Answer:
Explanation:
Given data:
initial construction co = 0.286 wt %
concentration at surface position cs = 0 wt %
carbon concentration cx = 0.215 wt%
time = 7 hr
for 0.225% carbon concentration following formula is used
where, erf stand for error function
from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815
from given table
x = 0.002395 mm
Answer:
True
Explanation:
The tensile forces are small in most arches and usually negligible.
Answer:
COP of the heat pump is 3.013
OP of the cycle is 1.124
Explanation:
W = Q₂ - Q₁
Given
a)
Q₂ = Q₁ + W
= 15 + 7.45
= 22.45 kw
COP = Q₂ / W = 22.45 / 7.45 = 3.013
b)
Q₂ = 15 x 1.055 = 15.825 kw
therefore,
Q₁ = Q₂ - W
Q₁ = 15.825 - 7.45 = 8.375
∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124
Answer:
Timing Diagrams 15 pts. A 10 MHz clock that generates a 0 to 5V pulse train with a 30% duty cycle is connected to input X of a two input OR gate that has a 20nS propagation delay. The clock also goes to an inverter with a 10 ns propagation delay. The output of the inverter goes to the Y input of the OR gate. a) Draw the circuit. 2 pts. b) Plot the output of the clock for two cycles. Show times and voltages. 5 pts. c) On the same page as part (b) plot the output of the inverter. Show times and voltages. 3 pts. d) On the same page as parts (b & c) plot the output of the OR gate. Show times and voltages. 5 pts.
