A rigid tank contains 5 kg of saturated vapor steam at 100°C. The steam is cooled to the ambient temperature of 25°C. (a) Sketch
the process with respect to the saturation lines on a T-vdiagram. Indicatepressure valuesat each state. (5pt) (b) Determinethe entropy change of the steam, in kJ/K. (10 pt) (c) Determinethe totalentropy change associated with this process, in kJ/K. (15pt)
1 answer:
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Complete Question:
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.
Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.
Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!
Answer & Explanation:
[Find the attachments]
Step 1 :
Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.
Emitter and base, collector and base are reverse biased then BJT in cut off region.
Three sketches one below the other is shown in Figure 1.
[find the figure in attachment]
Step 2:
Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.
Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.
[For Steps 3 4 5 6 and 7 find attachments]
Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
