Answer:
Explanation:
The most common HFC used in air conditioners is R-410A. This refrigerant is better than R-22 in terms of “Ozone Depletion” potential and energy efficiency, but it still causes global warming. A few more HFCs that are commonly used are: R-32 in Air Conditioners and R-134A in refrigerators.
Answer:
0.264 ; 0.079
Explanation:
Given that:
Sample size, n = 100
Probability of being active, p = 1% = 1/100 = 0.01
Using the binomial probability relation :
P(x =x) = nCx * p^x * (1 - p)^(n - x)
Probability that more than 1 user will be active
P(x > 1) = 1 - [p(x=0) + p(x = 1)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x > 1) = 1 - [0.366 + 0.370]
P(x > 1) = 0.264
2.)
Probability that more than 2 user will be active
P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185
P(x > 1) = 1 - [0.366 + 0.370 + 0.185]
P(x > 1) = 0.079
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Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.
