Answer:
a) 180 m³/s
b) 213.4 kg/s
Explanation:
= 1 m²
= 100 kPa
= 180 m/s
Flow rate
Volumetric flow rate = 180 m³/s
Mass flow rate
Mass flow rate = 213.4 kg/s
The correct question;
An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?
Answer:
h'_2 = 40 W/K.m²
Explanation:
We are given;
L1 = 1m
L2 = 5m
T_s = 400 K
T_(∞) = 300 K
V = 100 m/s
q = 20,000 W/m²
Both objects have the same shape and density and thus their reynolds number will be the same.
So,
Re_L1 = Re_L2
Thus, V1•L1/v1 = V2•L2/v2
Hence,
(h'_1•L1)/k1 = (h'_2•L2)/k2
Where h'_1 and h'_2 are convection coefficients
Since k1 = k2, thus, we now have;
h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)
Thus,
h'_2 = [20,000/(400 - 300)]•(1/5)
h'_2 = 40 W/K.m²
Answer:
Cost = $2527.2 per month.
Explanation:
Given that
Discharge ,Q = 130 L/min
So
Cost = $0.45 per cubic meter
1 month = 30 days
1 days = 24 hr = 24 x 60 min
1 month = 30 x 24 x 60 min
1 month = 43,200 min
Lets x
x = 0.13 x 43,200
So the total cost = 5616 x 045 $
Cost = $2527.2 per month.
Answer:
11 toes on one foot? and 5 one the other or just 11 toes?
your NO HINT threw me off
Explanation:
