Answer:
a) Δd(change in wood diameter) = 5%
b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter
C) new diameter (D2) = 10.5 in
Explanation:
Wood pole diameter = 10 inches
moisture content = 5%
FSP = 30%
A) The percentage change in the wood's diameter
note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter
Δd/d = 1/5(30 - 5)
Δd/d = 5%
Δd = 5%
B) would the wood swell or shrink
The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter
C) The new diameter of the wood
D2 = D + D(
)
D = initial diameter= 10 in , M1 = initial moisture content = 5%
therefore D2 = 10 + 10( 5/100 )
new diameter (D2) = 10.5 in
Answer:
option c is correct
47.2%
Explanation:
given data
consisting of refrigerant = 134 a
volume V = 0.01 m³/kg
pressure P = 1MPa = 1000 kPa
to find out
quality of the R 134a
solution
we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a
that is
Vf = 0.0008701 m³/kg
Vv = 0.0203 m³/kg
so we will apply here formula that is
quality = (V - Vf) / (Vv - Vf) ............1
put here value
quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )
quality = 0.4698
so quality is 47 %
SO OPTION C IS CORRECT
The answer is b i did the same thing and i got it right
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