Saygo, Joshua Gabriel C.
2CHE-B
THERMODYNAMICS ASSIGNMENT
2.18 Liquid water at 180 C and 1,002.7 kPa has an internal energy (on an arbitrary scale) of 762.0 kJ kg1 and a specific volume of 1.128 cm3 g-1
(a) What is its enthalpy?
(b) The water is brought to the vapor state at 300 C and 1,500 kPa, where its internal energy is 2,784.4
kJ kg-1 and its specific volume is 169.7 cm3 g-1. Calculate U and H for the process
(a) Enthalpy
𝐻 = 𝑈 + 𝑃𝑉
𝐻1 = 762 + (1002.7 𝑥 1.128 𝑥
10−6 𝑚3
𝑘𝐽
) = 763.131
−3
10 𝑘𝑔
𝑘𝑔
(b) ∆𝐻 𝑎𝑛𝑑 ∆𝑈
𝐻2 = 2784.4 + (1500 𝑥 169.7 𝑥
10−6 𝑚3
𝑘𝐽
) = 3038.99
−3
10 𝑘𝑔
𝑘𝑔
∆𝐻 = 𝐻2 − 𝐻1 = 3038.99 − 763.131 = 2275.86
∆𝑈 = ∆𝑈2 − ∆𝑈1 = 2784.4 − 762 = 2022.4
𝑘𝐽
𝑘𝑔
𝑘𝐽
𝑘𝑔
2.23 A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1 of
cool water at 25 C with 0.8 kg s-1 of hot water at 75 C. During mixing, the heat is lost to the
surroundings at the rate of 30 kJ s-1. What is the temperature of the heat is lost to the surroundings at
the rate of 30 kJ s-1. What is the temperature of the warm-stream water? Assume the specific heat of
water constant at 4.18 kJ kg-1 K-1.
𝑚2 𝐶𝑝 ∆𝑇 = 𝑚1 𝐶𝑝 ∆𝑇 + 𝑄
(0.8)(4.18)(75 − 𝑇) = (1)(4.18)(𝑇 − 25) + 30
𝑇 = 43.23°𝐶
2.25 Water at 28 C flows in a straight horizontal pipe in which there is no exchange of either heat or
work with the surroundings. Its velocity is 14 m s-1 in a pipe with an internal diameter of 2.5 cm until
it flows into a section where the pipe diameter abruptly increases. What is the temperature change of
the water if the downstream diameter is 3.8 cm? If it is 7.5 cm? What is the maximum temperature
change for an enlargement in the pipe?
𝑢1 𝑑12 = 𝑢2 𝑑22
Saygo, Joshua Gabriel C.
2CHE-B
𝐹𝑜𝑟 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 3.8 𝑐𝑚: (14)(2.5 𝑥 10−2 )2 = 𝑢2 (3.8 𝑥 10−2 )2
𝑢2 = 6.06
∆𝐻 +
4.18∆𝑇 +
𝑚
𝑠
∆𝑢2
𝑢22 − 𝑢12
= 0; 𝐶𝑝 ∆𝑇 +
=0
2
2
(6.06)2 − (14)2
= 0; ∆𝑇 = 0.027°𝐶
2
𝐹𝑜𝑟 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 7.5 𝑐𝑚: (14)(2.5 𝑥 10−2 )2 = 𝑢2 (7.5 𝑥 10−2 )2
𝑢2 = 1.56
𝑚
𝑠
(1.56)2 − (14)2
4.18∆𝑇 +
= 0; ∆𝑇 = 0.023°𝐶
2
2.26 Fifty (50) kmol per hour of air is compressed from P1 = 1.2 bar to P2 = 6.0 bar in a steady-flow
compressor. Delivered mechanical power is 98.8 kW. Temperatures and velocities are:
T1 = 300 K
T2 = 520 K
U1 = 10 m s-1
U2 = 3.5 m s-1
Estimate the rate of heat transfer from the compressor. Assume for air that Cp = 7/2 R and that
enthalpy is independent of pressure.
𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑎𝑖𝑟 = 50 𝑥 29 = 1450
𝑚 (∆𝐻 +
𝑘𝑔
1 ℎ𝑟
𝑘𝑔
𝑥
= 0.403
ℎ𝑟 3600 𝑠
𝑠
∆𝑢2
+ 𝑔∆𝑧) = 𝑄 + 𝑊
2
𝑛𝑜 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛; ∴ ∆𝑧 = 0
𝑚 ([𝐶𝑝 (𝑇2 − 𝑇1 )] +
𝑢22 − 𝑢12
)=𝑄+𝑊
2
(3.5)2 − (10)2
7 8.314
0.403 ([ 𝑥
𝑥(520 − 300)] +
) = 𝑄 + 98.8
2
29
2
𝑄 = −27.52
𝑘𝐽
𝑠
Saygo, Joshua Gabriel C.
2CHE-B
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 velocity is 20(ft)(s)-1. 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.
∆𝐻 +
∆𝑢2
𝑔∆𝑧 = 𝑄 + 𝑊
2
∆𝐻 +
∆𝑢2
=0
2
(20)(100)(𝑇2 )
𝑢𝑃
𝑢1 𝑃1 𝑢2 𝑃2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡;
=
; 𝑢2 =
= 0.311𝑇2
(20)(322)
𝑇
𝑇!
𝑇2
𝐶𝑝 ∆𝑇 +
𝑢22 − 𝑢12
=0
2
(0.311𝑇2 )2 − (20)2 1
7
(6.1324)(𝑇2 − 322) +
(
) (28)(0.0022) = 0
2
2
32.2
𝑇2 = 321.57 𝐾
2.28 Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As
it passes through the coil the water changes state from liquid at 200 kPa and 80 C to vapor at 100 kPa
and 125 C. Its entering velocity is 3 m s-1 and its exit velocity is 200 m s-1. Determine the heat
transferred through the coil per unit mass of water. Enthalpies of the inlet and outlet streams are:
Inlet: 334.9 kJ kg-1; Outlet: 2726.5 kJ kg-1.
∆𝐻 +
∆𝑢2
=𝑄
2
(𝐻2 − 𝐻1 ) +
(2726.5 − 334.9) +
𝑢22 − 𝑢12
=𝑄
2
(200)2 − (3)2
𝑘𝐽
𝑥10−3 = 𝑄; 𝑄 = 2411.6
2
𝑘𝑔
2.29 Steam flows at steady state through a converging insulated nozzle, 25 cm long and with an inlet
diameter of 5 cm. At the nozzle entrance (state 1), the temperature and pressure are 325 C and 700
kPa, and the velocity is 30 m s-1. At the nozzle exit (state 2), the stream temperature and pressure are
240 C and 350 kPa. Property values are: H1 = 3,112.5 kJ kg-1; V1=388.61 cm3 g-1; H2 = 2,945.7 kJ kg-1;
V2=667.75 cm3 g-1. What is the velocity of the stream at the nozzle exit, and what is the exit
diameter?
∆𝐻 +
∆𝑢2
+ 𝑔∆𝑧 = 𝑄 + 𝑊
2
Saygo, Joshua Gabriel C.
2CHE-B
∆𝐻 +
∆𝑢2
=0
2
(𝐻2 − 𝐻1 ) +
(2945.7 − 3112.5) +
𝑢22 − 𝑢12
=0
2
𝑢22 − 900
𝑥10−3 = 0; 𝑢2 = 578.4 𝑚/𝑠
2
𝑢1 𝑑𝑖2 𝑢2 𝑑𝑜2
=
𝑉1
𝑉2
(30)(102 )(5)2 (578.539)(102 )(𝑑𝑜2 ) 2
=
; 𝑑𝑜 = 1.49 𝑐𝑚
388.61
667.75
2.34 Carbon dioxide gas enters a water-cooled compressor at conditions P1 = 15(psia) and T1 = 50(F),
and is discharged at conditions P2 = 520(psia) and T2 = 200(F). The entering CO2 flows through a 4inch-diameter pipe with a velocity of 20(ft)(s)-1, and is discharged through a 1-inch-diameter pipe. The
shaft work supplied to the compressor is 5,360(Btu)(mol)-1. What is the heat-transfer rate from the
compressor in (Btu)(hr)-1?
H1=307(Btu)(lbm)-1
H2=330(Btu)(lbm)-1
V1=9.25(ft)3(lbm)-1
V2=0.28(ft)3(lbm)-1
𝑢1 𝑑12 𝑢2 𝑑22
=
𝑉1
𝑉2
(20)(4)2 (𝑢2 )(1)2
𝑓𝑡
=
; 𝑢2 = 9.69
9.25
0.28
𝑠
(𝐻2 − 𝐻1 ) +
(𝑢22 − 𝑢12 )
=𝑄+𝑊
2
(9.69)2 − (20)2
1
1 ′
𝐵𝑡𝑢
(330 − 307) +
(
) = 𝑄 + (5360 𝑥
) ; 𝑄 = −104.91
2
25.037
44.01
𝑙𝑏𝑚
𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
𝑢𝐴 (20)(0.0855)
𝑙𝑏𝑚
=
= 0.185
𝑉
9.25
𝑠
𝑅𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 = 𝑄 𝑥 𝑚 = −104.91 𝑥 0.185 = −19.41
𝐵𝑡𝑢 3600 𝑠
𝐵𝑡𝑢
𝑥
= −69870.06
𝑠
1 ℎ𝑟
ℎ𝑟
Saygo, Joshua Gabriel C.
2CHE-B