Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where = 101.3 kPa 
, 
, 
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = 
T = 298 K and = 1.18 
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Answer:
The architectural pattern i will use for the school management is the client-server pattern.
This pattern would consist of a server and many clients. wherein the server component would provide services to that of the clients and its components as specified and also there would be a client request service from that of the server.
Explanation:
Solution
A school management system would always involve the client server pattern as this pattern would have a server and many clients wherein the server component would give services to that of the clients and its components as specified and also there would be a client request service from that of the server. This server would share the appropriate services to such clients and also listen to the client's requests.
Such kind of pattern would mostly be used for for the online platforms or application like that of document.
Answer:
Explanation:
Hello!
In this case, we compute the heat output from coal, given its heating value and the mass flow:
Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:
Best regards!
Answer:
The required wall thickness is 
m
Explanation:
Given:
Fluid density 
Diameter of tank 
m
Length of tank 
m
F.S = 4
For A-36 steel yield stress 
MPa,
Allowable stress 
MPa
Pressure force is given by,
Pa
Now for a vertical pipe,
Where 
required thickness
m
Therefore, the required wall thickness is m
Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5
