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kotykmax [81]
3 years ago
12

Cimmaan08 for you! Thank you again :)

Engineering
2 answers:
fomenos3 years ago
5 0

Answer:

Your welcome!!! I hope that helped!!!

Explanation:

zepelin [54]3 years ago
4 0
Are u a fart bc u blew me away
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Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
zloy xaker [14]

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119.1°c

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Temperature can be referred to as a measure of the warmth or coldness of an object or substance with reference to some standard value. The temperature of two systems is the same when the systems are in thermal equilibrium.

To determine the tube surface temperature immediately after installation and after prolonged service.

We will apply to appropriate formula, but first lets define the terms in the question.

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4 years ago
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In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very diffe
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3 years ago
I WILL GIVE 20 POINTS!!
Alex777 [14]

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Use a resume header

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3 years ago
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For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh
Paul [167]

Answer:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere dp= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction \in =  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas \rho = 0.15 mol/dm ⁻³

viscosity of methane gas \mu = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  0.1 5 *\dfrac{16}{0.1^3}

Density =  2400

Density \rho_f =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

Re = \dfrac{dV \rho}{\mu}

Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}

Re=2276.317705

For Re > 1000

\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}

\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}

\Delta P=8575.755212*2.5

\Delta = 21439.38803 \ Pa

To atm ; we have

\Delta P = \dfrac{21439.38803 }{101325}

\Delta P =0.2115903087  \ atm

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

4 0
3 years ago
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