When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is
a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).Answer: 3.05 x 10-8 kg/m2-s
1 answer:
Answer:
See attachments for step by step explanation towards getting answer.
Explanation:
Given that;
College Engineering 10+5 pts
When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).
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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
From the refrigerant table A11-A13
sat vapor
m=
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:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
Temperature on the inside ofthe box
Explanation:
The power of the light bulb is the rate of heat conduction of the bulb,
The thickness of the wall, L = 1.2 cm = 0.012m
Length of the cube's side, x = 20cm = 0.2 m
The area of the cubical box, A = 6x²
A = 6 * 0.2² = 6 * 0.04
A = 0.24 m²
Temperature of the surrounding,
Temperature of the inside of the box,
Coefficient of thermal conductivity, k = 0.8 W/m-K
The formula for the rate of heat conduction is given by: