4-35 A well-insulated rigid tank contains 5 kg of a saturated liquid-vapor mixtu

Question

4-35 A well-insulated rigid tank contains 5 kg of a saturated liquid-vapor mixture of water at 100 kPa. Initially three-quarters of the mass is in the liquid phase. An electric resistor placed in the tank connected to a 100-V source, and a current of 8 A flows through the resistor when the switched is turn on. Determine how long it will take to vaporize all the liquid in the tank. Also, show the process on a T-v diagram with respect to saturation lines. (Note: I was trying to follow Cheggs/Cramster solution and I got lost when I tried to follow the interpolation, where did they get the values u2 = 2554.4 + (0.4243-0.4466)/(0.39248-0.4466) x (2559.1

Explanation / Answer

First of all since it is an insulated system, hence, Q = 0, and also , K.E. = P.E. = 0

By first law:

Win = U = m.(u2 - u1)

m = 5 Kg

Now, Win is only electrical energy (No Expansion work as Volume remains constant, hence P.dV = 0)

Win = V.I.t

Now at state 1, i.e. P1 = 100 Kpa and quality = 0.25

v1 = vf + 0.25 x vfg

   = (0.001043) + 0.25 x 1.693 = 0.4243 m^3/Kg

u1 = uf + 0.25 x ufg = 417.40 + 0.25 x 2088.2 = 939.05 KJ / Kg

[I hope u could find the values if vf , vfg, uf and ufg from the table. they have been taken directly from table]

Now v2 = v1 = 0.4243 m^3/kg

also do note that state 2 is a saturated vapour. ie. v2 = vg = 0.4243 m^3?kg

Now at Vg = 0.4466 m^3/kg u = 2559.1 KJ/Kg

and at Vg = 0.39248 m^3 /kg u = 2554.4 KJ/Kg

so vg = 0.424 3must be lying in b/w these two values and we have to find the corresponding u.

INTERPOLATION:

The rule which I am going to teach u now, stick to it in future also. It can help u out.

Interpolation can be imagined as a simple equation of line. Let it be:

y = m.x + c

now put , y= u and x = vg , to form 2 equations and find the value of m and c

2559.1 = m x 0.4466 + C (i)

2554.4 = m x 0.39248 + C (ii)

now on solving these 2 equations :

m = 86.84 and C = 2520.315

so line equation becomes :

y = 86.4 . X + 2520.31

Now we out X = 0.4243 to obtain thevalue of y which will be our U at Vg = 0.4243

U2 = y = 86.4 x 0.4243 + 2520.31

   = 2556.9 (approximately close)

V.I. dt = 5 x ( 2556.9 - 939.05) x 1000

110 x 8 x dt = 5 x 1617.85 x 1000

dt = 9186 s = 153.1 min

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