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3 years ago

Why are concrete masonry units (CMU) called breeze blocks?

Engineering

1 answer:

FromTheMoon [43]3 years ago

8 0

Answer:

A concrete masonry unit (CMU) is a standard-size rectangular block used in building construction. ... Those that use cinders (fly ash or bottom ash) as an aggregate material are called cinder blocks in the United States, breeze blocks (breeze is a synonym of ash) in the United Kingdom, and hollow blocks in the Philippines.

Explanation:

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In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium

Nimfa-mama [501]

Answer:

The diffusivity is given as $8.064\times 10^{-16} m^2/s$

Explanation:

From the given data as in the attached question found via the search (because the values were not clear in this one)

$D_o=2.0\times10^{-4}m^2/s\\Q_d=240.6 kJ/mol=2.406 \times 10^5 J/mol\\Gas Constant=R=8.314 J/mol K\\Temperature =830 C =830+273 =1103 K$

So the Diffusion coefficient is given as

$D=D_oe^{\frac{-Q_d}{RT}}\\D=(2.0\times 10^{-4})e^{\frac{-2.406\times 10^5}{8.314\times 1103}}\\D=8.0640\times 10^{-16} m^2/s$

So the diffusivity is given as $8.064\times 10^{-16} m^2/s$

4 0

4 years ago

The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large

SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0

3 years ago

Nitrogen at an initial state of 300 K,150 kPa and 0.2 m3is compressed slowly in an isothermal process to a final pressure of 800

Dmitry_Shevchenko [17]

Answer:

-50.22kJ

Explanation:

Given:

Initial state Temperature =300k

Pressure 1= 150kPa

Pressure 2 = 800kPa

Volume 1 = 0.2m^3

Work done = Pressure 1 × Volume 1 × (inverse of pressure 1 ÷ pressure 2)

= 150kPa × 0.2m^3 × (inv 150 kPa ÷800 kPa)

= - 50.22kJ

3 0

4 years ago

ISTH ME A LEMONITH ANSWERITH MEITH

e-lub [12.9K]

W h a t are you saying

3 0

3 years ago

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting,

ArbitrLikvidat [17]

Answer:

minimum separation distance between the two legs of the sting L = L 1 + L 2 therefore L = 9.48 + 4.68 = 14.16 m

L = 1.14 m

Explanation:

D ( diameter ) = 125 m

convection coefficient of h = 700 W/m^2

Calculate THE CROSS SECTIONAL AREA

Ac = $\frac{\pi }{4} * D^2$ = $\frac{\pi }{4} * ( 125 )^2$ = 0.79 * 15625 = 12343.75 m^2

perimeter

p = $\pi * D$ = 3.14 * 125 = 392.5 m

at 300k temperature the thermal conductivity of copper and constantan from the thermodynamic property table are :

Kcu = 401 w/m.k

Kconstantan = 23 W/m.k

To calculate the length of copper wire of the thermocouple junction

L 1 = 4.6 ( $\frac{Kcv Ac}{h P}$ ) ^ 1/2 = 4.6 $(\frac{401 *12343.75 }{700 *392.5})^\frac{1}{2}$

L 1 = 4.6 ( 4949843.75 / 274750 )^1/2

L 1 = 9.48 m

calculate length of constantan wire

L 2 = 4.6 $(\frac{kcons*Ac}{hp} )^\frac{1}{2}$

= 4.6 ( (23 * 12343.75) / ( 700 * 392.5) ) ^1/2

L 2 = 4.6 ( 283906.25 / 274750 ) ^ 1/2

L 2 = 4.68 m

I) therefore the minimum separation distance between the two legs of the sting L = L 1 + L 2

L = 9.48 + 4.68 = 14.16 m

ii) Evaluating the thermal conductivity of copper and constantan

Kc ( thermal conductivity of chromel) = 19 w/m.k

Ka ( thermal conductivity of alumel ) = 29 W/m.k

distance between the legs L = L 1 + L 2

THEREFORE

L = 4.6 ( (Kcn * Ac ) / ( hp ) )^1/2 + 4.6 ( (Kac * Ac)/(hp) )^1/2

L = 4.6 $(\frac{Ac}{hp} )^\frac{1}{2} [ (Kcn)^\frac{1}{2} + (Kal)^\frac{1}{2} ]$

L = 4.6 ( 12343.75 /( 700 * 392.5) )^1/2 * [ 19^1/2 + 29^1/2 ]

L = 4.6 ( 12343.75 / 274750 ) ^1/2 * 5.39

L = 1.14 m

4 0

4 years ago