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3 years ago

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50.38

1 answer:

inysia [295]3 years ago

4 0

Answer:

NOT OSHA

Explanation:

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4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

hodyreva [135]

Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out

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3 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

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2 years ago

The ignition temperature of 87-octane gasoline vapor is about 430 ∘C and, assuming that the working gas is diatomic and enters t

Flura [38]

Answer:

I have solved the problem below. I hope it will let you clear the concept.

For any inquiries ask me in the comments.

Explanation:

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3 years ago

Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligib

insens350 [35]

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m , Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( $\frac{60}{20}$ ) * 1706.0670

= 5118.201 liters ≈ 5.118 m^3/hr

4 0

2 years ago

134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.

vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = $- 10^{\circ}C$ = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = $60^{\circ}C$ = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T = $60^{\circ}C$ :

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at P = 140 kPa, T = $- 10^{\circ}C$ :

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = $s'_{s}$ and $h'_{s} = 278.06 kJ/kg.K$ at 700kPa

(a) Isentropic efficiency of the compressor, $\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}$

$\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%$

(b) The temperature of the environment, $T_{e} = 27^{\circ}C$ = 273 + 27 = 300 K

Availability at state 1, $\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg$

Similarly for state 2, $\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg$

Now, the efficiency of the compressor as per the second law;

$\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757$ = 67.57%

4 0

3 years ago