Q1: 1000 Kg of liquid at 30°C in a well stirred vessel has to be heated to 120°C, using immersed coils carrying condensing steam at 150°C. The area of the steam coils is 1.2 m² and the overall heat transfer co-efficient to the liquid is 1500 W/m².°C. Assuming negligible heat loss to the surrounding and specific heat capacity of the liquid to be 4 kJ/kg.°C, the time taken for the liquid to reach desired temperature will be

ANS:D - 51 min

To find the time taken for the liquid to reach the desired temperature, we can use the principle of energy balance. The rate of heat transfer from the condensing steam to the liquid can be expressed as: dtdQ​=UAΔT Where:

• dtdQ​ is the rate of heat transfer (W),
• U is the overall heat transfer coefficient (W/m²·°C),
• A is the surface area of the steam coils (m²),
• ΔT is the temperature difference between the steam and the liquid (°C).

The rate of heat transfer is also equal to the rate of temperature change of the liquid times its mass times its specific heat capacity: dtdQ​=mcdtdT​ Where:

• m is the mass of the liquid (kg),
• c is the specific heat capacity of the liquid (kJ/kg·°C),
• dtdT​ is the rate of change of temperature (°C/s).

Equating these two expressions, we get: mcdtdT​=UAΔT Rearranging for dtdT​: dtdT​=mcUAΔT​ Now, let's calculate the temperature difference ΔT: Δ=steam−liquid=150°−30°=120°ΔT=Tsteam​−Tliquid​=150°C−30°C=120°C Substitute the given values: =1500×1.2×1201000×4dtdT​=1000×41500×1.2×120​ =1800004000dtdT​=4000180000​  °C/sdtdT​=45°C/s Now, we know the rate of change of temperature. To find the time taken for the temperature to change from 30°C to 120°C, we'll integrate dtdT​ with respect to time: ∫30120​dT=∫0t​45dt 120−30=45t 90=45t 5t=4590​ 2 st=2s So, the time taken for the liquid to reach the desired temperature is 2 seconds. If there's any mistake or misinterpretation in your query, feel free to ask for clarification!