Mathematical modeling aids in technological design by simulating how.

A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa

Fed [463]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

4 0

3 years ago

Read 2 more answers

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value

OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^{-kt^n}$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^{-k^146^2.1}$

$0.50 = e^{-k306.6}$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26\times 10^{-3}$

for 86 % completion

$0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}$

$e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14$

$-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)$

$t^{2.1} = 869.96$

t = 25.10 sec

5 0

3 years ago

a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419

dusya [7]

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

E = 0 N/C

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

E = 0 N/C

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

E = 7.78 x10^5 N/C

4 0

3 years ago

The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 106 Btu/h. The comb

Veronika [31]

Answer:

Explanation:uhhhhhh all?

7 0

4 years ago

When they say in the United States that a car’s tire is filled “to 32 lb,” they mean that its internal pressure is 32 lbf/i

arsen [322]

Answer:

0.71 lbf

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

P = 32 lbf/in² + 14.7 lbf/in²

P = 46.7 lbf/in²

Absolute temperature is in Kelvin or Rankine:

T = 75 + 459.67 R

T = 534.67 R

Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:

PV = nRT

(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)

n = 0.02442 lb-mol

The molar mass of air is 29 lbm/lb-mol, so the mass is:

m = (0.02442 lb-mol) (29 lbm/lb-mol)

m = 0.708 lbm

The weight of 1 lbm is lbf.

W = 0.708 lbf

Rounded to two significant figures, the weight of the air is 0.71 lbf.

3 0

3 years ago