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3 years ago

Chemical engineering got is unofficial start around the time of the .

Engineering

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

Option A, World War II

Explanation:

During the period of industrial revolution around 1915-25, the chemical engineering has taken a new shape. During this period (i.e around the world war I), there was rise in demand for liquid fuels, synthetic fertilizer, and other chemical products. This lead to development of chemistry centre in Germany . There was rise in use of synthetics fibres and polymers. World war II saw the growth of catalytic cracking, fluidized beds, synthetic rubber, pharmaceuticals production, oil & oil products, etc. and because of rising chemical demand, chemical engineering took a new shape during this period

Hence, option A is the right answer

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A simple Brayton cycle using air as the working fluid has a pressure ratio of 10.9. The minimum and maximum temperatures in the

ladessa [460]

Air pressure at the end of the Turbine exit is : 730.57 K

<u>Given data :</u>

Pressure ratio ( p₂ / p₁ )= 10.9

minimum temperature ( T₁ ) = 280 K

maximum temperature ( T₃ ) = 1410 K

Assuming :

constant specific heat and efficiency of 100%

<h3 /><h3>Determine the Air pressure at the end of the turbine exit </h3>

For air :

Cp = 1.005 kJ/kg.k, Cv = 0.718 kJ/kg.k, v = 1.4

Given that efficiency for compressor and turbine is 100% the process ( 1-2 , 3 - 4 ) will all be isentropic

We will Apply the formula below

$\frac{T_{2} }{T_{1} } = ( \frac{p_{2} }{p_{1} } )^{\frac{v-1}{v} } = ( \frac{V_{1} }{V_{2} } )^{v-1}$

Insert values into equation ( 1 )

T₂ = 551.147 K ( temperature at compressor exit )

<u />

<u>Next : Determine the </u><u>value </u><u>of the temperature at </u><u>Turbine exit </u><u>( T₄ )</u>

T₃ / T₄ = 10.7^ $^{\frac{1.4-1}{1.4} }$

Therefore : T₄ = 1410 / 10.7^0.286

= 1410 / 1.93

= 730.57 K

Hence we can conclude that the Air pressure at the end of the Turbine exit is : 730.57 K .

Learn more about Turbine : brainly.com/question/894340

6 0

2 years ago

Mild neurotoxins can cause which of these physical effects?

joja [24]

Macroscopic manifestations of neurotoxin exposure can include widespread central nervous system damage such as intellectual disability, persistent memory impairments, epilepsy, and dementia. Additionally, neurotoxin-mediated peripheral nervous system damage such as neuropathy or myopathy is common.

5 0

3 years ago

Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,

77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

$L_{BC}^2 = L_{AB}^2 + L_{AC}^2$

$L_{BC}^2 = (3.5+2.5)^2+ 4^2$

$L_{BC}= \sqrt{(6)^2+ 4^2}$

$L_{BC}= \sqrt{36+ 16}$

$L_{BC}= \sqrt{52}$

$L_{BC}= 7.2111 \ m$

The cross -sectional of the cable is calculated by the formula :

$A = \dfrac{\pi}{4}d^2$

where d = 4mm

$A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2$

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection in length $\delta$ ; we can calculate for the force $F_{BC$ by using the formula:

$\delta = \dfrac{F_{BC}L_{BC}}{AE}$

$F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}$

where ;

E = modulus elasticity

$L_{BC}$ = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for $L_{BC}$ and 0.006 m for $\delta$ ; we have:

$F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}$

$F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN$ ---- (1)

Similarly; we can determine the force $F_{BC}$ using the allowable maximum stress; we have the following relation,

$\sigma = \dfrac{F_{BC}}{A}$

${F_{BC}}= {A}*\sigma$

where;

$\sigma =$ maximum allowable stress

Replacing 190 × 10⁶ Pa for $\sigma$ ; we have :

${F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN$ ------ (2)

Comparing (1) and (2)

The magnitude of the force $F_{BC} = 2.09676 \ kN$ since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

$\sum M_A = 0$

$3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0$

$3.5 P - 3.328 F_{BC} = 0$

$3.5 P = 3.328 F_{BC}$

$3.5 P = 3.328 *2.09676 \ kN$

$P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }$

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

4 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago

What does drivetrain mean in coding?

Rudik [331]

Answer:

A drivetrain allows a robot to be mobile by using wheels, tank treads, or another method. A drivetrain is sometimes referred to as a drive base. Identifying which kind of drivetrain to use is one of the first considerations when designing a robot.

Explanation:

Hope this helps! Brainliest plz!

4 0

3 years ago

Chemical engineering got is unofficial start around the time of the __________ __________ ________.

Chemical engineering got is unofficial start around the time of the .