Question

A closed, rigid tank contains 2 kg of water initially at 80 degree C and a quality of 0.6. Heat transfer occurs until the tank contains only saturated vapor at a higher pressure. Kinetic and potential energy effects are negligible. For the water as the system, determine the amount of energy transfer by heat, in kJ.

Answer 1

Answer:

Q=1847kJ

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

The heat that is being applied to water is given by the following equation using the first law of thermodynamics.

Q=m(h2-h1)

then to solve this problem we must find the enthalpy of state 1 and 2.

for state 1.

t=80C

x=0.6

h1=1720kJ/Kg

for state 2.

t=80C

X=1(saturated vapor)

h2=2643kJ/Kg

solving

Q=m(h2-h1)

Q=2(2643-1720)

Q=1847kJ

Answer 2

Answer:

$Q=1752.3kJ$

Explanation:

Hello,

In this case, the transferred heat is defined via the first law of thermodynamics as shown below as it is about a rigid tank which does not perform any work:

$Q_{in}=m_{H_2O}(u_2-u_1)$

The internal energy at the first state (80°C as a vapor-liquid mixture) is computed based on its quality as follows:

$u_1=334.97kJ/kg+0.6*2146.6kJ/kg=1622.93kJ/kg$

Now, the specific volume turn out into:

$v_1=0.001029m^3/kg+0.6*3.404271m^3/kg=2.0435916m^3/kg$

As the volume does not change due to the fact that this is about a rigid tank, we must look for a temperature at which the saturated vapor's volume matches with the previously computed volume. This turn out into a temperature of about 94.17 °C at which the internal energy of the saturated vapor is about (by interpolation):

$u_2=2499.1kJ/kg$

In such a way, the energy transfer by heat is:

$Q=2kg*(2499.1kJ/kg-1622.93kJ/kg)\\Q=1752.3kJ$

Best regards.