Answer:
The surface area of the primary settling tank is 0.0095 m^2.
The effective theoretical detention time is 0.05 s.
Explanation:
The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.
Volumetric flow rate = 0.570 m^3/s
Overflow rate = 60 m/s
Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2
Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate
Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3
Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s
Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is 
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation
Here
- M is the moment which is to be calculated
- I is the moment of inertia given as
Here
- b is the breath given as 0.75"
- d is the depth which is given as 8"
The yield moment is 400 k.in.
Part b:
The strain is given as
The stress at the station 2" down from the top is estimated by ratio of triangles as
Now the steel has the elastic modulus of E=29000 ksi
So the strain is
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as
The plastic moment is 600 ksi.
Explanation:
Step1
Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.
Step2
Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product. So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.
Answer:
The surface temperature of the ground is = 296.946K
Explanation:
Solution
Given
r₁= 0.05m
r₂= 0.08m
Tn =Ti = 77K
Ki = 0.0035 Wm-1K-1
Kg = 1 Wm-1K-1
Z= 2m
Now,
The outer type temperature (Skin temperature pipe)
Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)
Thus,
10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05
⇒ T₀ -77 = 231.72
T₀= 290.72K
The shape factor between the cylinder and he ground
S = 2πL/ln 4z/D
where L = length of pipe
D = outer layer of pipe
S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m
The heat gained in the pipe is = S * Kg * (Tg- T₀)
(10* 1) = 1.606 * 1* (Tg- 290.72)
Tg - 290.72 = 6.2266
Tg = 296.946K
Therefore the surface temperature to the ground is 296.946K
