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3 years ago

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For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste

stream under steady-state conditions with a first-order reaction: reactor volume = 280 m3, flow rate = 14 m3 · day−1, and reaction rate coefficient = 0.05 day−1.

1 answer:

andrew11 [14]3 years ago

3 0

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

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Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

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A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o

navik [9.2K]

The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

useful output energy = 2400 J

Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

h = 2400/800

h = 3 meters

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2 years ago

Do a summary what happen to titanic in the aspect of material(body) and the ductile brittle temperature (DBT) of the material.

vodka [1.7K]

Explanation:

A ductile material can convert into brittle material due to following reasons

1.At very low temperature

2.Due to presence of notch

In titanic ,the base of ship strike to the large ice cube and lower part of titanic ship material was made of steel .We know that steel is ductile material and when steel came with very low temperature of ice due to this ductile material converted in to brittle material and titanic ship failed.Brittle material does not show any indication before failure.

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3 years ago

Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,

Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take $p_o$ = 880 kg/m³ and $p_{glycerol$ = 1260 kg/m³

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

Explanation:

Given that;

Inlet velocity of Glycerin, $V_A$ = 6 m/s

Inlet velocity of oil, $V_B$ = 3 m/s

Density velocity of glycerin, $p_{glycerol$ = 1260 kg/m³

Density velocity of glycerin, Take $p_o$ = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, $d_A$ = 100 mm = 0.1 m

Diameter of oil pipe, $d_B$ = 80 mm = 0.08 m

Diameter of outlet pipe $d_C$ = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

$Q_A + Q_B = Q_C$

$A_Av_A + A_Bv_B = A_Cv_C$

π/4 × ( $d_A$ )² $v_A$ + π/4 × ( $d_B$ )² $v_B$ = π/4 × ( $d_C$ )² $v_C$

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)² $v_C$

0.04712 + 0.0150796 = 0.0113097 $v_C$

0.0621996 = 0.0113097 $v_C$

$v_C$ = 0.0621996 / 0.0113097

$v_C$ = 5.5 m/s

Now we apply the mass flow rate condition

$m_A + m_B = m_C$

$p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C$

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p

72.6412 = 0.06220335p

p = 72.6412 / 0.06220335

p = 1167.8 kg/m³

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

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3 years ago

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sashaice [31]

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity $= \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}$

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h = 300 W/ m^2.K

$T_i$ = 25 degree C = 298 k

$T_o$ = 60 degree C = 333 k

$T_{\infty}$ = 75 degree C = 348 L

diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

$Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226$

$\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}$

$\tau = 0.036 t$

$\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}$

at Bi = 0.226

Ai = 0.982

$\lambda = 0.876$

$\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}$

$0.3 = 0.982 e^{-0.2t}$

$0.305 = e^{-0.2t}$

-1.187 = - 0.02t

t = 59.37 s

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