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Roman55 [17]
3 years ago
14

The moisture in hot, humid, stagnant air surrounding a cold-water pipeline continually diffuses to the cold surface where it con

denses. The condensed water forms a liquid film around the pipe, and then continuously drops off the pipe to the ground below. At a distance of 10 cm from the surface of the pipe, the moisture content of the air is constant. Close to the pipe, the moisture content of the air approaches the vapor pressure of water evaluated at the temperature of the pipe
What is the simplest differential form of Fick's flux equation for water vapor, NA?
Engineering
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

The simplest differential form of Fick's flux equation for water vapor is

\frac{1}{r} \frac{d}{dr} (rN_{rA} )=0

Explanation:

Fick's law has to do with the density of the particles is proportional to their concentration gradient. For this question some assumptions will be taken into account:

-The pipe is considered to be very long and diffusion only occurs in the r direction.

-The reaction is not homogeneous, therefore RA = 0

-The concentration of the component A is constant

-No mixing, only diffusion

Due to the complexity of the terms used in solving the Fick equation, the solution is found in the attached file.

Download docx
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REVIEW QUESTIONS
babunello [35]
Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!
5 0
3 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

6 0
3 years ago
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical
stealth61 [152]

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

             C_D is the drag force constant.

                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

 When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

6 0
4 years ago
Does a food market have any rooms in particular? Also whats units?
makkiz [27]

Answer:

to be or not to be

Explanation:

Vivi is a drummer for a band. She burns 756756756 calories while drumming for 333 hours. She burns the same number of calories each hour.

8 0
3 years ago
Read 2 more answers
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