Question

Find the capacity of a field chopper with a throat size of \(25 \times 5\text{ cm}\) operating at 50 rpm, with the help of given information. 

• Number of knives in cutter head = 2 
• Bulk density of the material = \(500\text{ kg m}^{-3}\) 
• Length of cut = 2 cm 
• Field efficiency of the chopper = 70% 

• A. \(500\text{ kg}\cdot\text{ha}^{-1}\)
• B. \(525\text{ kg}\cdot\text{ha}^{-1}\)
• C. \(1000\text{ kg}\cdot\text{ha}^{-1}\)
• D. \(1050\text{ kg}\cdot\text{ha}^{-1}\)

Hint:

Direct calculation: \(0.0125 \times 0.02 \times 100 \times 60 \times 500 \times 0.70 = 525\text{ kg/h}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The theoretical capacity of a forage chopper depends on the cross-sectional area of the throat opening, the linear feed rate (length of cut times number of knife cuts per unit time), material bulk density, and operating field efficiency.
Key Formula or Approach:
\[ \text{Theoretical Capacity } (C_{\text{th}}) = A_{\text{throat}} \times L_{\text{cut}} \times (N \times K) \times \rho_{\text{bulk}} \times 60 \]
\[ \text{Effective Capacity } (C_{\text{eff}}) = C_{\text{th}} \times \eta_{\text{field}} \]

Step 2: Detailed Explanation:
Given data:
- Throat dimensions: \(25\text{ cm} \times 5\text{ cm} \implies A = 0.25\text{ m} \times 0.05\text{ m} = 0.0125\text{ m}^2\)
- Cutter head rotational speed: \(N = 50\text{ rpm}\)
- Number of knives: \(K = 2\)
- Length of cut: \(L = 2\text{ cm} = 0.02\text{ m}\)
- Bulk density of forage: \(\rho = 500\text{ kg/m}^3\)
- Field efficiency: \(\eta = 70\% = 0.70\)
Number of cuts per minute:
\[ \text{Cuts/min} = N \times K = 50 \times 2 = 100\text{ cuts/min} \]
Theoretical volume chopped per minute:
\[ V_{\text{min}} = A \times L \times (\text{cuts/min}) = 0.0125\text{ m}^2 \times 0.02\text{ m} \times 100 = 0.025\text{ m}^3\text{/min} \]
Theoretical volume chopped per hour:
\[ V_{\text{hour}} = 0.025 \times 60 = 1.5\text{ m}^3\text{/h} \]
Theoretical mass capacity:
\[ M_{\text{th}} = 1.5\text{ m}^3\text{/h} \times 500\text{ kg/m}^3 = 750\text{ kg/h} \]
Effective mass capacity with 70% efficiency:
\[ M_{\text{eff}} = 750 \times 0.70 = 525\text{ kg/h} \]

Step 3: Final Answer:
Therefore, the effective capacity is \(525\text{ kg}\cdot\text{ha}^{-1}\) (kg/h), corresponding to option (B).