50 tonnes of apple having specific heat of 0.80 kcal/kg °C is to be cooled from 28-15 °C in 24 hr. The heat of respiration for 24 hr is 745 kcal/tonne. Three men will work for 4 ho... 50 tonnes of apple having specific heat of 0.80 kcal/kg °C is to be cooled from 28-15 °C in 24 hr. The heat of respiration for 24 hr is 745 kcal/tonne. Three men will work for 4 hours and lighting load is estimated to be 125 watt. Air infiltration load is assumed as 980 kcal/24 hrs. The outside temperature is 36 °C and inside is maintained 5 °C. Ceiling area is 36 m² and brick wall area including all sides is 96 m². Calculate the plant capacity needed in terms of refrigeration. Given data: Thermal conductivity of bricks (Kb) = 0.45 kcal/hr/m°C, Thermal conductivity of cork (Kc) = 0.025 kcal/hr/m°C, Thermal conductivity of Cement plaster (Kp) = 0.25, Heat of respiration per person = 170 kcal/hr/m°C. Solve the problem. In a vapour absorption system heat is supplied to the generator at a temperature of 90 °C. The cooling in condenser and refrigeration evaporator takes place at 20 °C and -10 °C respectively. Find the COP of the system. A Carnot engine operates between 1000 K & 500 K temperatures. How much amount of heat energy will be absorbed from the source per second to perform 210 J work per second? Read more

Steps to Solve

1. Calculate Heat Load from Respiration

   • The heat load due to respiration can be calculated using: $$ Q_r = m \times R $$ where
   • ( m = 50 , \text{tonnes} = 50,000 , \text{kg} )
   • ( R = 745 , \text{kcal/tonne} ) Thus, $$ Q_r = 50000 , \text{kg} \times \frac{745 , \text{kcal}}{1000 , \text{kg}} = 37250 , \text{kcal} $$

2. Calculate Total Heat Load

   • Include other heat loads: lighting and infiltration.
   • Lighting load: $$ Q_l = \frac{125 , \text{watts} \times 24 , \text{hr} \times 3600 , \text{s/hr}}{1000} = 108 , \text{kcal} $$
   • Infiltration load is given as 980 kcal/24 hrs.
   • Total heat load: $$ Q_{total} = Q_r + Q_l + Q_{infiltration} = 37250 + 108 + 980 = 38238 , \text{kcal} $$

3. Heat Transfer through Walls

   • Calculate heat transferred through walls using the formula: $$ Q_w = K \cdot A \cdot \Delta T $$ where
   • ( K ) is the thermal conductivity (average of bricks and cement plaster),
   • ( A = \text{Total wall area} = 96 , \text{m}^2 )
   • ( \Delta T = T_{outside} - T_{inside} = 36 - 5 = 31^\circ C ) We need to consider ( K ): $$ K = K_b + K_c + K_p = 0.45 + 0.25 + 0.025 = 0.75 , \text{kcal/hr/m}^2,^\circ C $$ Then, $$ Q_w = 0.75 , \text{kcal/hr/m}^2,^\circ C \times 96 , \text{m}^2 \times 31^\circ C = 2242.5 , \text{kcal/hr} $$

4. Calculate Refrigeration Plant Capacity

   • The total heat load that the plant must counteract includes both the total heat load and heat transfer through walls or surfaces: $$ Q_{final} = Q_{total} + Q_w $$ Substituting the values gives: $$ Q_{final} = 38238 + 2242.5 \times 24 = 40446 , \text{kcal} $$

5. Find COP of the Vapor Absorption System

   • The Coefficient of Performance (COP) is calculated using: $$ COP = \frac{T_{cold}}{T_{hot} - T_{cold}} $$ Where:
   • ( T_{cold} = -10^\circ C = 263 , K )
   • ( T_{hot} = 20^\circ C = 293 , K ) So, $$ COP = \frac{263}{293 - 263} = \frac{263}{30} \approx 8.77 $$

6. Calculate Carnot Efficiency

   • The Carnot efficiency can be computed as: $$ \eta = 1 - \frac{T_{cold}}{T_{hot}} $$ Where:
   • ( T_{cold} = 1000 , K )
   • ( T_{hot} = 500 , K ) Thus, $$ \eta = 1 - \frac{500}{1000} = 0.5 $$