7.13 The contents of a tank containing 1 kg steam at 600 °C and 150 bar are allowed to

flow into an evacuated tank of equal capacity until the pressure in both tanks is the

same (Figure P7.13). The proses takes place isothermally.

(a)Calculate the final temperature and pressure in the combined tank system.

(b)Calculate the heat which must be supplied to allow the process to be isothermal.

Conclusion :

(Figure P7.13)

V1=V2

m = 1 kg steam awal = kosong

T1 = 600 °C akhir = P1’ = ?

P1 = 150 bar T1’ = ?

(a) Karena proses isothermal, maka T tetap = 600 °C

Dimana : V1 = 0,0249 m3/kg (dari tabel)

V2 = 0

Vakhir =V 1+V 2

2

= 0,0249+0

2

= 0,01245 m3/kg ( antara P = 250 bar dan P = 300 bar )

Interpolasi : P’ =

250+300−2500 ,0114−0 ,0141

(0 ,01245−0 ,0141)

Tanki 2Tanki 1

= 280,56 bar

(b) Proses isothermal → ∆U = 0

Q = ∆U + W

Q = W = P ∆V

Q = 280,56 bar x 0,01245 m3 x 100 kPa/bar

Q = 349,3 kPa.m3

Q = 349,3 KJ

Q – W = nRT ln P1/P2

= 1 kg

7. 20 A motor which supplies 750 kW power is used to compress stream at 500 psia and

900 °F adiabatically to 1000 psia and a temperature of 1140 °F. Calculate the flow

of steam which can be compressed assuming 100% efficiency.

P1 = 500 psia → 34,46 bar T1 = 900 °F → 482,22 °C

Pada 20 bar dan 482,22 °C

H =

3358+3467−3358500−450

( 482 ,22−450)

= 3428,24 Kj/Kg

Pada 40 bar dan 482,22 °C

H =

3331+3445−3331500−450

(482 ,22−450 )

= 3464,46 Kj/Kg

H1 =

3428 ,24+3464 ,46−3428 ,2440−20

(34 ,46−20 )

= 3411,05 Kj/Kg

P2 = 1000 psia → 68,93 bar T2 = 1140 °F → 615, 56 °C

Pada 60 bar dan 615,56 °C

H = 3657+3774−3657

650−600(615 ,56−600 )

= 3643,41 Kj/Kg

Pada 80 bar dan 615,56 °C

H = 3640+3754−3640

650−600(615 ,56−600)

= 3677,03 Kj/Kg

H2 = 3693 ,41+3677 ,03−3693 ,41

80−60(68 ,93−60 )

= 3686,1 Kj/Kg

dQdT - dwdT = F2 (H2 + gz2 + 1/2V2

2) – F1 (H1 + gz1 + 1/2 V12) dimana F1=F2

0 - dwdT = F2 (H2 + 0 + 0) – F1 (H1 + 0 + 0)

Kj/s = F(3686,1 – 3411,05) Kj/Kg

F = 2,7268 Kg/s

F = 981,4 Kg/h