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3 years ago

Calculate the x and y components of the given force vector.

Engineering

1 answer:

Sav [38]3 years ago

5 0

Answer:

sorry I didn't get it

Explanation:

could you please circle the question that I can identify the question and tell you tge answer.

I am an expert in maths. I am not praising myself becoz I know self praise is no praise. But I said I am an expert becoz I want to tell you people that if I an expert cannot identify the answer so nobody can identify.

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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

Who is the worst clown in rouge linage

Lerok [7]

All of them
explanation:
you don’t need one

4 0

3 years ago

What are the assumptions made for air standard cycle analysis?

diamong [38]

D i took this hope it helps

4 0

3 years ago

Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.

likoan [24]

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

print("Largest: " + string(number2))

else:

print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))

# END #

4 0

3 years ago

Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile

koban [17]

Answer:

Explanation:

Given that, .

Mass of car is

M = 1300kg

Velocity of car

V = 80km/h = 80 × 1000/3600

V = 22.22m/s

Calculate the kinetic energy of the vehicle as follows:

K.E = ½ MV²

K.E = ½ × 1300 × 22.22²

K.E = 320,987.65 J

Given that,

Enthalpy is 45MJ / kg

h = 45MJ / kg

Then, enthalpy is given as.

Enthalpy = Energy / mass

h = E / m

45 × 10^6 = 320,987.65 / m

m = 320,987.65 / 45 × 10^6

m = 7.133 × 10^-3 kg

m = 7.133 mg

Also, given that, density is 680kg/m³

Density is given as

Density = mass / Volume

ρ = m / v

Then, v = m / ρ

v = 7.133 × 10^-3 / 680

v = 1.049 × 10^-5 m³

We know that

1mL = 10^-6 m³

Therefore,

v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³

v = 10.49 mL

7 0

3 years ago

Calculate the x and y components of the given force vector.​

Calculate the x and y components of the given force vector.