# Archived Heat transfer

#### tweety1234

1. Homework Statement

In an experimental study, slurry is being concentrated in a continuous vacuum evaporator at a
rate of 70 kg/hour. The feed material has a temperature of 16°C and a total solids content of
11%. A product of 35% total solids is withdrawn at a temperature of 42°C. The evaporator
operates at a pressure of 7.375 kPa and removes vapour at 40°C. This vapour is fed to a
condenser, from which condensate at 40°C is removed.
i) Using a mass balance table over the evaporator, calculate the flowrates of the
product and vapour streams.
I have attached a diagram, for this question, I am stuck on how to calculate the flow rate.

there is 70kg/h of feed going into the evaporator, of which 11% is solids, so that's 7.7kg/h of solids, and so 62.3kg/h of liquid in the feed.

so now how do I calculate the amount of water in the vapour stream and the amount of solids in the product? I intially thought you just work out 35% of the 70kg/h feed for the product, since it says there is 35% solids in the product, but this is not correct.

Can someone please show/explain how to work out the flow rate of the product tream and vapour stream.

Thank you.

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#### MexChemE

Start out with a solids balance on the evaporator
$$0.11 \left(70 \frac{kg}{h} \right) = 0.35 F_3$$
$$F_3 = 70 \frac{kg}{h} \left( \frac{0.11}{0.35} \right) = 22 \frac{kg}{h}$$
This is the product stream flowrate. Now, we do an overall mass balance on the evaporator
$$F_1 = F_2 + F_3$$
$$F_2 = 70 \frac{kg}{h} - 22 \frac{kg}{h} = 48 \frac{kg}{h}$$
This is the vapor stream flowrate. There's not enough info to calculate the heat input of the evaporator (4), but we can calculate the mass flowrate of cooling water (F6 = F7). We get the latent heat of the process steam from saturated steam tables
$$h_{fg} = 2406.06 \frac{kJ}{kg}$$
Then, we calculate the mass flowrate of water using the following equation
$$Q = \dot{m} c \Delta T$$
$$F_6 = F_7 = \frac{h_{fg} F_2}{c \Delta T} = \frac{ \left(2406.06 \frac{kJ}{kg} \right) \left(48 \frac{kg}{h} \right)}{\left(4.184 \frac{kJ}{kg \cdot K} \right)(30°C - 10°C)} = 1380.15 \frac{kg}{h}$$

"Heat transfer"

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