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hoa [83]
3 years ago
12

Consider a 1000-W iron whose base plate is made of 0.5-cm-thick aluminum alloy 2024-T6 (rho = 2770 kg/m3 and cp = 875 J/kg·°C).

The base plate has a surface area of 0.03 m2. Initially, the iron is in thermal equilibrium with the ambient air at 22°C. Assuming 90 percent of the heat generated in the resistance wires is transferred to the plate, determine the minimum time needed for the plate temperature to reach 220°C.
Engineering
1 answer:
Lesechka [4]3 years ago
5 0

Answer: 79.98sec

Explanation:

Power received by plate = 0.9x1000

= 900W

Volume of metal = area x thickness

V = 0.5x10^-2 x 0.03 = 1.5x10^-4m3

Mass of metal = density x volume

= 2770kg/m3 x 1.5x10^-4 = 0.4155kg

Electrical heat energy = power x time = mass x specific heat capacity(Cp) x temperature change

900 x t = 0.4155 x 875 x (220-22)

900t = 71985.375

t = 79.98s

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What is the perimeter of 14-7 and 3-4
Goshia [24]

Answer:

If you mean two sides are 7 and two sides are 14 then you'd have 42

and for the second you'd have 14

Explanation:

7 + 7 = 14, 14 + 14 = 28, 14 + 28 = 42

3 + 3 = 6, 4 + 4 = 8, 8 + 6 = 14

5 0
3 years ago
A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu
bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

b.

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0
4 years ago
A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a
bazaltina [42]

Answer:

Q_{cv}=-339.347kJ

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

P_iV_i=m_iRT_i (i could be 1 for initial and 2 for the end)

State1

1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295

m_1=232kg

State2

8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350

m_2=11.946

So, the total mass of the aire entered is

m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

u_1=210.49kJ/kg

For Specific enthalpy

h_1=295.17kJ/kg

For the second state the Specific internal Energy (6bar, 350K)

u_2=250.02kJ/kg

At the end we make a Energy balance, so

U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

No work done there is here, so clearing the equation for Q

Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)

Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)

Q_{cv}=-339.347kJ

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0
3 years ago
You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0
3 years ago
The enforcement of OSHA standards is provided by federal and state
Gnoma [55]

Answer:

Explanation:

Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.

3 0
2 years ago
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