Engineering, 15.02.2020 05:24 asyamelissa01

A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is then heated until its temperature is 180°C. The final state is superheated water and the internal energy at this state should be obtained by interpolation. Calculate the heat transfer required for this process. Use data from the steam tables.

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Answer from: sonyalehardez

98.13kJ

Explanation:

Given that;

The rigid 10-L vessel initially contains a mixture of liquid water & vapor at

$T_1 =100^0C\\T_2 = 180^0C$

We are to calculate the heat transfer required during the process by obtaining our data from the steam tables.

In order to do that, let start with our Energy Balance

So, Energy Balance for closed rigid tank system is given as:

$\delta E_{system} = E_{in} - E_{out}$

Since the K.E and P.E are insignificant;

∴ K.E = P.E = 0

$Q_{in}= \delta U + W\\Q_{in} = m(u_2-u_1)+ W$

Where;

m = mass flow rate of the mixture

(u_2-u_1) = corresponding change in the internal energy at state point 2 and 1

However, since we are informed that the vessel is rigid, then there is no work done in the system, then W turn out to be equal to zero .i,e

W = 0

we have our above equation re-written as:

$Q_{in}= m (u_2-u_1)+0\\$

$Q_{in}= m(u_2-u_1)$

We were told to obtain our data from the steam table, so were going to do just that

∴ At inlet temperature T_1 = 100^0C , the given quality of mixture of liquid water and vapor (x_1) = 123% = 0.123

Using the equation:

$v_1 = v_f + x_1v_{fg}\\v_1 = v_f + x_1(v_g-v_f)$

where;

v_1 = specific volume at state 1

v_f = specific volume of the liquid

v_g = specific volume of the liquid vapor mixture

The above data from the steam table is given as;

v_f = 0.001043 m³/kg

v_g = 1.6720 m³/kg

so; we have

v_1 = 0.001043 + 0.123(1.6720-0.001043)

v_1 = 0.001043+0.123(1.670957)

v_1 =0.001043+0.205527711

v_1= 0.206570711m^3/kg

v_1=0.2065m^3/kg

At T_1 = 100°C and x_1=0.123 ;

the following steam data from the tables were still obtained for the internal energy; which is given as:

Internal Energy (u_1) at the state 1

$u_1= u_f + xu_{fg}$

where;

specific internal energy of the liquid (u_f) = 419.06 kJ/kg

The specific internal energy of the liquid vapor mixture $(u_{fg})$ = 2087.0 kJ/kg

∴ since ; $u_1= u_f + xu_{fg}$

(u_1) = 419.06 + (0.123 × 2087.0)

(u_1) = 675.761 kJ/kg

As the tank is rigid, so as the volume which is kept constant:

$v_2=v_1\\=0.2065 m^3/kg$

Now, let take a look at when T_2 = 180^0C from the data in the steam tables

Specific volume of the liquid (v_f) = 0.00113 m³/kg

specific volume of the liquid vapor mixture (v_g) = 0.19384 m³/kg

The quality of the mixture at the final state 2 can be determined by using the equation shown below:

$v_2=v_f+x_2v_{fg}$

$x_2=\frac{v_2-v_f}{v_{fg}}$

$x_2=\frac{v_2-v_f}{v_g-v_f}$

$x_2=\frac{0.2065-0.00113}{0.19384-0.00113}$

$x_2=\frac{0.20537}{0.19271}$

= 1.0657

From our usual steam table; we still obtained data for the Internal Energy when T_2=180^0C

Specific internal energy of the liquid (u_f) = 761.92 kJ/kg

Specific internal energy of the liquid vapor mixture $u_{fg}$ = 1820.88 kJ/kg

Calculating the internal energy at finsl state point 2 ; we have:

$u_2=u_f+u_{fg}$

= 761.92 + (1.0657 × 1820.88)

= 761.92 + 1940.511816

= 2702.431816

u_2 ≅ 2702.43 kJ/kg

Furthermore, let us calculate the mass in the system; we have:

$m= \frac{V_1}{v_1}$

where V₁ = the volume 10 - L given by the system and v₁ = specific volume at state 1 as 0.2065

V₁ = the volume 10 - L = 10 × ( 0.001 m³/L)

v₁ = 0.2065

∴

mass (m) = $\frac{10(0.001m^3/L)}{0.2065}$

= 0.04842 kg

Now, we gotten all we nee do calculate for the heat transfer that is required during the process:

$Q_{in}= m(u_2-u_1)$

$Q_{in}= 0.04842(2702.43-675.761)$

$Q_{in}= 0.04842(2026.669)$

$Q_{in}= 98.13 kJ$

Therefore, the heat transfer that is required during the process = 98.13 kJ

There you have it!, I hope this really helps alot!

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A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percen...