Question

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.

Answer 1

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ  = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

                   u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

                   u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

                 0.014m³ / 0.2066 m³/kg

              m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

                   u₁ = 419.06 kj / kg + 0.123  .  2087.0 kj/kg

                    u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ =  0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524  . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

Answer 2

Answer:

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!