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pav-90 [236]
3 years ago
12

A sheet of steel 1.5 mm thick has nitrogen (N2) atmospheres on both sides at 1200°C and is permitted to achieve steady-state dif

fusion condition. The diffusion coefficient for N2 in steel at this temperature is 6 ´ 10-11 m2 /s, and the diffusion flux is found to be 1.2 ´ 10-7 kg/m2 -s. Also, it is known that the concentration of N2 in the steel at the high-pressure surface is 4 kg/m3 . How far into the sheet from the high-pressure side will the concentration be 2.0 kg/m3 ? Assume a linear concentration profile.
Engineering
1 answer:
Gelneren [198K]3 years ago
3 0

Answer:

do the wam wam

Explanation:

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Option E

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3 years ago
When cutting a FBD through an axial member, assume that the internal force is tension and draw the force arrow _______ the cut s
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Problem-Solving Tip: When cutting an FBD through an axial member, assume that the internal force is tension and draw the force arrow directed away from the cut surface. If the computed internal force value turns out to be a positive number, then the assumption of tension is confirmed.

3 0
2 years ago
Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l
sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is  0.44 \frac{W}{m^{2} \times K} ,  the average convection heat transfer coefficient over the entire plate is  0.293 \frac{W}{m^{2} \times K}and the total heat flux transfer to the plate is 61.6 KJ.

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas  and surface. Further temperature  boundary layer will developed on flat plate in longitudinal direction.  

Hot carbon dioxide exhaust gas

physical properties

r= 1.05 \frac{kg}{m^{3}}

c_p = 1.02 \frac{kJ}{Kg \times K}

m= 231 \times 10^{7}  \frac{N \times s }{m^2}

υ = 21.8 \times 10^{6}  \frac{m^2}{s}

k = 32.5 \times 10^{3} \frac{W}{m \times K}

\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}

Pr = 0.725

Apart from these other data arr given below,

v= 3 \frac{m}{s}  \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

Nu = \frac{ h \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

where h= Average heat transfer coefficient

           L= Length of a plate

           k= Thermal Conductivity of carbon dioxide

           Re = Reynold's Number

           Pr  = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

    is referred as local convection heat transfer coefficient.

   

   To find convection heat transfer coefficient at 1 m from leading edge,

  Nu = \frac{ h_local \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

  Here, first we have to find Re and Pr,

   Re = \frac{r \times v \times L}{m}

   Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}

   Re = 20.63 \times  10^{-10}

   Pr number is take from physical property data and Pr is 0.725.

   Putting value of Re and Pr in main equation,

   we get

   Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   = 32.5 \times 10^{3} \times  0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   =  0.44 \frac{W}{m^{2} \times K}

(b)  To find average convection heat transfer coefficient,

      it can be find out as case (a), only difference is that instead of L=1 m,        L=1.5 m would come,  

   Therefore,

    Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    Finally,

      h  = \frac{0.44}{1.5}

      h  = 0.293 \frac{W}{m^{2} \times K}

(C) Total heat flux transfer to the plate is found out by,

     Q = h \times (T_g - T_s)

     Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140  \\ Q= 61.6 KJ

     

     

   

   

     

   

     

   

   

 

   

   

   

   

8 0
3 years ago
When handling chemicals and solvents, technicians are recommended to
Luda [366]

Answer:

  1. To wear PPE
  2. Have prior knowledge of explosive levels and elemental properties
  3. Know procedure to eliminate any threat
7 0
3 years ago
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