Answer:
Option E
Explanation:
All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.
Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is
.
Answer:
c. and d
Explanation:
As a whistle-blower, one of your aim is to guide against unethical dealings of other people , hence you are creating an environment that uphold ethical conduct,
In addition, whistle-blowing will disclose all imminent dangers to the software community thereby preventing security breaches.
Technical Drawings give a better understanding of what is needed and required in the project.
Explanation:
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