Ask question

Ask question

All categories

zloy xaker [14]

2 years ago

8

PTAC and window-unit air conditioners often use rotary compressors, which are effective but susceptible to failure due to high h

ead pressure. What is the most common cause of high head pressure in these units

1 answer:

icang [17]2 years ago

6 0

Answer:

Dirty filter.

Explanation:

You might be interested in

Refrigerant 134a is the working fluid in a vaporcompression heat pump that provides 35 kW to heat a dwelling on a day when the o

Dennis_Churaev [7]

Answer:

Hello, dear.

For the answer please see the explanation below.

Explanation:

The compressor power is:

2.212kW

(b) The refrigeration capacity is:

3.62 tons

(c) The coefficient of performance is:

5.75

5 0

4 years ago

Consider the formation of p-nitrophenol from p-nitrophenyl trimethyl acetate. The process is known as enzymatic hydrolysis and i

solniwko [45]

Solution :

cs=zeros(9001);

ca=zeros(9001);

cp=zeros(9001);

psi=zeros(9001);

t=[0:0.1:900];

cs(1)=0.5;

ce(1)=0.001;

cp(1)=0;

ca(1)=0;

psi(1)=0;

for i=1:1:9000

cs(i+1)=cs(i)-0.1*((0.015*cs(i))/(5.53+cs(i)));

cp(i+1)=cp(i)+0.1*((0.015*cs(i))/(5.53+cs(i))-0.0026*cp(i));

ca(i+1)=ca(i)+0.1*0.0026*cp(i);

psi(i+1)=((cp(i+1)-cp(i)))/((cs(i)-cs(i+1)));

end

plot(t,cs,t,cp,t,ca);

plot(t,psi);

6 0

3 years ago

A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

4 years ago

A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Know that car A has a mass o

posledela

Answer:

a) $d=236.280\,m$ , b) $F_{coupling} = -8848\,N$ The real force has the opposite direction.

Explanation:

a) Let assume that train moves on the horizontal ground. An equation for the distance travelled by the train is modelled after the Principle of Energy Conservation and Work-Energy Theorem:

$K_{A} = W_{brake}$

$\frac{1}{2}\cdot m_{train} \cdot v^{2} = F_{brakes}\cdot d$

$d = \frac{m_{train}\cdot v^{2}}{2\cdot F_{brakes}}$

$d = \frac{(51000\,kg)\cdot [(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (82000\,N)}$

$d=194.360\,m$

b) The acceleration experimented by both trains are:

$a = -\frac{v_{o}^{2}}{2\cdot d}$

$a = -\frac{[(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s})]^{2}}{2\cdot (194.360\,m)}$

$a = -1.608\,\frac{m}{s^{2}}$

The coupling force in the car A can derived of the following equation of equilibrium:

$\Sigma F = F_{coupling} - F_{brakes} = m_{A}\cdot a$

The coupling force between cars is:

$F_{coupling} = m_{A}\cdot a + F_{brakes}$

$F_{coupling} = (31000\,kg)\cdot(-1.608\,\frac{m}{s^{2}} )+41000\,N$

$F_{coupling} = -8848\,N$

The real force has the opposite direction.

8 0

3 years ago

Five Kilograms of continuous boron fibers are introduced in a unidirectional orientation into of an 8kg aluminum matrix. Calcula

Lunna [17]

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

7 0

3 years ago