Answer:
Explanation:
As a security professional, I will respond positively to the OSHA requirements overlap. OSHA guidelines are meant to provide general guidance to all members of various entities throughout the country, while local or state codes also ensure compliance with laws unique to their areas, taking into account workplace safety and security.
OSHA accepts the security codes of the state building To the degree that such codes comply with OSHA regulations, such as BOCA. All the codes and regulations for local, state-owned construction, electrical and life protection are under the same umbrella. Generally, all security protocols and specifications are in accordance with OSHA guidelines. Nonetheless, certain points will overlap, while localized codes will also be addressed to a particular community or state that may
Answer: a) Efficiency = 0.92
b) V = 319.19 m/s
c) S = 0.012 kj/kg.k
Explanation: please find the attached files for the solution
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
The answer to the question is 514.17 lbf
Explanation:
Volume of cylindrical tank = πr²h = 3.92699 ft³
Weight of tank = 125 lb
Specific weight of content = 66.4 lb/ft³
Mass of content = 66.4×3.92699 = 260.752 lb
Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg
=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N
To have an acceleration of 10.7 ft/s² = 3.261 m/s²
we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf
