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bonufazy [111]
1 year ago
5

This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and

cord-and-plug connected equipment required to be grounded. The point of this test is to make sure that the equipment grounding conductor is electrically continuous.
Engineering
1 answer:
Ray Of Light [21]1 year ago
5 0

Answer:

That would be Continuity test

Explanation:

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The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A
adelina 88 [10]

Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

COP=\frac{Q_{in}}{W}

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}

Where T_{house} = 24°C and T_{out} is temperature outside

To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump

Which has COP of:

COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}

So we equate the COP of our heater with COP of Carnot heater

\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}

Rearrange the equation

\frac{1.25}{4}(24-T_{out})^2-24=0

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

4 0
3 years ago
Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV  = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0
3 years ago
2. The device that varies incoming line pressure using centrif-
Kryger [21]

Answer:

I believe  reverse boost valve.

3 0
3 years ago
2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks
Amiraneli [1.4K]

Answer:

  • 100N
  • 25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

4 0
3 years ago
The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both
DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

                               L^2 = x_a^2 + y_b^2   ....... 1

                               y_b = sqrt( 20^2 - 16^2 )

                               y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

                               0 = 2*x_a*v_a + 2*y_b*v_b

                               0 = x_a*v_a + y_b*v_b   ..... 2

                               0 = 16*36 + 12*v_b

                               v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

                               0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

                               0 = 9 + 4*a_a/3 + 16 + a_b

                               4*a_a/3 + a_b = -25

                               4*a_a + 3*a_b = -75  .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

                               F_x = m_a*a_a

                               P - R_r*16/20 = m_a*a_a

                               

- For Slider B, Apply Newton's second law of motion in y direction:

                               F_y = m_b*a_b

                               - R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

                               P - 4*m_b*a_b / 3 = m_a*a_a

                               3P = 3*m_a*a_a + 4*m_b*a_b  ... 4

- Where,                  P = Is the force acting on slider A

                               P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.                    

5 0
3 years ago
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