Explain the differences between 1- Energy 2- Power 3- Work 4- Heat Your answer should explain the mathematica and physical meani
ng. A thermodynamic example with a nice sketch for each case is a must.
1 answer:
Answer:
Explanation:
Energy : Energy can be defined as both qualitative and quantitative property that can be given to an object in the form of work or heat.
For example - water stored in a tank at a height,h from the ground has a potential energy of U = ρgh.
Mathematically, Energy of water stored in a tank is
U = ρ.g.h
where, ρ is water density
g is acceleration due to gravity
h is height of water tank from ground
Power : Power is defined as rate at which work can be done.
For example - When a force F is applied to an object to move it through a distance d in time t seconds is an example of power delivered.
Mathematically, Power = Work done / Time
Work : When a force act on a body and the body moves, work is said to be done.
For example - Gas confined in a cylinder by a piston. The gas expands when it is heated, doing work on the piston.
Mathematically, Work,W = F X d
where, F is force
d is distance
Heat : Heat is a form of energy which is produced as a result of motion of atoms and particles.
For example - while increasing the temperature of water, heat is added to the water by means of gas burner or any other form.
Mathematically we can calculate heat release or heat absorb by
q = m.C.ΔT
where q is heat
m is mass
C is specific heat capacity
ΔT is change in temperature
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Answer:
A) W' = 178.568 KW
B) ΔS = 2.6367 KW/k
C) η = 0.3
Explanation:
We are given;
Temperature at state 1;T1 = 360 °C
Temperature at state 2;T2 = 160 °C
Pressure at state 1;P1 = 10 bar
Pressure at State 2;P2 = 1 bar
Volumetric flow rate;V' = 0.8 m³/s
A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;
Specific volume;v1 = 0.287322 m³/kg
Mass flow rate of water vapour at turbine is defined by the formula;
m' = V'/v1
So; m' = 0.8/0.287322
m' = 2.784 kg/s
Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific enthalpy;h1 = 3179.46 KJ/kg
Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific enthalpy;h2 = 3115.32 KJ/kg
Now, since stray heat transfer is neglected at turbine, we have;
-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2 - h1)
Plugging in relevant values, the work of the turbine is;
W' = -2.784(3115.32 - 3179.46)
W' = 178.568 KW
B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific entropy: s1 = 7.3357 KJ/Kg.k
Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific entropy; s2 = 8.2828 KJ/kg.k
The amount of entropy produced is defined by;
ΔS = m'(s2 - s1)
ΔS = 2.784(8.2828 - 7.3357)
ΔS = 2.6367 KW/k
C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;
h2s = 2966.14 KJ/Kg
Energy equation for turbine at ideal process is defined as;
Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Again, Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2s - h1)
W' = -2.784(2966.14 - 3179.46)
W' = 593.88 KW
the isentropic turbine efficiency is defined as;
η = W_actual/W_ideal
η = 178.568/593.88 = 0.3
