Efrigerant-134a is expanded isentropically from 600 kpa and 70°c at the inlet of a steady-flow turbine to 100 kpa at the outlet. the outlet area is 1 m2, and the inlet area is 0.5 m2. calculate the inlet and outlet velocities when the mass flow rate is 0.75 kg/s.

Inlet :

Outlet:  

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

, where represent the flow rate and the specific volume at the pressure and temperature given.

is the inlet area

pressure at the inlet area

temperature at the inlet area

is the outlet area

pressure at the outlet area

temperature at the outlet area

represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

and the entropy is

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

Our interest value would be given using interpolation like this:

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that , but since we have this:

If we solve from the velocity v we have this:

  (*)

And now we just need to replace the values into equation (*)

For the inlet case:

For the oulet case:

the largest known star in the universe is uy scuti, a hypergiant with a radius around 1,700 times larger than the

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.