Pulp & Paper Questions and Answers – Wood Moisture Content and Density

This set of Pulp and Paper Multiple Choice Questions and Answers (MCQs) focuses on “Wood Moisture Content and Density”.

1. The specific gravity (specific gravity) of wood’s the oven-dry weight of wood divided by the weight of _________ of H2O.
a) Mass
b) Density
c) Volume
d) Viscosity

Explanation: This produces a unitless number. The displaced volume of H2O can be calculated by the volume of the wood if it’s of even shape such as a rectangular block.

2. The basic sp. gr. of woods is commonly b/w 0.35 and 0.60, but can change from 0.2 – 0.7.
a) True
b) False

Explanation: The basic sp. gr. of cell wall material’s approx. 1.50. The basic specific gravity of woods is commonly b/w 0.35 and 0.60, but can vary from 0.2 to 0.7.

3. The viscosity of a material’s defined as the mass/unit volume.
a) True
b) False

Explanation: The density of a material is defined as the mass/volume. When the units of pounds or ounces are utilized one actually obtains a weight density. For wood, it’s customary to take the total mass / by the volume both at the same moisture content.

4. The _________of wood’s a measure of the H2O content relative to either the total wet weight of material or to the weight of oven dried wood material.
a) Over-dry weight
b) Specific gravity
c) Moisture content
d) Green weight

Explanation: The moisture content of wood’s a measure of the H2O content relative to either the net wet weight of material the green weight of wood or to the weight of oven dried wood material the oven-dry basis.

5. After converting 20% MCQD to MCQR the value would be _________
a) 10%
b) 15%
c) 5%
d) 17%

Explanation: One could come up with about 17% MCGR. Or else, one could solve it for 16.7% MCGR. 20% MCQD means 20 parts H2O to 100 parts dry wood; this, in turn, means 20 parts water for 120 parts wet wood or 16.7% MCQR.
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6. A sample of wet Douglas-fir wood 20 mm thick, 50 mm wide and 10 cm long weighs 90.21 g. It is then dried at 105°Celcius to constant weight and re-weighed after cooling in a desiccator. The oven-dry weight is 44.37 g. What are the MCGR, MCQD, and basic specific gravity of the wood sample?
a) 1%, 10%, and 0.333
b) 75.2%, 103.8%, and 1.555
c) 4.8%, 106.2%, and 6.254
d) 50.8%, 103.8%, and 0.444

Explanation:
MCGR= (90.21 – 44.37) / 90.21 x 100%= 50.8%
MCOD= (90.21 – 44.37) / 44.37 x 100%= 103.3%
sp. gr.= (44.37 / 2x5x10) / 1= 0.444

7. What is the name of the hidden part?

a) Overdry weight of wood
b) Wet weight of wood
c) Weight of wood in sodium
d) Weight of lignin

Explanation: Here moisture contents over 100 percent are possible and commonly encountered. The moisture content of wood (green basis) is typically 50 percent, but varies from 30 – 60 percent. This tie in to 0.43 to 1.5 kg H2O per kg dry wood, (43-150 percent MCQD).

8. What is the name of the hidden part?

a) 100%-MCGR
b) MCOD-100%
c) 100%+MCGR
d) MCOD+100%

Explanation: Here moisture contents over 100 percent are possible and commonly encountered. The moisture content of wood (green basis) is typically 50 percent, but varies from 30 – 60 percent. This tally to 0.43 to 1.5 kg H2O per kg dry wood, (43-150 percent MCQD).

9. What is the name of the hidden part?

MC GR = {(weight of water in wood) / (________)} x 100%

a) Overdry weight of wood
b) Wet weight of wood
c) Weight of wood in sodium
d) Weight of lignin

Explanation: Here moisture contents over 100 percent are possible and commonly encountered. The moisture content of wood (green basis) is typically 50 percent, but varies from 30 – 60 percent. This consonant to 0.43 to 1.5 kg H2O per kg dry wood, (43-150 percent MCOD).

10. What is the name of the hidden part?

MCGR = {(MCOD) / (_______)} x 100%

a) 100%-MCGR
b) MCOD-100%
c) 100%+MCGR
d) MCOD+100%

Explanation: Here moisture contents over 100 percent are possible and commonly encountered. The moisture content of wood (green basis) is typically 50 percent, but varies from 30 – 60 percent. This coincide to 0.43 to 1.5 kg H2O per kg dry wood, (43-150 percent MCOD). .

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