ A' = \pi D = \pi \times 0.02\,m = 0.0628\,m^2/m \] - Total area for \( N \) tubes of length \( L \): \[ A_{total} = N \times A' \times L \] 5. **Estimate \( U \):** Since convection dominates, \[ \frac{1}{U} \approx \frac{1}{h_{water}} \quad \text{(assuming negligible wall resistance)} \] \[ U \approx h_{water} = 10,000\,W/m^2 \cdot K \] (But to be precise, the overall heat transfer coefficient \( U \) considers the tube wall and fouling factors; here, assume ideal case with only convection resistance.) 6. **Calculate required total area:** \[ A_{total} = \frac{6,780,000\,W}{10,000\,W/m^2 \times 32\,K} \approx 21.19\,m^2 \] 7. **Calculate number of tubes \( N \):** Assuming a tube length \( L \), then: \[ N \times 0.0628 \times L = 21.19 \] To find the minimum number of tubes, choose a reasonable length. --- **Step 3: Determining the minimum number of tubes and tube length** 1. **Assume a tube length \( L \):** Suppose we select \( L = 3\,m \): \[ N = \frac{21.19}{0.0628 \times 3} = \frac{21.19}{0.1884} \approx 112.