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Question 4 A packed tower with diameter 100 mm contains spheres of 4 mm diameter of a slightly soluble material B. Pure water enters the top of the tower at 300 K with a flowrate of 4 litres per minute. The total surface area of the spheres in contact with water is 0.02 m². The shrinkage of the spheres may be considered negligible as the bed is operated only for a few days at pseudo-steady state. The correlation for mass transfer from a bed of spheres is $\text{Sh} = 2.5 \text{Re}^{0.33} \text{Sc}^{0.33}$ where velocity is calculated based on empty tower and particle diameter is used as characteristic length. i. Define the Schmidt, Sherwood and Reynolds numbers, describing the ratios they represent. (6) ii. Calculate the mass transfer coefficient for transfer of B to the water. (5) iii. Calculate the bulk concentration of B in the outlet water stream. (5) iv. Rewrite the given correlation for mass transfer as a heat transfer correlation for spheres in a packed tower and explain the analogy between heat and mass transfer. Does it extend to momentum transfer here? (4) Data: Viscosity of water stream: 0.9 mPa s Density of water: 990 kg m³ Diffusivity of B in water: 0.001 mm² s?¹ Saturation concentration of B in water: 0.03 kmol m?³ The physical properties of water may be assumed not to vary with concentration of B. 

          Question 4
A packed tower with diameter 100 mm contains spheres of 4 mm
diameter of a slightly soluble material B. Pure water enters the top of
the tower at 300 K with a flowrate of 4 litres per minute. The total
surface area of the spheres in contact with water is 0.02 m². The
shrinkage of the spheres may be considered negligible as the bed is
operated only for a few days at pseudo-steady state.
The correlation for mass transfer from a bed of spheres is
$\text{Sh} = 2.5 \text{Re}^{0.33} \text{Sc}^{0.33}$
where velocity is calculated based on empty tower and particle
diameter is used as characteristic length.
i.
Define the Schmidt, Sherwood and Reynolds numbers,
describing the ratios they represent.
(6)
ii.
Calculate the mass transfer coefficient for transfer of B to the
water.
(5)
iii.
Calculate the bulk concentration of B in the outlet water
stream.
(5)
iv.
Rewrite the given correlation for mass transfer as a heat
transfer correlation for spheres in a packed tower and explain
the analogy between heat and mass transfer. Does it extend to
momentum transfer here?
(4)
Data:
Viscosity of water stream: 0.9 mPa s
Density of water: 990 kg m³
Diffusivity of B in water: 0.001 mm² s?¹
Saturation concentration of B in water: 0.03 kmol m?³
The physical properties of water may be assumed not to vary with
concentration of B.
Show more…
Question 4 A packed tower with diameter 100 mm contains spheres of 4 mm diameter of a slightly soluble material B. Pure water enters the top of the tower at 300 K with a flowrate of 4 litres per minute. The total surface area of the spheres in contact with water is 0.02 m². The shrinkage of the spheres may be considered negligible as the bed is operated only for a few days at pseudo-steady state. The correlation for mass transfer from a bed of spheres is Sh = 2.5 Re^0.33Sc^0.33 where velocity is calculated based on empty tower and particle diameter is used as characteristic length. i. Define the Schmidt, Sherwood and Reynolds numbers, describing the ratios they represent. (6) ii. Calculate the mass transfer coefficient for transfer of B to the water. (5) iii. Calculate the bulk concentration of B in the outlet water stream. (5) iv. Rewrite the given correlation for mass transfer as a heat transfer correlation for spheres in a packed tower and explain the analogy between heat and mass transfer. Does it extend to momentum transfer here? (4) Data: Viscosity of water stream: 0.9 mPa s Density of water: 990 kg m³ Diffusivity of B in water: 0.001 mm² s?¹ Saturation concentration of B in water: 0.03 kmol m?³ The physical properties of water may be assumed not to vary with concentration of B.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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A packed tower with a diameter of 100 mm contains spheres of 4 mm diameter made of a slightly soluble material B. Pure water enters the top of the tower at 300 K with a flow rate of 4 liters per minute. The total surface area of the spheres in contact with water is 0.02 m². The shrinkage of the spheres may be considered negligible as the bed is operated only for a few days at pseudo-steady state. The correlation for mass transfer from a bed of spheres is: Sh = 2.5Re^0.33Sc^0.33 where velocity is calculated based on the empty tower and particle diameter is used as the characteristic length. i. Define the Schmidt, Sherwood, and Reynolds numbers, describing the ratios they represent. (6) ii. Calculate the mass transfer coefficient for the transfer of B to the water. (5) iii. Calculate the bulk concentration of B in the outlet water stream. (5) iv. Rewrite the given correlation for mass transfer as a heat transfer correlation for spheres in a packed tower and explain the analogy between heat and mass transfer. Does it extend to momentum transfer here? (4) Data: Viscosity of water stream: 0.9 mPa s Density of water: 990 kg/m³ Diffusivity of B in water: 0.001 mm²/s Saturation concentration of B in water: 0.03 kmol/m³ The physical properties of water may be assumed not to vary with the concentration of B.
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Question 4 Question 5 Question 6 Question 7

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Transcript

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00:01 So here in this fifth question we are given the value of d which is equals to 0 .1 meter.
00:05 This is basically the diameter of this sphere.
00:08 We are given the value of initial temperature of this sphere which is equals to 90 degree celsius.
00:12 U infinity is given which is equals to 2 meter per second and the temperature at infinity is given which is equals to 20 degree celsius.
00:20 Here we are also given the properties of water.
00:24 So in that properties firstly we are given the density of the water which is equals to 4000 kilogram per meter cube.
00:32 We are given the value of k infinity which is equals to 0 .6 weber per meter kelvin.
00:37 We are also given the value of mu w which is equals to 10 raised to the power minus 3 kilogram per meter second and we are also given the value of cpw which is equals to 4000 and 200 joule per kilogram kelvin and we are also given the value of ks which is equals to no we are given the value of beta.
01:01 So the value of beta from here is equals to 2 .28 multiplied by the 10 raised to the power minus 4 kelvin inverse.
01:12 So this is about the properties of water.
01:14 Now we are also given the properties of steel.
01:17 So in this properties we are given the density of this steel which is equals to 7800 kilogram per meter cube.
01:24 We are given the value of ks which is equals to 50 weber per mille kelvin.
01:29 We are given the value of cps which is equals to 700 joule per kilogram kelvin.
01:35 So here in the first part of this question we have to find out the force we are discussing about the forced heat transfer case where the value of u infinity is 2 meter per second.
01:46 So we need to find out the value of h.
01:49 H from here is representing the convective heat transfer coefficient due to convection.
02:04 So according to the vitekhar force pairs we can say that nu is equals to hd divided by the k plug in to the value this is equals to 2 plus 0 .4 multiplied by the re raised to the power 1 divided by 2 plus 0 .06 re raised to the power 2 divided by 3 multiplied by the pr raised to the power 0 .4 which is multiplied by the mu infinity divided by the mu s raised to the power 1 divided by 4.
02:32 Simplifying this term from here basically mu s is representing the viscosity of the flowing fluid at the surface temperature here it is 90 degree where mu s from here become equals to 0 .315 which is multiplied by the 10 raised to the power minus 3.
02:52 So re from here become equals to rho v infinity d which is divided by mu.
02:57 That from here is equals to 10 raised to the power 3 multiplied by the 2 multiplied by which is divided by the 10 raised to the power minus 3.
03:05 Simplifying this term we get the value of re which will become equals to 2 multiplied by the 10 raised to the power 5.
03:11 And the value of pr here become equals to mu multiplied by the cp which is divided by k that is equals to 10 raised to the power minus 3 multiplied by the 4200 which is divided by 0 .6.
03:25 Simplifying this term we get the value of pr which is equals to 7.
03:29 So, putting the values here to get nu, nu become equals to 2 plus 0 .4 multiplied by the 2 multiplied by the 10 raised to the power 5 raised to the power 0 .5 plus 0 .06 multiplied by the 2 multiplied by the 10 raised to the power 5 raised to the power 2 divided by 3 multiplied by the 7 raised to the power 0 .4 multiplied by the 10 raised to the power minus 3 which is divided by the 0 .315 multiplied by 10 raised to the power minus 3 it's raised to the power 1 divided by 4.
04:04 Simplifying this term we get the value of nu which is equals to 1118 .57.
04:09 So, ht divided by the kw here in this case become equals to 1118 .57.
04:15 So, simplifying this term we get the value of h which become equals to 1118 .57 multiplied by the 0 .6 which is divided by 0 .1.
04:25 Simplifying this term h become equals to 6711 .42 weber per meter square kelvin hence the answer to the this part of this question.
04:37 Now, we know for a lumped heat analysis that t minus t infinity which is divided by ti minus t infinity which is equals to e raised to the power minus hat divided by the rho s v multiplied by the cps.
04:50 Let's say this is equation number 1 for any time we get the value of temperature t which is equals to 50 degree celsius.
04:56 So, plug into the value that is 50 minus 20 which is divided by 90 minus 20 is equals to e raised to the power minus 6711 .42 which is further multiplied by the 3t which is divided by 7800 multiplied by 0 .05 multiplied by the 700...
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