To solve the problem, make assumptions for missing data and justify. Given:

The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h

diamong [38]

Answer:

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

Explanation:

We have to combine the following formula to find the mass yield:

$M=JAt$

$M=-DAt(ΔC/Δx)$

The diffusion coefficient : $D=6.0*10^{-8} m/s^{2}$

The area : $A=0.25 m^{2}$

Time : $t=3600 s/h$

ΔC: $(0.64-3.0)kg/m^{3}$

Δx: $3.1*10^{-3}m$

Now substitute the values

$M=-DAt(ΔC/Δx)$

$M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]$

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

8 0

3 years ago

If you are unsure about holding a piece of wood to be drilled, then you should always use a

alisha [4.7K]

C I took construction class

4 0

3 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0

3 years ago

Need help fast I been stuck in this for the longest

gavmur [86]

Answer:

3rd and 4rth

Explanation:

a geologist studies the earth and both of these have something to do with the earth

6 0

3 years ago

Deidre has just moved from the sales department into the finance department. On her first day in her new role, she receives an e

melisa1 [442]

Answer:

d. The company uses role-based access control and her user account hasn't been migrated into the correct group(s) yet

Explanation:

Since Deidre is accessing her e-mail there appears to be nothing wrong with her account or password. Since her role is new, most likely the problem is associated with her new role.

3 0

3 years ago