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3 years ago

Which is a common refrigerant for domestic air conditioners?

Engineering

2 answers:

mihalych1998 [28]3 years ago

6 0

Answer:

The most common refrigerants used for air conditioning over the years include

Explanation:

The most common refrigerants used for air conditioning over the years include.

Reptile [31]3 years ago

4 0

Answer:

Explanation:

The most common HFC used in air conditioners is R-410A. This refrigerant is better than R-22 in terms of “Ozone Depletion” potential and energy efficiency, but it still causes global warming. A few more HFCs that are commonly used are: R-32 in Air Conditioners and R-134A in refrigerators.

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Can you help me with this

Fiesta28 [93]

the function is to provide sealed combustion so that the loss of gas is minimized

6 0

1 year ago

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000 $\frac{ft lbm}{s^{2} }$

We know that moment of inertia is given as

$I = \frac{bt^{3} }{12}$

$I = \frac{5 (1.5^{3} )}{12}$

I = 1.40625 $in^{4}$

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = $\frac{PL^{3} }{48EI}$

$0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}$

E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0

2 years ago

You are ordering steel cable for a 250 foot long zip-line you are building in your back yard. The cable can be ordered in diamet

Margarita [4]

Answer:

If i am correct It should be 1/4 of an inch

Explanation:

Sorry but i can't quite explain

3 0

3 years ago

Read 2 more answers

An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure

Tamiku [17]

Answer:

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

Explanation:

Given that:

The mass of the oxygen $m_{O_2}$ = 35 kg

The mass of the nitrogen $m_{N_2}$ = 40 kg

The cooling temperature of the mixture T = 84 K

The cooling pressure of the mixture P = 0.1 MPa

From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.

i.e.

Liquid percentage of $O_2$ = 70% = 0.70

Vapor percentage of $O_2$ = 34% = 0.34

The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively

Thus, the number of moles of each component is:

number of moles of oxygen = 35/32

number of moles of oxygen = 1.0938 kmol

number of moles of nitrogen = 40/28

number of moles of nitrogen = 1.4286 kmol

Hence, the total no. of moles in the mixture is:

$N_{total} = 1.0938+1.4286$

$N_{total} = 2.5224 \ kmol$

So, the total no of moles in the whole system is:

$N_f + N_g = 2.5224 --- (1)$

The total number of moles for oxygen in the system is

$0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)$

From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)

∴

0.7(2.5224 - N_g) + 0.34 N_g = 1.0938

1.76568 - 0.7 N_g + 0.34 N_g = 1.0938

1.76568 - 0.36 N_g = 1.0938

1.76568 - 1.0938 = 0.36 N_g

0.67188 = 0.36 N_g

N_g = 0.67188/0.36

N_g = 1.866

From equation (1)

$N_f + N_g = 2.5224$

N_f + 1.866 = 2.5224

N_f = 2.5224 - 1.866

N_f = 0.6564

Thus, the mass of oxygen in the liquid and vapor phases is:

$m_{fO_2} = 0.7 \times 0.6564 \times 32$

$m_{fO_2} = 14.703 \ kg$

The mass of oxygen in liquid phase = 14.703 kg

$m_{g_O_2} = 0.34 \times 1.866 \times 32$

$m_{g_O_2} = 20.302 \ kg$

The mass of oxygen in the vapor phase = 20.302 kg

8 0

2 years ago

In a production facility, 3 cm thick large brass plates (k = 110 W/mC, α = 33.9 × 10-6 m2 /s) that are initially at a uniform

zysi [14]

Answer:

Explanation:

Given.

Thickness of brass plate $t = 3 cm$

Thermal conductivity of brass $k = 110 W/m.°C$

Density of brass $\rho = 8530 kg/m^3$

Specific heat of brass $C_p =380J/kg.°C$

Thermal diffusivity of brass $\alpha = 33.9\times 10^{-6} m^2/s$

Temperature of oven $T_{\infty} = 700°C$

The initial temperature $T_i= 25°C$

Plate remain in the oven $t =10 min$

Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane.

The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

The Biot number for this process $Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109$

The constants $\lambda_1$ and $A_1$ corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

$\lambda_1=0.1039$ and $A_1=1.0018
$

The Fourier number $\tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2$

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

$\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C$

3 0

2 years ago

Read 2 more answers