(a) Draw the Moore finite state machine (FSM) of an electronic combination lock with a RESET button, two number buttons (0 and 1
), and an unlock output. The combination should be 01011.
1 answer:
Answer:
Explanation:
There are a total of 6 states and 3 bits in this problem. Whenever the Reset button is pressed, RESET state is called otherwise the state according to the diagram is called. For the combination to be "01011", the input sequence has to be in the same order. If 0 is pressed instead of 1 in state "010", the last state of output ending with 0 will be called and likewise in all the states that follow.
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Answer:
The mass flow rate of refrigerant is 0.352 kg/s
Explanation:
Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:
<u>At State A</u>:
From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:
ha = 244.5 KJ/kg
Sa = 0.93788 KJ/kg.k
<u>At State B</u>:
Since, the process from state A to B is isentropic. Therefore,
Sb = Sa = 0.93788 KJ/Kg
From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:
hb = 256.85 KJ/kg (By interpolation)
<u>At State C</u>:
From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:
hc = 107.34 KJ/kg
Now, from the diagram it is very clear that:
Heat Loss = m(hb = hc)
m = (Heat Loss)/(hb - hc)
where,
m = mass flow rate = ?
Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)
Heat Loss = 52.753 KW
Therefore,
m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)
<u>m = 0.352 kg/s</u>
The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
