All categories

3 years ago

What is the lehr and what purpose does it serve?

Engineering

1 answer:

levacccp [35]3 years ago

5 0

Explanation:

Step1

Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.

Step2

Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product. So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.

You might be interested in

Select the correct answer.

boyakko [2]

I think the answer is b

6 0

3 years ago

Read 2 more answers

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio

Mama L [17]

Answer:

a) 570 kWh of electricity will be saved

b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

$P_{incandescent}$ = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find the total power consumption by the fluorescent lamp

$P_{fluorescent}$ = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

$P_{saved}$ = $P_{incandescent}$ - $P_{fluorescent}$

$P_{saved}$ = 750 - 180

$P_{saved}$ = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved = 570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

3 0

3 years ago

Given an array of words representing your dictionary, you test words to see if it can be made into another word in dictionary. T

katen-ka-za [31]

Answer:

Detailed solution is given below:

8 0

3 years ago

A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show

aleksandrvk [35]

Answer:

$GM$

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter $d=2m$

Length $l=2.5m$

Weight $W=22kN$

Specific weight of sea water $\mu= 10.25kN/m^3$

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

$W=(pwg)Vd$

Where

$V_d=\pi/4(d)^2y$

Therefore

$22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m$

Therefore

Center of Bouyance B

$B=\frac{y}{2}=0.26m\\\\B=0.75$

Center of Gravity

$G=\frac{I.B}{2}=2.6m$

Generally the equation for\BM is mathematically given by

$BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\$

Therefore

$BG=2.6-0.476\\\\BG=0.64m$

Therefore

$GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\$

Therefore

$GM$

So the bouy does not float with its axis vertical

4 0

3 years ago

Write a script named dif.py. This script should prompt the user for the names of two text files and compare the contents of the

Ainat [17]

Answer:

Created on Nov 3, 2018 @author: ASLand

7import atexit

#Read, nanes of both files

Rrintll"Enter tvo files to be compared below

userliamel input ("Enter the nome of the first file: ")

userliame2 input("Enter the name of the second file: ")

ROpen each file

f1 - open(userNamel, r')

@17 f2 = opan(useriame 2, )

tread all the lines into a list

d1 f1.readlines ()

d2 f2.readlines()

re equivalent, print "Yes" else pri

oiterate, and conpare

#11

the

if dl == d2:

print("Yes")

atexit

elif for i in range(@, min(len (d1), len(d2))):

if di[i]!=d2[i]:

PCint("No")

print(d1[i])

pcint(d2[])

7 0

3 years ago