2.7.
◦
A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 24.1 C for four-phase equilibrium of allotropic solid forms of the exotic chemical β -miasmone. -miasmone. Evaluate the claim. Jawab
Jadi dalam penentuan Quadraple point, hanya membutuhkan salah satu syarat. P (tekanan) atau T (suhu) saja. Misalnya menggunakan T (suhu) 24,1
o
C, pada suhu
tersebut memiliki P yang mungkin berbeda dari 10,2 Mbar. Jadi pada suhu tertentu memiliki tekanan tertentu yang spesifik. Jadi tidak bisa menggunakan keduanya. 2.12.
Heat on the amount of 7,5 kJ is added to a closed system while its internal energy decreases by 12 kJ. How much energy is transferred as work? For a process causing the same change of the state but for which the work is zero, how muchheat is transferred? Jawab
Kondisi pertama: ∆U = Q + W
∆U = -12 kJ Q = 7,5 kJ maka W = ∆U – W = -12 kJ – 7,5 kJ = -19,5 kJ
Kondisi kedua: ∆U = Q + W
∆U = -12 kJ W = 0 maka Q = ∆U = -12 kJ
2.17.
A hydroturbine operates with a head of 50 m of water. Inlet and outlet conduits are 2 m in diameter. Estimate the mechanical power developed by the turbine for an outlet velocity -1
of 5 ms . Diketahui -1
h2 = 50 m
v = 5 ms
h1 = 0 m
d=2m
Jawab
()
() 2.22.
An incompressible (ρ = constant) liquid flows steadily through a conduit of circular cross-section and increasing diameter. At location 1, the diameter is 2,5 cm and the velocity is 2 m/s. At location 2, the diameter is 5 cm (a) What is the velocity of location 2? (b) What is the kinetic energy-change (J/kg) of the fluid between locations 1 and 2? Jawab
D1 = 2,5 cm
D2 = 5 cm
v1 = 2 m/s
(a) Untuk fluida yang bersifat incompressible (ρ = konstan) berlaku persamaan v1A1 = v2A2 v2 =
v1
karena A =
2 D , maka persamaannya dapat diubah menjadi:
2 ) v1 2 ( ) x 2 m/s
v2 = ( =
2
= ( 0,5 ) x 2 m/s = 0,25 x 2 m/s = 0,5 m/s 2.27. Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter of 1.5(in). A pressure drop results from flow through a partially opened valve. Just upstream ◦ from the valve the pressure is 100(psia), the temperature is 120( F), and the average -1 velocity is 20(ft)(s) . If the pressure just downstream from the valve is 20(psia). What is the temperature? Assume for nitrogen that PV/T is constant, Cv = (5/2)R and Cp = (7/2)R. (Values for R are given in App. A.) Diketahui :
d = 1,5 inch
P1 = 100 psia = 698 . 103 Pa o
P2 = 20 psia Cv =
R
T1 = 120 F = 321,89 K
Cp =
V1 = 20 ft/s = 6,098 m/s
R = 8,314 (m) (Pa)(mol) (K)
Mr = 28 gm/mol Jawab
=
C
R 3
-1
-1
= =
Q=A.v= Persamaan kontinuitas : Q 1 = Q2 Karena Nitrogen mengalir dalam kran yang sama, maka A 1 = A2 dan t1 = t2. Sehingga persamaan konyinuitasnya menjadi : = v2 =
v1
v2 =
v1
∆v2 = v22 - v12 2
– v12
=(
v1 )
=(
- 1 ) v1
2
dalam soal ini, energi kinetik yang dilakukan oleh Nitrogen dihasilkan oleh energi panas. Ek = - ∆H
½ . m. ∆v2 = 2
½.( ½(
2
. Cp . ∆T
- 1 ) v1 . Mr = - Cp . ( T2 – T1 ) 2
2
) - 1 ) (6,096) . 28 = -
8,314 . (T2 – 321,89)
T2 = 188 K 2.32.
Find the equation for the work of a reversible, isothermal compression of 1 mol of gas in piston/cylinder assembly if the molar volume of the gas is given by
Where b and R are positive constants. V= v/n = RT + b P RT = v – b = v – bn P n n P=nRT V – bn ∂W = - P dV n= 1 mol
∫∂W = - ∫ n R T dV V – bn V2 Wkompresi = - RT ∫ V1
2.37.
1 dV V – bn
Wkompresi = RT ln V1 – b V2 – b ◦ ◦ The condition of a gas change in a steady-flow process from 20 C and 1,000 kPa to 60 C and 100 kPa. Devise a reversible nonflow process (any number of steps) for accomplishing this change of state, and calculate ∆U and ∆ H for the process on the basis 1 mol of gas. Assume for the gas that PV/T is constant, Cv = (5/2)R, and Cp = (7/2)R Diketahui
⁰ = 60 ⁰C = 333 K
: T1 = 20 C = 293 K P1 = 1000 kPa T2
P2 = 100 kPa
Diasumsikan bahwa
= konstan, dan basis mol gas adalah 1 mol.
Cp = 7/2 R
Cv = 5/2 R
Ditanyakan
: Berapa ∆U dan ∆H dari proses tersebut?
Jawab
:
Nilai R yaitu 8,314
= konstan Ta2 = T1.
Ta2 = 293 K .
Ta2 = 29,3 K
∆ Ta = Ta2 – T1
∆ Tb = T2 – Ta2
= 29,3 K – 293 K = - 263,7 K
= 333 K – 29,3 K =303,7 K
∆ Hb = Cp . ∆ Tb
. 303,7 K = 8837, 3663 = . 8,314
∆Ua
= Cv . ∆ Ta
. (-263,7 K) = - 5481, 0045 = . 8,314
PV = nRT V1 = n .
V2 = n .
= 1 mol . 8,314
. 293 K
= 1 mol . 8,314
1000 kPa -3
3
∆ Ha
= ∆Ua + V1(P2-P1)
= - 7673,4045
-3 3 + 2,436 10 m . (100 - 1000) kPa
= ∆Hb – P2 (V2 - V1) = 8837, 3663 = 6312,9663
3
= 27,68 10 m
= - 5481, 0045
∆Ub
100 kPa -3
= 2,436 10 m
. 333 K
-3 3 + 100 kPa (27,68 – 2,436) 10 m
∆U = ∆Ua + ∆Ub = (- 5481,0045 + 6312,9663) = 831,9618
∆H = ∆Ha + ∆Hb
= (-7673,4045 + 8837, 3663) = 1163, 9618
