Is This Mass Exchanger Design Feasible?

[Ciocolatta]

Homework Statement

An evaporator is required to move 1000kgh-1 of water from a milk stream.
Steam is available for heating the evaporator (by condensation) at a pressure of 70.1 kPa absolute and a temperature of 90 decrees celcius.
The vapour, into which the water is evaporated from the milk, is at a pressure of 47.4kPa absolute and a temperature of 80 degrees celcius.

Number of tubes: 50
Tube inside diameter: 50mm
Tube length: 3m
tube thickness: 3mm

do one pass through calculation to

a) estimate an overall mass transfer coefficient for the evaporation
b) estimate the driving force for mass transfer
c) compare the mass transfer area required wit hthe original mass transfer area from the specification of the design variables.

Homework Equations

antoine equation:
pressure(Pa) = 133.3 exp (18.3036 - 3816.44/(T(in celcius) + 277.02))

heat of vaporisation for water/steam = 2200 kJ/kg

properties for steam (both condensing and evaporating):
K (W/mK) = 0.023
density (kg/m3) = 0.45
Diff (m2/s) = 1.5 x 10^5
viscosity (kg/ms) = 1 x 10^-5

thermal conductivity of tube material: 45 W/mK
condensation heat transfer coefficient for steam at 90 degrees celcius: 10,000 W/m2K

overall mass transfer coefficient (m/s) from gas phase perspective:
K = 1 / [1/k(gas) + H(gas->liquid)/k(liquid)]
k(gas) = gas film mass transfer coeff
k(liquid) = liquid film mass transfer coeff
H(gas->liquid) = henry's law constant between liquid and gas (dimensionoless)

N = K * A * deltaC(LMTD)
N = mass transfer rate (kg/s)
A = mas transfer area (m2)
deltaC(LMTD) = log mean concentrarion driving force (kg/m3) = deltaC1-deltaC2/ln(deltaC1/deltaC2))
C1 and C2 are concentration driving forces at one and the other end of the mass exchanger.

The Attempt at a Solution

this is a design question...so we're allowed to make assumptions where necessary. But the problem is...I'm not sure if I'm allowed to assume some mass transfer coefficients or if I can calculate them with the information above.
I don't really understand how to do this.

im assuming the gas MT coeff = 20, and liquid MT coeff = 200 which fall in the typical ranges


and i think henry's constant for gas --> liquid = gas mole frac/ liquid mole frac
where gas mole frac = absolute pressure of water in steam / absolute pressure of steam = 47.4/70.1

im really stuck... =/

 

[mark726]

Thank you for your post. I am a scientist and I would be happy to assist you with your questions.

To estimate the overall mass transfer coefficient for the evaporation, we can use the equation provided in the homework, which is K = 1 / [1/k(gas) + H(gas->liquid)/k(liquid)]. As you mentioned, we can assume the gas mass transfer coefficient (k(gas)) to be 20 and the liquid mass transfer coefficient (k(liquid)) to be 200. However, we do not need to assume the Henry's law constant (H(gas->liquid)) as it can be calculated using the information provided.

H(gas->liquid) = gas mole frac/ liquid mole frac = (absolute pressure of water in steam / absolute pressure of steam) = (47.4/70.1) = 0.676

Using this value of H(gas->liquid), we can now calculate the overall mass transfer coefficient (K) as follows:

K = 1 / [1/20 + 0.676/200] = 0.023 m/s

To estimate the driving force for mass transfer, we can use the equation N = K * A * deltaC(LMTD). Here, N is the mass transfer rate, K is the overall mass transfer coefficient calculated above, A is the mass transfer area, and deltaC(LMTD) is the log mean concentration driving force. We can calculate deltaC(LMTD) using the information provided as follows:

deltaC(LMTD) = log mean concentration driving force (kg/m3) = deltaC1-deltaC2/ln(deltaC1/deltaC2))
C1 and C2 are concentration driving forces at one and the other end of the mass exchanger.

Since we are given the mass flow rate (1000 kg/h), we can calculate the concentration driving force using the density of water (1000 kg/m3) and the mass flow rate as follows:

deltaC = mass flow rate / density = (1000 kg/h)/(1000 kg/m3) = 1 kg/m3

Therefore, deltaC1 = deltaC2 = 1 kg/m3

Plugging these values into the equation for deltaC(LMTD), we get:

deltaC(LMTD) = ((1-1)/ln(1/1)) = 0

Now, we