Answer:
T(water)=50.32℃
T(air)=3052.6℃
Explanation:
Hello!
To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.
The equation is as follows!
Q = heat
h = heat transfer coefficient
Ts = surface temperature
T = fluid temperature
a = heat transfer area
The surface area of a cylinder is calculated as follows
Where
D=diameter=20mm=0.02m
L=leght=200mm)0.2m
solving
For water
Q=2Kw=2000W
h=5000W/m2K
a=0.01319m^2
Tα=20C
solving for ts
for air
Q=2Kw=2000W
h=50W/m2K
a=0.01319m^2
Tα=20C
Answer:
Manufacturer’s Recommendations means the instructions, procedures, and recommendations which are issued by the manufacturer of any equipment used at the Facility relating to the operation, maintenance, or repair of such equipment, and any revisions or updates thereto from time to time issued by the manufacturer.
Manufacturer’s Recommendations means the instructions, procedures and recommendations which are issued by any manufacturer of the Equipment relating to the operation, maintenance and repair of the Equipment and any revisions to such instructions, procedures and recommendations agreed to by any manufacturer of the Equipment and which are valid at the time such operation, repair and maintenance is being carried out.
Manufacturer’s Recommendations means the written instructions, procedures and recommendations which are issued by the original equipment manufacturer of any plant or equipment used at the Utility Plant relating to the operation, maintenance and repair of such plant or equipment and any revisions thereto issued by the manufacturer, which are valid and applicable at the time such operation, maintenance or repair is undertaken. Notwithstanding the above, Manufacturer’s Recommendations shall not include any instructions, procedures or recommendations of a manufacturer of any plant or equipment that the Owner and the Operator have agreed in writing to exclude from this definition or have agreed in writing should not be followed.
Explanation:
Answer:
simple projects bovonhztisgx
Answer:
material remove in 3 min is 16790.4 mm³/s
Explanation:
given data
length L = 80 cm = 800 mm
width W = 30 cm
height H = 15 cm
make grove length = 80 cm
width = 8 cm
depth = 10 cm
mill toll diameter = 4 mm
axial cutting depth = 20 mm
to find out
How much material removed in 3 minutes
solution
first we find time taken for length of advance that is
time = 
here advance is given as 0.001166 mts / sec
so time = 
time = 686.106 seconds
now we find material remove rate that is
remove rate = mill toll rate × axial cutting depth × advance
remove rate = 4 × 20×0.001166 ×1000
remove rate = 93.28 mm³/s
so
material remove in 3 minute = 3 × 60 = 180 sec
so material remove in 3 min = 180 × 93.28
material remove in 3 min is 16790.4 mm³/s
Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s
