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.
Answer:
D is the correct choice.
Explanation:
I'm assuming that this is probably a phase in the textbook or progarm you are studying, and this is just a matter of reading thoroughly.
Engineers usually benefit from catching a mistake, and would also benfit from keeping record of a misstep in order to remain clear of that mistake in the future.
Have a great day, and mark me brainliest if I am most helpful!
:)
Answer
No;
The two flows are not dynamically similar
Explanation: Given
T∞,1 = 800k
V∞,1 = 200m/s
p∞,1 = 1.739kg/m³
T∞,2 = 200k
V∞,2 = 100m/s
p∞,2 = 1.23kg/m³
Size1 = 2 * Size2 (L1 = 2L2) Assumptions Made
α ∝√T
μ∝√T Two (2) conditions must be met if the two flows are to be considered similar.
Condition 1: Similar Parameters must be the same for both flows
Condition 2: The bodies and boundaries must be genetically true. Condition 2 is true
Checking for the first condition...
Well need to calculate Reynold's Number for both flows
And Check if they have the same Reynold's Number Using the following formula
Re = pVl/μ
Re1 = p1V1l1/μ1 Re2 = p2V2l2/μ2 Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2
Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2
Re1/Re2 = p1V1l1μ2/p2V2l2μ1
Re1/Re2 = p1V1l1√T2 / p1V1l1√T1
Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)
Re1/Re2 = 9837.2/3479
Re1/Re2 = 2,828/1
Re1:Re2 = 2.828:1
Re1 ≠ Re2,
So condition 1 is not satisfied Since one of tbe conditions is not true, the two flows are not dynamically similar
Answer:
a. 78.4 pieces/hr
b. $0.1806/min
c. $1.34/piece
Explanation:
(a) With a cycle time Tc = 45 sec = 0.75 min.
Production rate, Rp = 60/0.75 = 80 pieces/hr.
Let us factor in the 98% proportion uptime, so Production rate, Rp = 0.98*(80) = 78.4 pieces/hr
Quantity of product annualy = 6000 *(78.4) = 470,400 pieces/yr
(b) Equipment cost rate, Ceq = 500,000(1.30)/(60 x 10 x 6000) = $0.1806/min.
(c) Cost per piece of mould, Ct = 100,000/1,000,000 = $0.10/piece
Cost rate of labour ,CL = 18.00(0.20) = $3.60/hr = $0.06/min
Conclusively, cost per piece, Cpc = 1.20(0.88) + (0.06 + 0.1806)(0.75) + 0.10 = $1.34/piece
Answer:
False
Explanation:
COP is Coeficient of performance, it relates useful heat moved with work consumed, it is used in refrigerators and heat pumps.
For refrigerators the useful heat is the one extracted from the cold reservoir. Therefore:
This often exceeds 1.
