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2 years ago

On-site oil storage containers must be marked "Used Oil." True. False.

Engineering

1 answer:

Scrat [10]2 years ago

7 0

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

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A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

Molodets [167]

Answer:

Explanation:

Mean temperature is given by

$T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}$

Tmean = (Ti + T∞)/2

$T_mean = 107.5^{0}$

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

$\mu$ = 221.6 × 10⁻⁷N.s/m²

$\kappa$ = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

= 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/ $\mu$

Substituting for Pv

Re = 4m/(πD $\mu$ )

= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

= 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0

2 years ago

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$