Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.


Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get

where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now

= 4.7476 m/sec
= 4.75 m/s
I hope my answer helped u under stand better
Answer:
life (N) of the specimen is 117000 cycles
Explanation:
given data
ultimate strength Su = 120 kpsi
stress amplitude σa = 70 kpsi
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 120
Se = 60 kpsi
and we know strength of friction f = 0.82
and we take endurance limit Se is = 60 kpsi
so here coefficient value (a) will be
a =
......................1
put here value and we get
a =
a = 161.4 kpsi
so coefficient value (b) will be
b =
b =
b = −0.0716
so here number of cycle N will be
N = 
put here value and we get
N = 
N = 117000
so life (N) of the specimen is 117000 cycles