Two rigid tanks are connected by a valve. Tank A contains 0.2 m3 of water at 400 kPa and 80 percent quality. Tank B contains 0.5 m3 of water at 200 kPa and 250°C. The valve is now opened, and the two tanks eventually come to the same state. Determine the pressure and the amount of heat transfer when the system reaches thermal equilibrium with the surroundings at 20°C.

Pressure = 3.1698 KPa

Amount of heat transfer = 2,170KJ

Explanation:

In this question, we are asked to calculate the pressure and the amount of heat transferred Q when a system at a previous temperature reaches thermal equilibrium with its surroundings at a new temperature.

To answer this question, we make use of the saturated water pressure table. It is from here that we select values for saturated pressure, specific volume, saturated vapor, internal energy and evaporation at the initial temperature.

It is from here that we use a mix of mass transfer calculations get the amount of heat transferred. Please do check attachment for explicit calculations

A)Final pressure = 3.17KPa

B) The amount of heat transfer is 2170.5Kj

Explanation:

A) From energy balance equation in thermodynamics ;

Ein - Eout = ΔEsyst

Now, let's determine the specific volume for the tank A.

at P = 400 KPa and x1 = 0.8,using tables 4 to 6 i attached, by interpolation, we trace that;

Specific volume of liquid; vfa= 0.001084 m³/kg

Specific volume of vapour; vg = 0.46242 m³/kg

Internal energy of liquid; ufa = 604.22 Kj/kg

Internal energy of vapour; ufg = 1948.9 Kj/kg

From the question, we know that quality; qa = 0.8

Thus, let's calculate total specific volume in tank A;

It's given by;

va = vfa + qa(vg - vfa)

va = 0.001084 + 0.8(0.46242 - 0.001084) = 0.001084 + 0.8(0.461336) = 0.37015 m³/kg

Now, we can find the mass of tank A

Mass is given as;

M = V/va

Where V = volume of tank A = 0.2 m³

Thus, Ma = 0.2/0.37015 = 0.54032 kg

Now, in the same way, let's calculate internal energy of tank A.

ua = uf + qa(ug)

ua = 604.22 + (0.8 x 1948.9) = 2163.34 Kj/kg

Now, for Tank B;

from table A6 among the tables i attached and looking at pressure of 200 kPa and temperature of 250∘C, we have;

Specific volume; vb = 1.1989 m³/kg

Internal energy; ub = 2731.4 kj/kg

volume of tank B = 0.5 m³

Thus;

Mass of tank B = Vb/vb = 0.5/1.1989 = 0.417 kg

Now, the final pressure will be rhe saturation pressure at temperature of 25°C. From the table A4 attached, Psat = 3.17 KPa

Thus, Final pressure = 3.17KPa

B) To determine the amount of heat transfer, first of all, we have to find the quality of the mixture at the final state from;

q = (v(total) - v(liquid))/(v(vap) - v(liquid))

Again, from the tables I attached, at 25°C;

v(liquid) = vf = 0.001003 m³/s

v(vap) = vg = 43.34 m³/kg

Now, v(total) = V(total)/M(total)

V(total) = Va + Vb = 0.2 + 0.5= 0.7m³

m(total) = 0.54032 + 0.417 = 0.95732 kg

Thus, v(total) = 0.7/0.95732 = 0.7312 m³/kg

Hence,

q = (0.7312 - 0.001003)/(43.34 - 0.001003) = 0.0168

Internal energy (u) is given as;

u = u(liquid) + q(u(evap))

From table A-4 attached and at 25°C;

u(liquid) = uf = 104.83 Kj/kg

u(evap) = ufg = 2304.3 Kj/kg

Thus;

u = 104.83 + (0.0168 x 2304.3) = 143.54 Kj/kg

I've already talked about energy balance at the beginning. Thus, in this question;

Amount of heat transfer is given as; ΔQ = - ΔU

Thus,

ΔQ = - [(m(total) x u) - (ma x ua) - (mb x ub)]

= - [(0.95732 x 143.54) - (0.54032 x 2163.34) - (0.417 x 2731.4)

= -137.412 + 1168.896 + 1138.994 = 2170.478 Kj or approximately 2170.5Kj

answer:

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