Answer:
Distribution factor P = =38.33
V = 7.826 ml
Explanation:
given details:
BOD =230 mg/l
DO inital = 8.0mg/l
DO final = 2.0mg/l
we know
BOD = [DO inital -DO final] * distribution factor
230 = [8 - 2] D.F
Distribution factor P 
Distribution factor P = =38.33
THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is
distribution factor 
V = 7.826 ml
Answer:
t= 4.5 mm
Explanation:
Given that
P = 520 KPa ( gauge)
Maximum allowable normal stress ,σ= 150
d= 2.6 m
Wall thickness = t
The normal stress for pressure vessel given as
( hoop stress)
We always take maximum stress for safe design.
Now by putting the values
t= 4.5 mm
So the minimum thickness, t, of the wall is 4.5 mm
Answer:
There are 50 ASE certification tests, covering almost every imaginable aspect of the automotive repair and service industry.
Explanation:
yww <33
They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.
Answer:
The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3
Heat flux is 9.67×10^7 Btu/hrft^2
Explanation:
Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr
Area (A) = πD^2/4
Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft
A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2
Volume (V) = A × Length
L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft
V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3
Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3
Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2
