Sure! Let’s break down **System 6** step by step: --- ## **Given Data- **Brine NaCl concentration**: 13% (by weight) - **Initial cheese NaCl concentration**: .25% - **Initial cheese moisture (wet basis)**: 66% - **Convective mass transfer coefficient (\( k_c \))**: \( 3 \times 10^{-7} \) m/s - **Diffusivity of NaCl in cheese (\( D \))**: \( 1.1 \times 10^{-10} \) m²/s - **Cheese dimensions**: \( 10 \) cm x \( 7 \) cm x \( 3 \) cm - **Midpoint for analysis**: \( x = 5 \) cm - **Equilibrium distribution coefficient (\( K \))**: \( 1.35 \) kg brine / kg cheese - **Brine tank volume**: 50 L - **Number of cheese blocks**: 10 --- ## **Step 1: Model the Salt Diffusion** This is a classic **unsteady-state diffusion problem** (Fick’s 2nd law) with convection at the surface. ### **Assumptions** - 1D diffusion (through the 10 cm thickness) - Constant brine concentration at the surface (large volume, or well-stirred) - The cheese is initially free (or nearly free) of salt ### **Relevant equation** For a slab of thickness \( 2L \) (here, \( L = 5 \) cm = .05 m is the half-thickness), the solution for concentration at \( x = L \) (center) is: \[ \frac{C(x,t) - C_i}{C_s - C_i} = 1 - \frac{4}{\pi} \sum_{n=}^\infty \frac{1}{2n+1} \exp \left[ -\frac{D (2n+1)^2 \pi^2 t}{4L^2} \right] \cos \left( \frac{(2n+1)\pi x}{2L} \right) \] Where: - \( C(x,t) \) = salt concentration at position \( x \) and time \( t \) - \( C_i \) = initial concentration in cheese (.25%) - \( C_s \) = surface concentration (equilibrium with brine) - \( D \) = diffusivity - \( L \) = half-thickness = .05 m - \( x \) = position (midpoint, so .05 m) - \( t \) = time (in seconds) --- ## **Step 2: Find Surface Concentration \( C_s \)** The equilibrium is set by the partition coefficient \( K \): \[ K = \frac{C_{brine}}{C_{cheese,surface}} \implies C_{cheese,surface} = \frac{C_{brine}}{K} \] Given \( C_{brine} = 13\% = .13 \) (mass fraction), \( K = 1.35 \): \[ C_s = \frac{.13}{1.35} = .0963 \text{ (or 9.63%)} \] --- ## **Step 3: Plug in the Values** - \( C_i = .25\% = .0025 \) - \( C_s = 9.63\% = .0963 \) - \( L = .05 \) m - \( x = .05 \) m (center) - \( D = 1.1 \times 10^{-10} \) m²/s ### **Time points (convert hours to seconds):** - 1 h = 360 s - 3 h = 10800 s - 10 h = 36000 s - 24 h = 86400 s - 72 h = 259200 s --- ## **Step 4: Calculate Concentration at the Center** For \( x = L \), \( \cos[\frac{(2n+1)\pi x}{2L}] = \cos(\frac{(2n+1)\pi}{2}) = \) for all odd \( n \), alternating signs. But for \( x = L \): \[ \frac{C(L,t) - C_i}{C_s - C_i} = 1 - \frac{4}{\pi} \sum_{n=}^\infty \frac{1}{2n+1} \exp \left[ -\frac{D (2n+1)^2 \pi^2 t}{4L^2} \right] \cos \left( \frac{(2n+1)\pi}{2} \right) \] But at \( x = L \), \( \cos[(2n+1)\pi/2] \) alternates between and \(\pm 1\). - \( n=: \cos(\pi/2) = \) - \( n=1: \cos(3\pi/2) = \) So, at the center, the sum simplifies to only terms where \( \cos \) is nonzero. But for a slab with both sides exposed, we use the above formula, and at the midpoint, the cosine term alternates as stated. For simplicity, **let’s use the first term (n=) for an approximation**: \[ \frac{C(L,t) - C_i}{C_s - C_i} \approx 1 - \frac{4}{\pi} \exp \left[ -\frac{D \pi^2 t}{4L^2} \right] \cdot \cos \left( \frac{\pi}{2} \right) \] But \( \cos(\pi/2) = \), so the center is initially unaffected; as more terms are added, the approximation improves. However, in practice, at the center: \[ \frac{C(L,t) - C_i}{C_s - C_i} = 1 - \frac{4}{\pi} \sum_{n=}^\infty \frac{1}{2n+1} \exp \left[ -\frac{D (2n+1)^2 \pi^2 t}{4L^2} \right] \cdot (-1)^n \] So, the first few terms: - For \( n = \): \( (-1)^ = 1 \) - For \( n = 1 \): \( (-1)^1 = -1 \) - For \( n = 2 \): \( (-1)^2 = 1 \) - etc. \[ \frac{C(L,t) - C_i}{C_s - C_i} = 1 - \frac{4}{\pi} \left[ \frac{1}{1} e^{-D\pi^2 t/4L^2} - \frac{1}{3} e^{-9D\pi^2 t/4L^2} + \frac{1}{5} e^{-25D\pi^2 t/4L^2} - ... \right] \] --- ## **Step 5: Plug in Numbers** Let’s compute the exponential term: \[ \alpha = \frac{D \pi^2}{4L^2} \] \[ D = 1.1 \times 10^{-10} \, \text{m}^2/\text{s} \] \[ L = .05 \, \text{m} \] \[ \alpha = \frac{1.1 \times 10^{-10} \times (\pi^2)}{4 \times (.05)^2} \] \[ = \frac{1.1 \times 10^{-10} \times 9.8696}{.01} = \frac{1.086 \times 10^{-9}}{.01} = 1.086 \times 10^{-7} \] So for each \( t \) value: \[ e^{- \alpha t} \] where \( \alpha = 1.086 \times 10^{-7} \) #### **Example for 1 hour (\( t = 360 \) s):** \[ \alpha t = 1.086 \times 10^{-7} \times 360 = 3.9096 \times 10^{-4} \] \[ e^{-3.9096 \times 10^{-4}} \approx .99961 \] Plug into the series: \[ \frac{C(L,t) - C_i}{C_s - C_i} \approx 1 - \frac{4}{\pi} [.99961 - \frac{1}{3} e^{-9\times3.9096\times10^{-4}} + \frac{1}{5} e^{-25\times3.9096\times10^{-4}} - ... ] \] \[ e^{-.00352} \approx .9965, \quad e^{-.00977} \approx .9903 \] So, \[ \frac{C(L,t) - C_i}{C_s - C_i} \approx 1 - \frac{4}{\pi} [.99961 - .3322 + .1981] \] \[ = 1 - \frac{4}{\pi} [.8655] = 1 - 1.101 [.8655] = 1 - .953 = .047 \] So after 1 hour, about 4.7% of the way from initial to equilibrium at the center. \[ C(L,1h) = C_i + .047 \times (C_s - C_i) = .0025 + .047 \times (.0963 - .0025) = .0025 + .047 \times .0938 = .0025 + .00441 = .00691 \text{ (or .691%)} \] --- ### **Repeat for other times:** You can repeat the above process for each time by recalculating \( \alpha t \) and the exponentials. --- ## **Step 6: Plot and Discuss Relationship** Plot \( C(L, t) \) vs. \( t \) (on linear or log scale). You will see a typical diffusion curve: rapid initial uptake, slowing as equilibrium is approached. --- ## **Step 7: Salt Loss from Brine** - Each cheese block: \( 10 \times 7 \times 3 = 210 \) cm³ = .21 L - 10 blocks: 2.1 L cheese - Initial brine: 50 L at 13% = \( .13 \times 50 = 6.5 \) kg NaCl At equilibrium, cheese will have \( C_s = 9.63\% \) NaCl. - Total in cheese: \( 2.1 \) L \(\approx\) 2.1 kg (assuming density ~1) - \( 2.1 \) kg \(\times\) .0963 = .202 kg If initial cheese had \( .25\% \): \( 2.1 \times .0025 = .00525 \) kg - Net uptake: \( .202 - .00525 \approx .197 \) kg NaCl moved into cheese So, salt in brine will decrease by \(\approx .197\) kg. --- ## **Step 8: Comments** - Salt uptake follows a diffusion curve, rapid at first, slows as equilibrium is reached. - The brine loses a small amount of salt relative to its total content. - The equilibrium is set by the distribution coefficient, limiting ultimate salt uptake in cheese. --- # **Final Answers Summary** ### **1. Center concentration of salt at various times:** - Use the solution above; at 1 hour, center NaCl ≈ **.691%**. - Repeat for other times using the formula. ### **2. Plot the concentration vs. time** - The graph would show a rising curve, leveling off as time increases. ### **3. Salt loss from brine (with 10 cheese blocks):** - About **.197 kg** of salt is lost from the brine (assuming density ~1 kg/L). --- **If you need the exact values for each time point or a plot, let me know!**