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Chapter 7

1. The vapor pressure of water at 373.15 K is 101.325 kPa. Given the enthalpy of vaporization is 40.7 kJ/mol, determine the vapor pressure at 360 K and 400 K.

𝑃𝑃 = 𝐶𝐶𝑒𝑒  

101325𝑃𝑃𝑃𝑃 = 𝐶𝐶𝑒𝑒/ ∙

. / ∙  × .  𝐶𝐶 = 5.0493×10  

𝑃𝑃 = 5.0493×10 𝑒𝑒/ ∙

. / ∙  × = 62  748  𝑃𝑃𝑃𝑃  

𝑃𝑃 = 5.0493×10 𝑒𝑒/

. / ∙  × = 244  436  𝑃𝑃𝑃𝑃   2. Water and ice are at equilibrium at 0°C and 101.325 Pa. Calculate the saturation pressure at -10°C using the data below. State all assumptions. Δhsl = 6016.6 J/mol vl = 18.0 cm3/mol vs = 19.7 cm3/mol

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=∆𝐻𝐻

𝑇𝑇(𝑣𝑣 − 𝑣𝑣 )

𝑑𝑑𝑑𝑑∆𝐻𝐻

=𝑑𝑑𝑑𝑑

𝑇𝑇(𝑣𝑣 − 𝑣𝑣 )

1∆𝐻𝐻

𝑃𝑃 =1

(𝑣𝑣 − 𝑣𝑣 )ln  (𝑇𝑇)

𝑃𝑃 − 𝑃𝑃∆𝐻𝐻

=ln 𝑇𝑇

𝑇𝑇(𝑣𝑣 − 𝑣𝑣 )

𝑃𝑃 =ln 𝑇𝑇

𝑇𝑇 ∆𝐻𝐻

(𝑣𝑣 − 𝑣𝑣 )  +  𝑃𝑃

𝑃𝑃 =ln 263𝐾𝐾

273𝐾𝐾 6016.6𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚(18𝑐𝑐𝑐𝑐 /𝑚𝑚𝑚𝑚𝑚𝑚 − 19.7𝑐𝑐𝑐𝑐 /𝑚𝑚𝑚𝑚𝑚𝑚)(1𝑚𝑚/100𝑐𝑐𝑐𝑐)

 +  101325𝑃𝑃𝑃𝑃

𝑃𝑃 = 132  MPa

3. Find the specific entropy a saturated water-steam mixture at 40°C and 50% quality.

s = xsg + (1− x)s f = 0.5 × 8.2570 + 0.5 × 0.5725 = 4.415 kJ/kgK

4. If water is at 130°C and 300kPa, what phase is it in and what is its specific enthalpy?

At 0.3 MPa, the saturation temperature is 133.52°C. Since T>Tsat, the water is superheated steam. From the superheated steam tables h=3069.3 kJ/kg.

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5. Steam at a pressure of 200kPa has a specific volume of 0.05 m3/kg. What are the internal energy and entropy for 1kg of steam under these conditions?

At 200kPa, from the saturated steam table, vf = 0.001061 m3 /kg , vg = 0.8857 m3 /kg Since vf<v<vg we conclude that the steam is a saturated mixture of liquid and vapour.

v = xvg + (1− x)v f = v f + x(vg − v f )

⇒ x =v − vfvg − vf

= 0.05 − 0.0010610.8857 − 0.001061

= 0.05532

u = xug + (1− x)u f = 0.05532 × 504.49 kJ/kg + (1− 0.05532)× 2529.5 kJ/kg = 616.50 kJ/kg

h = hf + xh fg = 504.71+ 0.05532 ⋅ 2201.6 = 626.50kJ /kg

6. Saturated steam at 100°C is heated in a rigid tank until its temperature reaches 500°C. What is the final pressure?

Initial specific volume v1=vg=1.6729 m3/kg Constant volume process v2= v1=1.6729 m3/kg Final Temperature T2=500°C From superheated steam tables at 500°C,

at P=0.2 MPa, v=1.7814 m3/kg at P=0.3 MPa, v=1.1867 m3/kg

T2 − 200 kPa300 kPa-200 kPa 

= 1.6729 m3 /kg −1.7814 m3 /kg1.1867 m3 /kg −1.7814 m3 /kg

Interpolating, P2=218.2 MPa

7. A closed container in the shape of a vertical cylinder is 0.1 m high. At the bottom of the container is a 1 cm deep layer of saturated water and the rest is filled with saturated steam. If this system is in equilibrium at 300 kPa, what is its quality?

At 300 kPa vf=0.001073 m3/kg, vg=0.6058 m3/kg

mf =Vf

vf= 0.01 m ×1 m2

0.001073 m3 /kg= 9.3197 kg

mg =Vgvg

= 0.99 m ×1 m2

0.6058 m3 /kg= 1.6342 kg

x =mg

mf +mg

= 1.6342 kg9.3197 kg+1.6342 kg

= 0.149

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8. A rigid tank containing 1 kg of water at 250°C, 1.4 MPa is cooled until the pressure in it is 100 kPa. Find the amount of heat removed.

At P1=1.4 MPa, T1=250°C, v1=1.6350 m3/kg u1=2698.3 kJ/kg

Constant volume process, so v2= v1=1.6350 m3/kg.

At P2=100 MPa vf=0.001043 m3/kg vg=1.694 m3/kg

x2 =v2 − vfvg − vf

= 1.6350 kJ/kg − 0.001043 kJ/kg1.694 kJ/kg − 0.001043 kJ/kg

= 0.9651

At P2=100 MPa uf=417.36 kJ/kg ug=2506.136 kJ/kg

u2 = x2ug + (1− x2 )u f = 0.9651× 2506.1 kJ/kg + (1− 0.9651)× 417.36 kJ/kg = 2519.7 kJ/kg

Q = m(u2 − u1) = 1 kg(2519.7 kJ/kg − 2698.3 kJ/kg)=178.6 kJ

9. An insulated cylinder with a frictionless piston contains 2 L of saturated liquid water at 100 kPa. The water is heated until its quality is 80%. Find the amount of heat added.

At 100 kPa, vf=0.001043 m3/kg

m = Vv= 2 ×10−3 m3

0.001043 m3 /kg= 1.918 kg

At 100 kPa, h1=hf=417.46 kJ/kg hg=2675.5 kJ/kg

h2 = xhg + 1− x( )hf = 0.8 × 2675.5 kJ/kg + 0.2 × 417.46 kJ/kg=2223.5 kJ/kg

Q = m(h2 − h1) = 1.918 kg(2223.5 kJ/kg − 417.46 kJ/kg) = 3464 kJ

10. A rigid tank with a volume of 0.6 m3 contains 10 kg of water at a pressure of 2 MPa. Heat is added to the tank until the pressure is equal to 3.5 MPa. Determine the final temperature of the water and the total heat transfer.

Solution:

v1 = 0.6m3

10kg= 0.06m3 /kg

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v f < v1 < vg ⇒ water is in saturation at the initial state.

x1 =v1 − vgv f − vg

= 0.06 − 0.0011770.099587 − 0.001177

= 0.5977

u1 = 0.5977 ⋅ 2599.1+ 0.4023 ⋅ 906.12 =1918.0kJ /kg

v2 = v1 = 0.06m3 / kg

P2 = 3.5Mpa

⎫⎬⎪

⎭⎪⇒

T2 = 256.4C

u2 = 2638.7kJ / kg

⎧⎨⎪

⎩⎪

ΔV = 0⇒ Q = ΔU = m(u2 − u1) =10kg(2638.7 −1918.0)kJ /kg = 7207kJ 11. One kilogram of a saturated water-steam mixture at 200 °C expands against a piston in a cylinder. It stops when it reaches a pressure of 100 kPa, when the steam quality is 60%. Assuming a reversible and adiabatic process, determine the initial pressure of the water and the work done during the process. Solution:

P2 =100kPax2 = 0.6

⎫ ⎬ ⎭ ⇒

u2 = 0.6 ⋅ 2506.6 + 0.4 ⋅ 417.40 =1670.9kJ /kgs2 = 0.6 ⋅ 7.3589 + 0.4 ⋅1.3028 = 4.9365kJ /(kgK)

⎧ ⎨ ⎩

Isentropic process

s1 = s2 = 4.9365kJ /(kgK)

T1 = 200Csaturated

⎫ ⎬ ⎭ ⇒

sf = 2.3305kJ /(kgK)sg = 6.4302kJ /(kgK)Ps =1554.9kPa

⎧

⎨ ⎪

⎩ ⎪

Initially, the water steam is saturated as well

⇒ P1 = PS =1554.9kPa

x1 =s1 − sfsg − sf

= 4.9365 − 2.33056.4302 − 2.3305

= 0.6357

u1 = 0.6357 ⋅ 2594.2 + 0.3643 ⋅ 850.46 =1959.0kJ /(kgK) 1st Law in control mass

w = Δu = u2 − u1 =1670.9 −1959.0 = −288.1kJ /kg 12. A closed, partially insulated cylinder is divided into two parts by a movable, insulated piston. One part is filled with air and completely insulated, while the other part is filled with a saturated water-steam mixture and not insulated. Initially, the system is at equilibrium and the volume of each section is 1 m3, with the air at 40 °C and the steam at 80 °C and 30% quality. Heat is added to the steam until the system pressure reaches 400kPa. Determine the amount of heat transfer. Solution: Initially, the water is at 80 °C and with quality of 30 % ⇒ Pw,1 = 47.416kPa ,

vw,1 = 0.3 ⋅3.4052 + 0.7 ⋅0.001029 = 1.022m3 / kg ,

uw,1 = 0.3 ⋅2481.6 + 0.7 ⋅334.97 = 978.9kJ / kg

⇒ mw =Vw,1 / vw,1 =

11.022

= 0.9785kg

For the air, initially,

⇒ Pa,1 = Pw,1 = 47.416kPa

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⇒ ma =

Pa,1Va,1

RTa,1

=47.416kPa ⋅1m3

0.2870kJ / (kgK ) ⋅ (40 + 273)K= 0.5278kg

For the air, the process is isentropic, therefore,

Pa,1Va,1

γ = Pa,2Va,2γ ,

γ =

cp

cv

= 1.400

⇒Va,2 =Pa,1Pa,2

γ ⋅Va,1 =47.416400

1.4 ⋅1= 0.2180m3 ,

Ta,2 = Pa,2Va,2

maR= 400kPa ⋅ 0.2180m3

0.5278kg ⋅ 0.287kJ /(kgK)= 575.7K

⇒Vw,2 = (2 − 0.2180)m3 =1.7120m3

⇒ vw,2 =Vw,2 /mw = 1.71200.9785

=1.750m3 /kg

vw,2 =1.750m3 /kgPw,2 = 400kPa

⎫ ⎬ ⎭ ⇒ uw,2 ≈ 4563.7kJ /kg

ΔUa = maCv (Ta,2 −Ta,1) = 0.5278kg ⋅ 0.7165kJ /(kgK) ⋅ (575.7K − 313K) = 99.34kJ

ΔUw = mw (uw,2 − uw,1) = 0.9785kg(4563.7 − 978.9)kJ /kg = 3508kJ Take Air and water as one system,

Q = ΔU = ΔUa + ΔUw = 99.34 + 3508 = 3607kJ

13. A rigid tank with 0.1 m3 volume contains superheated steam at 4 MPa, 350°C. A safety valve in the tank opens if the pressure exceeds 4 MPa, venting steam to the atmosphere. The tank is heated until the steam temperature reaches 600°C. Find the mass of steam that escapes and the heat added.

At P1=4 MPA, T1=350°C, v1=0.06645 m3/kg u1=2826.7 kJ/kg h1=3092.5 kJ/kgK

At P2=4 MPA, T2=600°C, v2=0.09885 m3/kg u2=3279.1 kJ/kg h2=3674.4 kJ/kgK

Assume that the enthalpy of the escaping steam is the average value

he =h1 + h2

2= 3092.5 kJ/kg+3674.4 kJ/kg

2= 3383.5 kJ/kg

m1 =Vv1

= 0.1 m3

0.06645 m3 /kg= 1.505 kg

m2 =Vv2

= 0.1 m3

0.09885 m3 /kg= 1.012 kg

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Δm = m1 −m2 = 1.505 kg −1.012 kg = 0.493 kg

Energy Balance

Q −mehe = m2u2 −m1u1

Q = mehe +m2u2 −m1u1 = 0.493 kg × 3383.5 kJ/kg+1.012 kg × 3279.1 kJ/kg −1.505 kg × 2826.7 kJ/kg Q=732.3 kJ

14. Steam at 850 kPa and 300°C enters a nozzle with a mass flow rate of 0.8 kg/s and negligible velocity and is discharged at 550 kPa. Assuming isentropic expansion, determine the exit velocity.

Solution: 1) P1 = 850 kPa T1 = 300°C H1 = 3055.6 kJ/kg S1 = 7.2070 kJ/kgK P2 = 550 kPa S2 = 7.2070 kJ/kgK H2 = 2961.69 kJ/kg

𝑄𝑄  +  𝑤𝑤 =  𝑚𝑚 ∆ℎ +  𝑣𝑣 −  𝑣𝑣

2+  𝑔𝑔∆𝑧𝑧

0 =  0.8𝑘𝑘𝑘𝑘𝑠𝑠

2961.69𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘

+  𝑣𝑣

2(1000𝐽𝐽/𝑘𝑘𝑘𝑘)−  3055.6

𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘

v2 = 387.6 m/s

15. Superheated steam at 200°C and 1000 kPa expands through a nozzle to 200 kPa. Assuming the process is reversible and adiabatic, determine the exit enthalpy of the steam.

Solution: T1 = 200°C P1 = 1000 kPa P2 = 200 kPa Reversible & adiabatic isentropic H1 = 2828.3 kJ/kg S1 = 6.6956 kJ/ kg·K S1 = S2 = 6.6956 kJ/kg·K Since value of S2 is less than entropy of saturated vapour (sg) at 200 kPa, the final state is in the saturated water region.

P2 = 200kPasaturation

⎫ ⎬ ⎭ ⇒

sf =1.5302kJ /(kgK)sg = 7.1270kJ /(kgK)

⎧ ⎨ ⎩

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x =s− sfsg − sf

= 6.6956 −1.53027.1270 −1.5302

= 0.9229  

h2 = 0.9229 × 2706.3+ 0.0771× 504.71= 2536.6kJ /kg  

16. Steam enters a turbine at 1 MPa, 400°C with a velocity of 50 m/s and leaves at 100 kPa with a velocity of 200 m/s. Determine the work output of the turbine per kg of steam passing through it, assuming the process is reversible and adiabatic.

Solution:

Pi = 1Mpa

Ti = 400C

⎫⎬⎪

⎭⎪⇒

si = 7.4670kJ / (kgK )hi = 3264.5kJ / kg

⎧⎨⎪

⎩⎪

The process is reversible and adiabatic isentropic process

se = si = 7.4670kJ /(kgK)

se = 7.4670kJ /(kgK)Pe =100kPa

⎫ ⎬ ⎭ ⇒ he ≈ 2717.9kJ /kg

q + w = (he − hi) + Ve2 −Vi

2

2+ g(Ze − Zi)

w = (he − hi )+Ve

2 −Vi2

2= (2717.9 − 3264.5) kJ/kg + 2002 − 502

2 ⋅1000 kJ/kg = −527.9 kJ/kg

17. Steam enters a turbine at 1MPa and 500 °C. It expands in a reversible, adiabatic process and exits at a pressure of 50 kPa. The power output of the turbine is 800 kW. Determine the mass flow rate of steam, neglecting changes in kinetic and potential energy. Solution:

The process is reversible and adiabatic isentropic process

se = si = 7.7642kJ /(kgK)

18. An adiabatic steam turbine produces 3.25 MW of power when steam enters it at 2,400 kPa and 500°C and leaves as saturated vapor at 50 kPa. Find the mass flow rate of steam through the turbine and the turbine efficiency.

Solution: Given:    P1  =  2400  kPa     T1  =  500°C  From  Tables:    H1  =  3464.9  kJ/kg   S1  =  7.3471  

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Give:    P2  =  50  kPa,  saturated  vapour  From  Tables:  Hv  =  2545.2  kJ/kg  Hl  =  340.54  kJ/kg    

𝑄𝑄  +  𝑤𝑤 =  𝑚𝑚 ∆ℎ +  𝑣𝑣 −  𝑣𝑣

2+  𝑔𝑔∆𝑧𝑧  

−3250  𝑘𝑘𝑘𝑘 =  𝑚𝑚 2545.3 − 3463.9  𝑚𝑚 = 3.54𝑘𝑘𝑘𝑘/𝑠𝑠   If  process  was  isentropic,  what  would  the  equality  of  the  final  state  be?  S1  =  S2  Sl  =  1.0912  kJ/  kg·∙K   Sv  =  7.4931  kJ/kg·∙K  7.3471  =  x(7.4931)  +  (1  –  x)1.0912  x  =  0.977    Finding  the  final  state  enthalpy  if  process  was  isentropic  H2’  =  xHv  +  (1  –  x)Hl  

H2’  =  0.977(2545.2)  +  (1  –  0.977)340.54  H2’  =  2494.49  kJ/kg    Calculating  efficiency  of  turbine  

𝜂𝜂 =  𝐻𝐻 − 𝐻𝐻  𝐻𝐻 −  𝐻𝐻  

=  2545.1 − 3463.92494.49 − 3463.9

 

𝜂𝜂 = 0.948