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2 years ago

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Link OB is 20 mm wide and 10 mm thick and is made from low-carbon steel with Sy = 200 MPa. The pin joints are constructed with s

ufficient size and fit to provide good resistance to out-of-plane bending. Determine the factor of safety for out-of-plane buckling.

1 answer:

Paha777 [63]2 years ago

4 0

Answer:

F = 2671.04 N

Explanation:

The solved solution is in the attached document.

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Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

From the tables air property

$\mu_{394 k}=2.27 \times 10^{-5}$

Ideal gas law is used to calculate the density:

ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

ρ = 1.97 Kg / $m^{3}$

ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

q is expressed in the unit m/s

$q=\frac{2.45}{1.97}$

q = 1.24 m/s

Re = $\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}$

Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

$\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24$ $+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}$

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0

2 years ago

How does Accenture generate value for clients through Agile and DevOps?

Nana76 [90]

By permanently locking in stakeholder requirements during a project's planning phase. -through highly detailed process documentation that is updated following every work cycle.

7 0

1 year ago

Read 2 more answers

A thin, flat plate that is 0.2 m × 0.2 m on a side is oriented parallel to an atmospheric airstream having a velocity of 40 m/s.

Leto [7]

Answer:

The rate of heat transfer from both sides of the plate to the air is 240 W

Explanation:

Given;

area of the flat plate = 0.2 m × 0.2 m = 0.04 m²

velocity of atmospheric air stream, v = 40 m/s

drag force, F = 0.075 N

The rate of heat transfer from both sides of the plate to the air:

q = 2 [h'(A)(Ts -T∞)]

where;

h' is heat transfer coefficient obtained from Chilton-Colburn analogy

$h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}$

Properties of air at 70°C and 1 atm:

ρ = 1.018 kg/m³, cp = 1009 J/kg.K, Pr = 0.7, v = 20.22 x 10⁻⁶ m²/s

$\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K$

Finally;

q = 2 [ 30(0.04)(120 - 20) ]

q = 240 W

Therefore, the rate of heat transfer from both sides of the plate to the air is 240 W

4 0

2 years ago

Read 2 more answers

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

dezoksy [38]

Answer:

(a). Entropy change of refrigerant is = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space $dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is dS = 0.0232 $\frac{KJ}{K}$

Explanation:

Given data

Saturation pressure = 140 K pa

Saturation temperature from property table

$T_{sat}$ = - 18.77 °c = - 18.77 + 273 = 254.23 K

(a). Entropy change of refrigerant is given by

$dS_{ref} = \frac{Q}{T_{sat}}$

Since heat absorbed by refrigerant Q = 180 KJ

$dS = \frac{180}{254.23}$

dS = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space

$dS_{space} = - \frac{Q}{T_{space}}$

$T_{space}$ = - 10 °c = 263 K

$dS_{space} = - \frac{180}{263}$

$dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is given by

$dS = dS_{ref} + dS_{space}$

dS = 0.7077 - 0.6844

dS = 0.0232 $\frac{KJ}{K}$

This is the value of total entropy change.

4 0

2 years ago

air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying

notsponge [240]

Answer:

a) 251.31 m/s and 55.29 m/s

b) The mass flow rate is 0.0396 kg/s

c) The rate of entropy production is 0.0144 kW/K

Explanation:

a) The steady state is:

mi = mo

$\frac{A_{i}V_{i} }{v_{i} } =\frac{A_{o}V_{o} }{v_{o} } \\V_{i} =V_{o}(\frac{v_{i}}{v_{o}} )=V_{o}(\frac{T_{i}P_{o} }{T_{o}P_{i} } )=V_{o}(\frac{330*120}{300*600}) =0.22V_{o}$

The energy balance is:

$h_{i}+\frac{V_{i}^{2} }{2} =h_{2}+\frac{V_{o}^{2} }{2} \\h_{i}-h_{o}+(\frac{V_{i}^{2}-V_{o}^{2} }{2} )=0\\V_{o}=\sqrt{\frac{2(h_{i}-h_{o})}{0.9516} } =\sqrt{\frac{2*(330.24-300.19)x10^{3} }{0.9516} } =251.31m/s$

Vi = 0.22 * 251.31 = 55.29 m/s

b) The mass flow rate is:

$m=\frac{A_{o}V_{o}}{v_{o}} =\frac{\pi d^{2}V_{o}P_{o} }{4RT_{o}} =\frac{\pi *(0.012^{2})*251.31*120x10^{3} }{4*287*300} =0.0396kg/s$

c) The entropy produced is equal to:

$\frac{Q}{T} +S_{gen} =m(s_{2} -s_{1} )\\0+S_{gen} =m(s_{2} -s_{1} )\\S_{gen} =m(s_{2} -s_{1} )\\S_{gen}=0.0396*(c_{p} ln\frac{T_{o}}{T_{i}} -Rln\frac{P_{o}}{P_{i}} )=0.0396*(1.004ln\frac{300}{330} -0.287ln\frac{120}{600} )=0.0144kW/K$

3 0

2 years ago