Answer:
Given:
high temperature reservoir
low temperature reservoir
thermal efficiency
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as
The thermal efficiency of second heat engine can be written as
The temperature of intermediate reservoir can be defined as
To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.
By definition the entropy change would be defined as
Using the Boyle equation we have
Where,
= Specific heat at constant pressure
= Initial temperature of gas
= Final temprature of gas
R = Universal gas constant
= Initial specific Volume of gas
= Final specific volume of gas
According to the statement, it is an isothermal process and the tank is therefore rigid
The equation would turn out as
<em>Therefore the entropy change of the ideal gas is 0</em>
Into the surroundings we have that
Where,
Q = Heat Exchange
T = Temperature in the surrounding
Replacing with our values we have that
<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>
Answer:
mass of LNG: 129501.3388 kg
quality: 0.005048662
Explanation:
Volume occupied by liquid:
400 m^3*0.9 = 360 m^3
Volume occupied by vapor
400 m^3*0.1 = 40 m^3
A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present
For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg
From specific volume definition:
vf = liquid volume/liquid mass
liquid mass = liquid volume/vf
liquid mass = 360 m^3/0.002794 m^3/kg
liquid mass = 128847.5304 kg
vg = vapor volume/vapor mass
vapor mass = liquid volume/vg
vapor mass = 40 m^3/0.06118 m^3/kg
vapor mass = 653.8084341 kg
total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg
Quality is defined as the ratio between vapor mass and total mass
quality = 653.8084341 kg/129501.3388 kg = 0.005048662
