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3 years ago

Why is the process for making flat glass called the float process?

Engineering

1 answer:

mario62 [17]3 years ago

5 0

Explanation:

Step1

Float glass is the process of glass manufacturing on the flat surface of metal like tin. In this method molten glass is allowed to float on the surface of metal.

Step2

Float glass gives the uniform and flat surface of glass product. The thickness of the glass produced is uniform throughout. This process of glass making is very cheap and has negligible distortion. Flat glass process is called the float process because of producing high quality flat surface.

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A stream of liquid n-pentane flows at a rate of 50.4 L/min into a heating chamber, where it evaporates into a stream of air 15%

allochka39001 [22]

Answer:

(a) the fractional conversion of pentane achieved in the furnace is 90% conversion

(b) the volumetric flow rates (Umin) of the feed air is 256 x 10³ 1/m

Explanation:

a) Molecular weight of pentane = 72.15 g/mol

density of liquid pentane = 626 kg/m3

Flow rate of feed liquid nitrogen = 50.4 l/min

= 626*50.4*10-3

= 31.55 kg/min

= 31.55/72.15 kmol/min

= 0.4372 kmol/min

Pentane existng the burner = 3.175 kg/min

Fractional conversion = (31.55 - 3.175)/ 31.55

= 0.9 = 90% conversion

C₂H₅ + 8O₂ -----------> 5CO₂ + 6H₂O

From the Stoichiometric reaction,

8 mol of O2 are used for combustion of 1 mol of pentane

for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required

= 3.49 kmol/min of O2

amount of air will be = 3.49/0.21 = 16.62 kmol/min

15% excess air = 16.62*1.15 = 19.12 kmol/min

assuming air to be ideal gas

V = nRT / P.........(1)

V = 19.12 X 8.314 X 336 / 208.6

= 256 m³ = 256 x 10³ 1/m

c) Oxygen:

Amount of pentane consumed = (31.55 - 3.175) = 28.375 kg/min = 28.375/72.15 = 0.3932 kmol/min......(2)

Amount of O2 consumed = 8*0.3932 = 3.146 kmol/min

Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim

unused O2 left = 4.0215 - 3.146

= 0.8755 kmol/min = 19.75*103 l/min............ (using (1))

Carbon Dioxide:

1 mol of pentane = 5 mol of CO2

0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)

= 1.966 kmol/min

= 44.362*103 l/min..........................(using (1))

Nitrogen:

v = 0.79 x 19.12 x 8.314 x 275 / 100.325

= 340.8 m³ / min = 340.8 x 10³ 1/min

Total volumetric flow rate of gases leaving the condenser = (340.8 + 44.36 + 19.75) x 103 l/min

= 404.9 x 103 l/min

6 0

3 years ago

Why are today’s homes more likely to have dangerous levels of carbon monoxide than older homes?

Margarita [4]

Answer:

Carbon monoxide has no odor, color or taste,so it cannot be detected by our senses. This means that dangerous concentrations of the gas can build up indoors and humans have no way to detect the problem until they become ill.

4 0

3 years ago

A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i

kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

= 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

= 25050 ft

using super elevation formula

$e + f = \dfrac{v^2}{rg}$

$0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}$

$r = \dfrac{95.33^2}{32.2\times 0.18}$

r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

$L = \dfrac{\piR\Delta}{180}$

$L = \dfrac{\pi\times 1574\times 38}{180}$

L = 1044 ft

Tangent of the curve calculation

$T = R tan(\dfrac{\Delta}{2})$

$T = 1574 tan(\dfrac{38}{2})$

T = 542 ft

The station PC and PT are

PC = PI - T

PC = 25050 - 542

= 24508 ft

= 245 + 8 ft

PT = PC + L

= 24508 + 1044

=25552

= 255 + 52 ft

the middle ordinate calculation

$MO = R(1-cos\dfrac{\Delta}{2})$

$MO = 1574\times (1-cos\dfrac{38}{2})$

MO = 85.75 ft

degree of the curvature

$D = \dfrac{5729.578}{R}$

$D = \dfrac{5729.578}{1574}$

D = 3.64°

8 0

4 years ago

Engineers please help im not good when it comes to drawing

kakasveta [241]

Answer:

I'm good at drawing and computer-animated design

Explanation:

3 0

3 years ago

Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the cycle is modified with regeneration that

Elza [17]

Answer:

The correct answer is OPTION A (the turbine work output will decrease.)

Explanation:

In open feedwater heaters, there is a mixing of cold feedwater and steam which raise the temperature of feed water causing the fluid average temperature to rise and thereby increasing the thermodynamic efficiency of the cycle.

In a simple ideal Rankine cycle, heating of the water is converted into work done by the cycle in potential and kinetic energy forms. If there is a fixed boiler flow rate, and the simple ideal Rankine cycle is modified with regeneration, turbine work output, and heat supplied, will decrease.

3 0

3 years ago