Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters
at 400 kPa and 300 K. Stream 3 leaves the control volume at 150 kPa and 270 K. The control volume does 3 kW of work on the surroundings while losing 5 kW of heat. Find the mass flow rate of stream 2. Neglect changes in kinetic and potential energy.
1 answer:
Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
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Answer:
The strength coefficient is K = 591.87 MPa
Explanation:
We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by
where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.
Solving for strength coefficient
From the strain hardening equation we can solve for K
And we can replace values
Thus we get that the strength coefficient is K = 591.87 MPa
Answer:
Solution 2:
The thermal energy of any particle is given as =
where m= Effective mass of the particle,
v= Velocity of the particle
The average kinetic energy is given as =
where K= Boltzmann Constant
T= Temperautre of the particle
At equilibrium,
1/2 mv² = 3/2 KT
Hence, v=
As, effective mass ratio is calculated with respect to rest mass of electron,
melectron = 0.11 * 9.11 *10-31 kg
mhole = 0.35 *9.11 *10-31 kg
K = 1.38 × 10-23 m2 kg s-2 K-1
Plotting, over a range of temperature of 0 K to 400 K, we obtain the attached Graphs (attached).
Solution 3:
The above two curves plotted are not identical as seen. From the same value of Temperature, and under identical conditons, the Thermal velocity solely depends on effective mass ratio. And it is inversely proportional to effective mass ratio. More the effective mass ratio, less the thermal velocity and flatter the slope of the curve and vice versa. As, the mass ratio of holes is more than that of electrons, the curve of electrons has a steeper slope than that of holes. Hence, the curves are not just identical.
