Bioprocess Engineering Basic Concepts 2nd Edition Solution 🆒

Solution: Mass of medium = 1000 L * 1000 g/L = 1,000,000 g or 1000 kg. Energy required = 1000 kg * 4.2 kJ/kg°C * (37°C - 20°C) = 1000 * 4.2 * 17 = 71,400 kJ.

Solution: Using the equation for sterilization, N(t) = N0 * e^(-kt), where N0 is the initial number of spores, k is the death rate constant, and t is time. N(15) = 10^6 * e^(-0.5*15) = 10^6 * e^(-7.5).

Solution: Power per unit volume = 2 kW / 2000 L = 0.001 kW/L or 1 W/L. Bioprocess Engineering Basic Concepts 2nd Edition Solution

5.1. A medium is sterilized at 121°C for 15 minutes. If the initial number of spores is 10^6 per mL and the death rate constant is 0.5 min^-1, what is the final number of spores per mL?

Solution: The main goals of bioprocess engineering are to develop efficient, cost-effective, and safe methods for producing valuable products using biological systems. Solution: Mass of medium = 1000 L *

Solution: Annual production = 200 kg/day * 300 days/year = 60,000 kg/year.

2.1. A bioreactor contains 1000 L of medium with an initial cell concentration of 1 g/L. If 500 L of medium is added, what is the new cell concentration? N(15) = 10^6 * e^(-0

3.1. A bioprocess requires heating 1000 L of medium from 20°C to 37°C. If the specific heat capacity of the medium is 4.2 kJ/kg°C and the density is 1 g/mL, what is the energy required?