Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.
The specific gravity of the wood chips is 0.494.
The average volume of a log is
The weight of one log is
To provide 3617 ton/day of wood chips, we need
The feed rate of logs is 14.3 logs/min.
Answer:
Speed of Sound
Explanation:
Speed of sound, fluid phases
m/s
notes
WEL 206
use 323 27 °C
CRC 323 27 °C
WEL 319
Electrons are orbiting around the nucleus in a specific energy level as described in Bohr's atomic model. There are 7 energy levels all in all; 1 being the strongest and nearest to the nucleus, and 7 being the weakest and farthest away from the nucleus. Electron can transfer from one energy level to another. If it increases energy, it absorbs energy. If it goes down an energy level, it emits energy in the form of light. This light can be measure in wavelength through the Rydberg equation:
1/λ =R(1/n₁² -1/n₂²), where
λ is the wavelength
R is the Rydberg constant equal to 1.097 × 10⁻7<span> per meter
n</span>₁ and n₂ are the energy levels such that n₂>n₁
In the Paschen series is an emission spectrum of hydrogen when the energy level is at least n=4. So, this covers n=4 to n=7.
1/λ =(1.097 × 10⁻7)(1/4² -1/7²)
λ = 216.57 ×10⁻⁶ m or 216.57 μm
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
