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Answer:
W18 * 106
Explanation:
Given that the section modulus of the wide flange beam is 200 in^3 the lightest beam possible that can satisfy the section modulus must have a section modulus ≥ 200 in^3. also the value of the section modulus must be approximately closest to 200in^3
From wide flange Beam table ( showing the section modulus )
The beam that can satisfy the condition is W18 × 106 because its section modulus ( s ) = 204 in^3
Answer:
1.25 cm/day
Explanation:
An air thickness , (l) = 0.15 cm
Air Temperature =
Mass Diffusion coefficient (D) =
If the air pressure
We are to determine how fast will the water level drop in a day.
From the property of air at T = 20° C
from saturated water properties.
The mass flow of can be calculated as:
where:
R(constant) = 8.314 kJ/mol.K
Since 1 mole = 18 cm ³ of water
will be:
Again:
Converting the above value to cm/day: we have:
= 1.25 cm/day
∴ the rate at which the water level drop in a day = 1.25 cm/day
The effectiveness for a heat exchanger is 0.6428.
<u>Explanation:</u>
a) Sketch the temperature profiles for a 50% and 80% effectiveness heat exchanger.
The heat transfer type was not mentioned.
Assume the heat exchanger which is counter flow and the heat exchanger performance is high.
In the cold fluid stream, the 80% of effective heat exchanger compares the 50% of heat exchanger.
The diagram was attached below.
b) The effectiveness for a heat exchanger is calculated.
Given data
= 2 lb/s
= 160 F
= 4 lb/s
= 110 F
= 20 F
ε =
From tables,
= 1.001 Btu / lb ° F
×
= 1.001 × 2
= 2.002
= 0.24 Btu / lb ° F
= ×
= 0.24 × 4
= 0.96
ε =
=
ε = 0.6428