Answer:
The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v
Explanation:
Solution
Given that:
V+ = 20v
Re = 2kΩ
Rc = 1kΩ
Now we will amke use of the method KVL in the loop.
= - Ve + IE . Re + VEB + VB = 0
Thus
IE = V+ -VEB -VB/Re
Which gives us the following:
IE = 20-0.7 - 10/2k
= 9.3/2k
so, IE = 4.65 mA
IB = IE/β +1 = 4.65 m /101
Thus,
IB = 0.046039 mA
IB = 46.039μA
IC =βIB
Now,
IC = 100 * 0.046039
IC is 4.6039 mA
Now,
VB = 10v
VE = VB + VEB
= 10 +0.7 = 10.7 v
So,
Vc =Ic . Rc = 4.6039 * 1k
=4.6039 v
Finally, this is the table summary from calculations carried out.
Summary Table
Parameters IE IC IB VE VB Vc
Unit mA mA μA V V V
Value 4.65 4.6039 46.039 10.7 10 4.6039
Answer:
The heat transfer is 29.75 kJ
Explanation:
The process is a polytropic expansion process
General polytropic expansion process is given by PV^n = constant
Comparing PV^n = constant with PV^1.2 = constant
n = 1.2
(V2/V1)^n = P1/P2
(V2/0.02)^1.2 = 8/2
V2/0.02 = 4^(1/1.2)
V2 = 0.02 × 3.2 = 0.064 m^3
W = (P2V2 - P1V1)/1-n
P1 = 8 bar = 8×100 = 800 kPa
P2 = 2 bar = 2×100 = 200 kPa
V1 = 0.02 m^3
V2 = 0.064 m^3
1 - n = 1 - 1.2 = -0.2
W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ
∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ
Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ
Answer:
n = 2r³/Rd²
Explanation:
See the attached file for the derivation.
