Manual | Introduction To Food Engineering Solutions

$$Q = \dotm m c p,m (T_m,out - T_m,in)$$ $$Q = (0.5)(3900)(72 - 4) = 0.5 \times 3900 \times 68 = 132,600 \text W$$

$$\ln(0.01803) = -5.2275 X \Rightarrow -4.015 = -5.2275 X \Rightarrow X = 0.768$$ $$Fo_cyl = 0.768 = \frac\alpha tR^2 = \frac(1.5\times10^-7) t(0.04)^2$$ $$t = \frac0.768 \times 0.00161.5\times10^-7 = 8192 \text s$$ Introduction To Food Engineering Solutions Manual

Let $X = Fo_cyl$: $$0.02083 = 1.155 \exp\left[-(4.2025)X - (2.3104)(0.444X)\right]$$ $$0.01803 = \exp\left[-(4.2025 + 1.025)X\right] = \exp(-5.2275 X)$$ $$Q = \dotm m c p,m (T_m,out - T_m,in)$$ $$Q = (0

$$0.02083 = [1.10 e^-(2.05)^2 Fo_cyl] \times [1.05 e^-(1.52)^2 Fo_slab]$$ But $Fo_cyl = \frac\alpha tR^2$, $Fo_slab = \frac\alpha tL^2 = Fo_cyl \times \fracR^2L^2 = Fo_cyl \times \frac0.04^20.06^2 = 0.444 Fo_cyl$ $$Q = \dotm m c p