Three return steam lines in a chemical processing plant enter a collection tank operating at steady state at 1 bar. Steam enters
inlet 1 with flow rate of 0.8 kg/s and a quality of 0.9. Steam enters inlet 2 with flow rate of 2 kg/s at 200 C. Steam enters inlet 3 with flow rate of 1.2 kg/s at 95 C. Steam exits the tank at 1 bar. The rate of heat transfer from the collection tank is 40 kW. Neglecting kinetic and potential engery effects, determine for the steam exiting the tank
(a) the mass flow rate, in kg/s
(b) the temperature, in degrees C
(a) the mass flow rate, in kg/s
(b) the temperature, in degrees C
1 answer:
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Answer:
See calculation below
Explanation:
Given:
W = 30 kN = 30x10³ N
d = 75 mm
p = 6 mm
D = 300 mm
μ = tan Φ = 0.2
<u><em>1. Force required at the rim of handwheel </em></u>
Let P₁ = Force required at the rim of handwheel
Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm
Mean diameter of screw: *d = = (75 + 69) / 2 = 72 mm
and
tan α = p / πd = 6 / (π x 72) = 0.0265
∴ Torque required to overcome friction at he threads is T = P x d/2
T = W tan (α + Ф) d/2
T =
T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2
T = 245,400 N-mm
We know that the torque required at the rim of handwheel (T)
245,400 = P1 x D/2 = P1 x (300/2) = 150 P1
P1 = 245,400 / 150
P1 = 1636 N
<u><em>2. Maximum compressive stress in the screw</em></u>
30x10³
Qc = W / Ac = -------------- = 8.02 N/mm²
π/4 * 69²
Qc = 8.02 MPa
Bearing pressure on the threads (we know that number of threads in contact with the nut)
n = height of nut / pitch of threads = 150 / 6 = 25 threads
thickness of threads, t = p/2 = 6/2 = 3 mm
bearing pressure on the threads = Pb = W / (π d t n)
Pb = 30 x 10³ / (π * 72 * 3 * 25)
Pb = 1.77 N/mm²
Max shear stress on the threads = τ = 16 T / (π dc³)
τ = (16 * 245,400) / ( π * 69³ )
τ = 3.8 M/mm²
*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72
∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))
τmax = 5.5 Mpa
<u><em>3. efficiency of the straightener</em></u>
<u><em></em></u>
To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm
∴Efficiency of the straightener is η = To / T = 28,620 / 245,400
η = 0.116 or 11.6%