Question

A 3-ft3 rigid tank initially contains saturated water vapor at 300°F. The tank is connected by a valve to a supply line that carries steam at 200 psia and 400°F. Now the valve is opened, and steam is allowed to enter the tank. Heat transfer takes place with the surroundings such that the temperature in the tank remains constant at 300°F at all times. The valve is closed when it is observed that one-half of the volume of the tank is occupied by liquid water. Find (a) the final pressure in the tank, (b) the amount of steam that has entered the tank, and (c) the amount of heat transfer.

Answer 1

A) 67.028 psia

B) 85.728 lbm

C) 80,910.64 Btu

Step-by-step explanation:

Volume of tank (V) = 3 ft³

Temperature in tank (T1) = 300°F

Temperature in steam Line (TL) = 400°F

Steam pressure in line(PL) = 200 psia

Volume of liquid water (Vf) = 0.5V = 0.5 x 3 = 1.5 ft³

A) From the energy balance equation ;

ΔE(sys) = E(in) - E(out)

Thus, M(i)h(L) - Q(out) = m2u2 - m1u1

Also, the mass balance is given by;

M(in) - M(out) = ΔMsys

Thus the final pressure in the tank (P2) will be equal to the saturated pressure at 300°F temperature.

Thus,looking at the first table i attached,

It is seen that the Pressure P2 at 300°F = 67.028 psia

B) Looking at the first table, at T = 300°F, we have the following ;

v1 = vg = 6.4663 ft³/lbm

u1 = ug = 1099.8 Btu/lbm

vf = 0.01745 ft³/lbm

uf = 269.51 Btu/lbm

From second steam table attached, at P(L) = 200 psia and T(L) = 400 °F

We have; h(L) = 1210.9 Btu/lbm

Initial mass in the tank is gotten by;

m1 = V/v1

Thus, m1 = 3/6.4663 = 0.464 lbm

Ow, let's calculate final mass(m2) ;

m2 = Vf/vf + Vg/vg

Plugging in the relevant values to get ;

m2 = 1.5/0.01745 + 1.5/6.4663 = 85.96 + 0.232 = 86.192 lbm

Hence, the amount of steam that has entered the tank will be;

Mi = M2 - M1 = 86.192 lbm - 0.464 lbm = 85.728 lbm

C) The internal energy in the final state will be given as;

U2 = mfuf + mgug = (85.96 x 269.51) + (0.232 x 1099.8) = 23422.23 Btu

Now, the heat transfer is given as;

Q' = mihL + m1u1 - U2 = (85.74 x 1210.9) + (0.464 x 1099.8) - 23422.23 = 80,910.64 Btu

Answer 2

Step-by-step explanation:

`Given\\V=3ft^3\\T_(1) =300^oF\\T_(L) =400^oF\\P_(L) =200psia\\V_(f) =0.5V`

We assume kinetic and potential energy changes are negligible and there is no work interactions.

a) Taking tank as a system, The energy balance can be define as

`E_(in)- E_(out)= E_(sys)`

`m_(i) h_(L) -Q_(out)=m_(2) u_(2)-m_(1) u_(1)`

The mass balance could be written as

`m_(in)- m_(out)= m_(sys) \\m_(i)= m_(2)- m_(1)`

The final pressure in the tank could be defined as following

`P_(2) =P_(sat300^oF)`

from standard steam table we know at

`T_(2)= T_(1)=300^oF\\ P_(2)=67.028psia`

b)

From steam table at

`T_(1) =300^oF\\v_(1)= v_(g)=6.4663ft^3/lbm\\v_(1)= v_(g)=1099.8Btu/lbm\\ v_(f) =0.01745ft^3/lbm\\v_(f) =269.51Btu/lbm\\`

`at\\P_(L) =200psia\\T_(L)=400^oF\\h_(L)=1210.9Btu/lbm`

initial mass in the tank could be define as

`m_(1)=(V)/(v_(1) ) \\m_(1) =(3)/(6.4663) =0.464lbm\\`

Final mass in the tank could be define as

`m_(2)=(V_(f) )/(v_(f) )+(V_(g) )/(v_(g) ) \\ m_(2) =(1.5)/(0.01745) +(1.5)/(6.4663) =86.2lbm`

The amount of steam that has entered the tank

`m_(i)=m_(2)- m_(1)\\ m_(i)=86.2-0.464=85.74lbm`

c)

The internal energy in final state could be defined as following

`U_(2)=m_(f) u_(f)+ m_(g) u_(g)\\ U_(2) =85.96*269.51+0.232*1099.8=23422Btu\\`

The heat transfer could be defined as following

`Q_(out)=m_(i) h_(L)+m_(1) u_(1)-m_(2) u_(2)\\ Q_(out)=85.74*1210.9+0.464*1099.8-23422=80910Btu`