Let's break down the step-by-step solution for each part of the problem, focusing primarily on the **first question** about the **evaporation rate**, as that's the initial quantitative task. --- ## 1. **Calculate the Evaporation Rate** Given data: | Temperature (K) | 313 | 323 | 333 | 343 | 353 | 363 | 373 | |-----------------|-----|-----|-----|-----|-----|-----|-----| | Moisture % (wb) | 42.5| 38.2| 31.6| 25.4| 18.2| 12.8| 7.5 | **Step 1: Convert Moisture Content to Dry Basis** Moisture content on a wet basis (wb) is given. \[ \text{Moisture % (wb)} = \frac{\text{mass of water}}{\text{mass of wet sample}} \times 100 \] Moisture content on a dry basis (db): \[ \text{Moisture % (db)} = \frac{\text{mass of water}}{\text{mass of dry solids}} \times 100 \] Let \( x_{wb} \) be the moisture fraction (wb): \[ x_{wb} = \frac{\text{mass of water}}{\text{mass of water} + \text{mass of dry solids}} \] So, \[ \text{mass of dry solids} = \text{mass of wet sample} \times (1 - x_{wb}) \] \[ \text{mass of water} = \text{mass of wet sample} \times x_{wb} \] \[ x_{db} = \frac{x_{wb}}{1 - x_{wb}} \] **Step 2: Calculate the Amount of Water Removed** Given: - Initial feed = **850 kg/h** - Initial moisture (wb) = **let's use the first value, 42.5%** - Final moisture (db) = **2.5%** Let’s find the mass of water removed. **A. Find mass of dry solids in the feed:** Initial moisture (wb): \( x_{wb,1} = .425 \) Let \( M_f = 850 \) kg/h (mass of wet feed). \[ \text{Mass of dry solids} = 850 \times (1 - .425) = 850 \times .575 = 488.75\ \text{kg} \] **B. Find final mass of product:** Final moisture (db): \( x_{db,2} = .025 \) \[ x_{wb,2} = \frac{x_{db,2}}{1 + x_{db,2}} = \frac{.025}{1 + .025} = \frac{.025}{1.025} = .0244 \] So, mass of final wet product: \[ \text{Final mass} = \text{Mass of dry solids} + \text{Final mass of water} \] \[ \text{Final mass of water} = 488.75 \times .025 = 12.22\ \text{kg} \] \[ \text{Final mass} = 488.75 + 12.22 = 500.97\ \text{kg} \] **C. Amount of water removed:** \[ \text{Initial water in feed} = 850 - 488.75 = 361.25\ \text{kg} \] \[ \text{Final water in product} = 12.22\ \text{kg} \] \[ \text{Water removed} = 361.25 - 12.22 = 349.03\ \text{kg} \] **D. Water removed per hour:** \[ \boxed{349.03\ \text{kg/h}} \] Or, per kg of initial feed: \[ \frac{349.03}{850} = .41\ \text{kg water/kg feed} \] --- ## 2. **Evaporation Rate at Different Temperatures** You can use the moisture values for each temperature to plot the drying curve (change of moisture content with time/temperature) and estimate the evaporation rate during each interval (difference in water content divided by time or mass flow). --- ## **Final Answers:** 1. **Water removed per hour:** \[ \boxed{349.03\ \text{kg/h}} \] 2. **Water removed per kg of feed:** \[ \boxed{.41\ \text{kg water/kg feed}} \] --- If you want solutions for the other sub-questions (humidity ratio, C_out(t), Antoine equation, etc.), let me know which part to solve next!

# Evaporation Rate Calculation and Analysis To solve the problem regarding the evaporation rate and moisture content, let’s first verify and lay out the correct data in a structured format. ## 1. **Given Data** ### Temperature and Moisture Content | Temperature (°C) | 55 | 57.5 | 59 | 56.5 | 55 | 53 | 57.5 | |-------------------|-----|-------|-----|------|-----|-----|------| | Moisture % (wb) | 42.5| 38.7 | 35.6| 25.4 | 18.8| 13.2| 5.5 | ### Initial Conditions - Initial feed: **850 kg/h** - Final moisture content on a dry basis: **2.1%** ## 2. **Calculating Evaporation Rate** ### Step 1: Convert Moisture Content to Dry Basis Given the moisture content on a wet basis (wb), we can convert it to dry basis (db) using the formula: \[ \text{Moisture % (db)} = \frac{x_{wb}}{1 - x_{wb}} \times 100 \] Where \( x_{wb} \) is the moisture fraction. #### Example Calculation for 42.5% wb: \[ x_{wb} = 0.425 \] \[ x_{db} = \frac{0.425}{1 - 0.425} = \frac{0.425}{0.575} \approx 0.7391 \quad (73.91\% \text{ on dry basis}) \] ### Step 2: Calculate the Mass of Water Removed #### A. Mass of Dry Solids in the Feed Using the first moisture value (42.5% wb): \[ \text{Mass of dry solids} = 850 \times (1 - 0.425) = 850 \times 0.575 = 488.75\ \text{kg} \] #### B. Mass of Water in the Final Product Using the final moisture content on dry basis (2.1%): \[ x_{db,2} = 0.021 \] Convert to wb: \[ x_{wb,2} = \frac{0.021}{1 + 0.021} = \frac{0.021}{1.021} \approx 0.0206 \] Calculating final mass of water: \[ \text{Final mass of water} = 488.75 \times 0.0206 \approx 10.06\ \text{kg} \] #### C. Initial and Final Water Mass Initial water mass in the feed: \[ \text{Initial water} = 850 - 488.75 = 361.25\ \text{kg} \] Final water in product: \[ \text{Final water} = 10.06\ \text{kg} \] ### Step 3: Amount of Water Removed \[ \text{Water removed} = \text{Initial water} - \text{Final water} = 361.25 - 10.06 \approx 351.19\ \text{kg} \] ### Step 4: Water Removed per Hour \[ \text{Evaporation rate} = \frac{351.19}{850} \approx 0.413\ \text{kg water/kg feed} \] ## Final Results 1. **Total water removed per hour:** \[ \boxed{351.19\ \text{kg/h}} \] 2. **Water removed per kg of feed:** \[ \boxed{0.413\ \text{kg water/kg feed}} \] ## 3. **Further Analysis** If required, I can proceed with other calculations such as: - Humidity ratio and its relationship with temperature. - The vapor pressure using the Antoine equation. - Solubility curves for crystallization processes. Let me know if you would like to explore any of these additional topics!