Question

1. An important aspect of food processing is being able to follow the flow of materials as they travel through each step of the operation. You need to know where your losses are occurring and how large they are. This is particularly important if you are going to be able to make any improvements to improve the overall yield by reducing these losses. In the following example, we will examine a hypothetical process and perform some mass balance and heat calculations on it in order to gain a better understanding of how it is operating.
   
   A processor has 800 kg of fresh cranberries which are to be made into cranberry sauce. Approximately 6% of the total weight of the cranberries is stems plus other waste material or inferior quality berries that will be removed from the cranberry sauce during its processing. The processor likes the finished cranberry sauce to be somewhat lower in moisture content than the initial 87% moisture of the cranberries. 80% moisture is viewed as optimal since much of this product will be used in applications where a lower moisture content is more desirable.

   Consider the following fictitious process:         

   i).      The cranberries are sent to a temporary storage area where they are held at 8⁰C. Assume that there is no weight change in this step of the process.

   ii).     The entire 800 kg batch of cranberries is washed in a large vat of water at 8⁰C and checked for quality. The stems and other waste material which float to the surface (plus any inferior berries) are then skimmed off and immediately discarded. The remaining cranberries are dried gently to remove the excess surface moisture from the washing step.

   iii).    The remaining cranberries are mashed to expose the contents of the berries. They are then heated to their boiling point of 103⁰C (sugars present in the cranberries raise its boiling point beyond that of pure water).

   iv).    The desired amount of water is boiled off (the mixture is still at 103⁰C).

   v).     The remaining product is cooled to 25⁰C and filled into plastic containers.

   Assumptions:            No water is added to the initial mixture

   All parts of the cranberry contain 87% moisture.  

   This is actually not true, but it will simplify the problem for you.

   Note:   This is NOT a terribly difficult question from a mathematical point of view. However, it is rather involved and needs to be approached in a structured or disciplined manner. I would urge you (actually, I beg you) to draw a flow diagram showing where the various materials go as they are being processed. Each step in the process may be represented by a separate square on your diagram. You can include temperatures etc. on this diagram and show where materials are added or removed to help you follow the process. Take your time with this problem and don’t become frustrated with it.

   Please allow me to “ramble on” a little bit about my past experiences with questions such as this.

   The key to solving this problem is to organize the information in such a way as to allow you to understand what is happening. I have given problems similar to this in Assignment 2 for quite a few years and I always emphasize that the problem cannot be solved unless you sit down and draw a flow diagram to illustrate what is happening. Invariably, I end up with a flood of e-mails saying that they cannot do the problem or that the problem is too weird to do. When I ask to see a copy of their flow diagram, they tell me that they haven’t done one. It seems to me that they consider drawing a diagram as a sign of some intellectual weakness and that they should be able to follow the flows of materials in their heads without the use of such trivial aids as drawing a picture. You cannot possibly calculate the weights of water and solids throughout the process if you have no idea where things are going. The flow diagram is actually your road map to the final solution. Once you understand where things are going, everything else should start to fall into place.

   Thank you for indulging me and letting me share my frustrations with you.

   Now you can get to work on the problem. (18 marks total)

   First, draw a process flow diagram to show what is happening

   Then, calculate the following:  

   a).    The weight of stems and other waste material that is removed. Express your answer to one decimal place.

   b).    The amount of water removed from the cranberry sauce (after the stems and other waste material have been removed).  

   c).     The amount of cranberry sauce obtained from the initial 800 kg of cranberries.  

   d).    The amount of heat that must be added to cook the cranberry sauce and drive off the desired amount of water. For simplicity, use 2,257kJ/kg as the heat required to evaporate 1 kg of water at 103⁰C. While this is not the exact number for this temperature, it is close enough for the purpose of this problem.

   e).    The amount of heat that must be removed from the finished cranberry sauce to cool it to the desired final temperature. Recognizing that the final cranberry sauce has a lower water content than the initial cranberries, we need to reflect this in the value of the specific heat capacity for the finished product. I have taken the liberty of calculating the Cp of the final 80% moisture cranberry sauce as being approximately 3.517 kJ/kg K. Please use this value in your calculations if needed.   

Answer 1

The flow diagram is like this

Raw material (800 kg) (washing and sorting at 8 C) -->  berries to be crushed (752 kg) _-->heating to 103 for removal of water ---> cooling at 25 C and packing

First consider the data given

The weight of the raw unprocessed cranberries: 800 kg

The waste material: 6% of 800 kg = 48 kg

Final weight undergoing process: 800 – 48 = 752 kg

The moisture content initially is 87% of the initial weight, thus weight of water initially present = 654.24 kg

The mass of the cranberry: 752 – 654.24 = 97.76 kg

The final weight is using 20% = 97.76 kg, = 488.8 kg

The moisture content finally is 80% of the final weight , the weight of water finally present = 391.04 kg

The amount of water removed : 654.24 – 391.04 = 263.2 kg

Heat required for the purpose: 263.2 x 2257 = 294042.4 kJ

The amount of heat required to remove, to lower the temp from 103 C to 25 C is calculated by
Q = m. Cp . dT = 391.04 x 3.517 x {(103 – 25) + 273.15} = 482,932.27 kJ