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I have done a permeability experiment on LDPE in the presence of air. At $\pu{30^\circ C}$, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

$$ \begin{align} \text{Volume of bag}&= \pu{0.2064 m^3}\\ \text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\ \text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\ \text{Time }&= \pu{5 \times 60s} = \pu{300s}\\ \end{align}$$$$ \begin{align} \text{Volume of bag}&= \pu{0.2064 m^3}\\ \text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\ \text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\ \text{Time }&= 5 \times \pu{60s} = \pu{300s}\\ \end{align}$$

Is it correct if I convert barrer to SI units in the following manner?

$$\pu{1 barrer = 7.5005\times 10^{-18} m^{2}s^{-1}Pa^-1}$$$$\pu{1 barrer} = 7.5005\times10^{-18}\ \mathrm{m^2\ s^{-1}\ Pa^{-1}}$$

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make any sense at all!

$\displaystyle\text{Permeability} = \frac{\text{Volume}\times\text{thickness}}{\text{Area}\times\text{time}\times\text{Change in Pressure}}$

I have done a permeability experiment on LDPE in the presence of air. At $\pu{30^\circ C}$, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

$$ \begin{align} \text{Volume of bag}&= \pu{0.2064 m^3}\\ \text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\ \text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\ \text{Time }&= \pu{5 \times 60s} = \pu{300s}\\ \end{align}$$

Is it correct if I convert barrer to SI units in the following manner?

$$\pu{1 barrer = 7.5005\times 10^{-18} m^{2}s^{-1}Pa^-1}$$

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make any sense at all!

$\displaystyle\text{Permeability} = \frac{\text{Volume}\times\text{thickness}}{\text{Area}\times\text{time}\times\text{Change in Pressure}}$

I have done a permeability experiment on LDPE in the presence of air. At $\pu{30^\circ C}$, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

$$ \begin{align} \text{Volume of bag}&= \pu{0.2064 m^3}\\ \text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\ \text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\ \text{Time }&= 5 \times \pu{60s} = \pu{300s}\\ \end{align}$$

Is it correct if I convert barrer to SI units in the following manner?

$$\pu{1 barrer} = 7.5005\times10^{-18}\ \mathrm{m^2\ s^{-1}\ Pa^{-1}}$$

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make any sense at all!

$\displaystyle\text{Permeability} = \frac{\text{Volume}\times\text{thickness}}{\text{Area}\times\text{time}\times\text{Change in Pressure}}$

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How to apply the Conversion of a permeability value found from an experiment?barrer to SI unit

I have done a permeability experiment on LDPE in the presence of air. At 30'C$\pu{30^\circ C}$, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

Volume of bag= 0.2064 m^3
Approximate area = 1.35 m^2 (this is just a quick approximate)
Thickness of LDPE = 0.042 *10^-3 m
Time = 5 * 60s = 300s

$$ \begin{align} \text{Volume of bag}&= \pu{0.2064 m^3}\\ \text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\ \text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\ \text{Time }&= \pu{5 \times 60s} = \pu{300s}\\ \end{align}$$

Is it correct if I convert barrer to SI units in the following manner?

 1 barrer = 7.5005*10^-18 m^2.s^-1.Pa^-1

$$\pu{1 barrer = 7.5005\times 10^{-18} m^{2}s^{-1}Pa^-1}$$

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make any sense at all!!

Permeability = Volume*thickness/Area*time*Change_in_Pressure

Any help or suggestions will be greatly appreciated!!$\displaystyle\text{Permeability} = \frac{\text{Volume}\times\text{thickness}}{\text{Area}\times\text{time}\times\text{Change in Pressure}}$

How to apply the permeability value found from an experiment?

I have done a permeability experiment on LDPE in the presence of air. At 30'C, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

Volume of bag= 0.2064 m^3
Approximate area = 1.35 m^2 (this is just a quick approximate)
Thickness of LDPE = 0.042 *10^-3 m
Time = 5 * 60s = 300s

Is it correct if I convert barrer to SI units in the following manner?

 1 barrer = 7.5005*10^-18 m^2.s^-1.Pa^-1

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make sense at all!!

Permeability = Volume*thickness/Area*time*Change_in_Pressure

Any help or suggestions will be greatly appreciated!!

Conversion of a permeability value from barrer to SI unit

I have done a permeability experiment on LDPE in the presence of air. At $\pu{30^\circ C}$, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

$$ \begin{align} \text{Volume of bag}&= \pu{0.2064 m^3}\\ \text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\ \text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\ \text{Time }&= \pu{5 \times 60s} = \pu{300s}\\ \end{align}$$

Is it correct if I convert barrer to SI units in the following manner?

$$\pu{1 barrer = 7.5005\times 10^{-18} m^{2}s^{-1}Pa^-1}$$

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make any sense at all!

$\displaystyle\text{Permeability} = \frac{\text{Volume}\times\text{thickness}}{\text{Area}\times\text{time}\times\text{Change in Pressure}}$

1
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How to apply the permeability value found from an experiment?

I have done a permeability experiment on LDPE in the presence of air. At 30'C, the permeability was found to be 1.846 barrer.

Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:

Volume of bag= 0.2064 m^3
Approximate area = 1.35 m^2 (this is just a quick approximate)
Thickness of LDPE = 0.042 *10^-3 m
Time = 5 * 60s = 300s

Is it correct if I convert barrer to SI units in the following manner?

 1 barrer = 7.5005*10^-18 m^2.s^-1.Pa^-1

By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make sense at all!!

Permeability = Volume*thickness/Area*time*Change_in_Pressure

Any help or suggestions will be greatly appreciated!!