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2 years ago

What are the nominal dimensions for a 1x2 stick of lumber, a 2x4 stick of lumber and a standard sheet of plywood?

1 answer:

6 0

3/4 x 1 1/2 inches (19 x 38 mm) is the actual size for 1x2 stick of lumber,

1 1/2 x 3 1/2 inches (38 x 89 mm) is the actually size for a 2x4 stick of lumber,

Plywood is usually sold in 4 x 8-foot sheets. The most common nominal thicknesses of plywood are 1/2 inch and 3/4 inch, but once again the actual sizes are slightly different. A sheet of 1/2-inch plywood is really 15/32 inch thick, while a 3/4-inch sheet is 23/32 inch thick.

Hopefully this answers your question, I apologize if it doesn’t :)

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A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar

Westkost [7]

Answer:

a) $m_{2} = 0.753\,kg$ , b) $Q_{in} = 2122.963\,kJ$

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.02726\,\frac{m^{3}}{kg}$

$u = 1192.94\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.2$

State 2 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.06643\,\frac{m^{3}}{kg}$

$u = 1718.12\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.5$

The model for the rigid tank is created by using the First Law of Thermodynamics:

$Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}$

Initial and final masses are:

$m_{1} = \frac{V_{1}}{\nu_{1}}$

$m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }$

$m_{1} = 1.834\,kg$

$m_{2} = \frac{V_{2}}{\nu_{2}}$

$m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }$

$m_{2} = 0.753\,kg$

a) The final mass within the tank is:

$m_{2} = 0.753\,kg$

b) The total amount of heat transfer is:

$Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}$

$Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )$

$Q_{in} = 2122.963\,kJ$

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3 years ago

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Answer:

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What is a reasonable approximation for an inductor at DC steady state?

abruzzese [7]

Answer:

Short circuit

Explanation:

An inductor works on the principle of electromagnetic induction.

When a current passes through an inductor, a magnetic field is generated inside it; if the magnitude of this magnetic field changes, then according to Faraday-Newmann-Lenz law, an emf (electromotive force) is induced in the inductor itself, according to:

$\epsilon=-\frac{d\Phi}{dt}$

where $\Phi$ is the magnetic flux linkage through the inductor. Therefore, as a result of this induced emf, a current is also induced in the inductor, and the direction of this current is such that it opposes to the change in magnetic flux linkage.

Therefore, the intensity of this current depends on the rate of change of magnetic flux linkage through the inductor.

If we have a DC (direct current), then the current in the inductor is steady; as a result, its magnetic field is also steady, and therefore, there is no change in magnetic flux, so no induced current. Therefore, the inductor will oppose no resistance to the flow of original current, and so it can be approximated to a short circuit (zero resistance).

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2 years ago