In equation 17.9, show that units of Pa for pressure and m^3 for volume are compatible with J fo

In equation 17.9, show that units of Pa for pressure and m^3...

Key Concepts

Energy (Joule)

The Joule (J) is the SI unit of energy, work, or heat. One Joule is equivalent to one Newton meter (N·m), linking mechanical work directly to energy. This unit is fundamental in expressing changes in energy across physical systems.

Units Compatibility

The compatibility of units is demonstrated by multiplying pressure (in Pascals, or N/m²) by volume (in m³), which yields Newton meters (N·m), or Joules. This shows that when pressure and volume are appropriately combined, the resulting unit aligns with the energy unit used in thermodynamic and mechanical calculations.

Pressure (Pascal)

A Pascal (Pa) is the SI unit of pressure defined as one Newton per square meter (N/m²). It represents the force exerted per unit area and establishes a direct relationship between the applied force and the surface over which it acts.

Volume (Cubic Meters)

Cubic meters (m³) measure the space occupied by a substance. In SI units, volume is an essential physical quantity that, when combined with pressure, helps in calculating work or energy in various physical processes.

Transcript

00:01 A reactor which is filled with water at a given state is placed inside a containment room.

00:06 The containment room is well insulated and evacuated.

00:10 The reactor then ruptures and the water fills the entire room.

00:14 If we are given the room volume, we want to find the final pressure using trial and error.

00:21 So we can take our control volume as the containment room and the reactor.

00:26 And we can write the continuity equation for the mass of water, m2, is equal to m1 so the mass stays constant and this mass is equal to the volume of the reactor divided by the specific volume of the state v1 and we go to our tables we can find the specific volume so that's 1 over 0 .001823 and so we get the mass of water which is constant to be 548 0 .5 kg.

01:09 Now we can write the energy equation from the first law as well.

01:17 M into u2 minus u1.

01:21 The change in internal energy within the room must equal to the heat transfer q minus the work done w.

01:29 But both these terms are zero.

01:31 So this is equal to 0 minus zero which is and therefore we can see that the internal energy is also constant u2 is equal to u1 this is 1 ,702 .8 kilojoules per g so the specific internal energy does not change now the total volume and mass gives us v2 which is the total volume of the room divided by the total mass m2 the final mass.

02:15 And if we put our values in here, we get the specific volume to be 0 .1823 cubic meters per kg.

02:30 Now note that in the vicinity of this specific volume 0 .1823 cubic meters per kg, crossing the saturated vapor line, the internal energy is about 2 ,585 kilojoules per kg.

02:47 But at the actual state 2, we found that the internal energy u2 is equal to 1 ,702 .8 kilojoules to kg.

02:57 Therefore, state 2 must be in the two -phase region.

03:01 So we need to use trial and error...

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