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:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
T₁ = 300 K
= 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + 
(T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
= 1400×
= 1007.6 K
T₅ = T₄ + ( - T₄)/
= 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + (T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
= 1400× = 1007.6 K
T₇ = T₆ + ( - T₆)/ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. 
= cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. 
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Answer:
Explanation:
The options are:
- In an isometric drawing, multiple angles and axes can be shown in one sketch.
- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.
- Only one sketch will be needed since all other previous designs will no longer be necessary.
- Computer programs will not be necessary to create the exact dimensions of the design.
Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view. Isometric drawings are perspective views from different angles.
So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."
Answer is your company’s address
Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
