Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.

Answer: hello your question lacks some data below is the missing data

Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol

H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.

H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg

Answer :

a) 34.98 lit/min

b) 1432.53 m^3/min

Explanation:

a) Calculate how much water is produced

density of water = 1 kg/liter

First we will determine the mass of condensed water using the relation below

inlet mass - outlet vapor mass =  0.0339508 * n * 18/1000 ----- ( 1 )

where : n = 57241.57

hence equation 1 = 34.98 Kg/min

∴ volume of water produced =  mass of condensed water / density of water

                                                =  34.98 Kg/min / 1 kg/liter

                                                = 34.98 lit/min

b) calculate the Volumetric flow rate of air entering the unit

applying the relation below

Pv = nRT

101325 *V = 57241.57 * 8.314 * 305  

∴ V = 1432.53 m^3/min

Answer: Power

Explanation:

The rate at which work is done or the amount of work done based on a period of time is referred to as power.

Power can also be defined as the amount of energy that is being transferred per unit time. The unit of power is one joule per second or simply called the watt.

Answer:

Engine

Explanation:

Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.

Semi-independent suspension give the front wheels some individual movement.

This suspension only used in rear wheels.

Thus, the correct option is Semi-independent suspension

Learn more about Semi-independent suspension

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Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

Answer:

letse see

Explanation:

Answer:

the liquid woulriekwvhrnsshsnekwb ndrhwmoadi

Answer:

True

Explanation:

During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system

Hence, the given statement is true

Complete question:

If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.

Answer:

the flow rate of the water from the tank is 0.05 m³/min

Explanation:

Given;

volume of water in the tank, v = 30 m³

length of the waterproof faucet, L = 2cm = 0.02 m

duration of water flow through the tank, t = 10 hours

The flow rate of the water from the tank is calculated as;

[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]

Therefore, the flow rate of the water from the tank is 0.05 m³/min

Answer:

the surface temperature of the plates when they come out of the oven is approximately 445 °C

Explanation:

Given the data in the question;

thickness t = 3 cm = 0.03 m

so half of the thickness L = 0.015 m

thermal conductivity of brass k = 110 W/m°C

Density p = 8530 kg/m³

specific heat [tex]C_p[/tex] = 380 J/kg°C

thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s

Temperature of oven T₀₀ = 700°C

initial temperature T[tex]_i[/tex] = 25°C

time t = 10 min = 600 s

convection heat transfer coefficient h = 80 W/m².K

Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.

So, using analytical one-term approximation method, the Fourier number > 0.2.

now, we determine the Biot number for the process

we know that; Biot number Bi =  hL / k

so we substitute

Bi =  hL / k

Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109

Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )

The interpolation method used to find the

λ₁ = 0.1039 and A₁ = 1.0018

so

The Fourier number т = ∝t/L²

we substitute

Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²

т = 0.02034 / 0.000225

т = 90.4

As we can see; 90.4 > 0.2

So,  analytical one-term approximation can be used.

∴ Temperature at the surface will be;

θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation

θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )

so we substitute

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )

θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998

θ(L,t)[tex]_{wall[/tex] = 0.3775

so we substitute into equation 1

θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)

0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )

0.3775 = ( T(L,t) - 700 ) / ( - 675 )

0.3775 × ( - 675 ) = ( T(L,t) - 700 )

- 254.8125 = T(L,t) - 700

T(L,t) = 700 - 254.8125

T(L,t) = 445.1875 °C ≈ 445 °C

Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C