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Vilka [71]
3 years ago
7

6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

ameter of 0.01 m and containing N2 at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of CO2 at one end is 456 mm Hg and 76 mm Hg at the other end. The diffusivity Dan is 1.67 x 10-5 m2/s at 298 K. Calculate the flux of CO2 in cgs and SI units for equimolar counterdiffusion

Engineering
1 answer:
zzz [600]3 years ago
4 0

Answer:

Heat flux of CO₂ in cgs

                 = 170.86 x 10⁻⁹ mol / cm²s

SI units

       170.86 x 10⁻⁸ kmol/m²s  

Explanation:

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The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.
OlgaM077 [116]

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

0+F(t_{2} -t_{1} )=0.2v_{2}

if F=2 kN and t2-t1=2x10^-3 s. Replacing

0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

T_{2} =\frac{1}{2} mv_{2}^{2}  \\T_{3} =0\\U_{2---3} =-F_{res} x

Replacing:

\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N

3 0
3 years ago
Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D
Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ  F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ  F$

Change in temperature, $\Delta T = T_2-T_1$

                                         $\Delta T = -2.43-87.745$

                                           $\Delta T=-90.175^\circ  F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

   $=11.401+0.32198(82.898)$

   $=38.09297 \ Btu/lbm$

Change in internal energy  

$\Delta u= u_2-u_1$

   = 38.09297-40.5485

  = -2.4556        

5 0
2 years ago
An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel
Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

3 0
3 years ago
You are designing the access control policies for a Web-based retail store. Customers access the store via the Web, browse produ
PolarNik [594]

Answer:

Object-Oriented Software Engineering Using UML, Patterns, and Java, 3e, shows readers how to use both the principles of software engineering and the practices of various object-oriented tools, processes, and products.

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2 years ago
How does Hydro Technology and Hydro Energy Help with Global Water Issues?
GrogVix [38]
It would have environmental and societal impacts
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3 years ago
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