(SP 1)

Given:

In a closed rigid tank,

State 1: 𝑚1,𝑖𝑐𝑒 = 1 𝑘𝑔, 𝑚1,𝑔 = 0.05 𝑘𝑔

P1= 0.0381 kPa, T1= -30oC

State 2: the liquid – vapor equilibrium line, either saturated liquid or saturated vapor

Find:

(a) The volume of the tank

(b) P1 in bar

(c) a saturated liquid or a saturated vapor

(d) T2

(e) P-T and P-V diagrams

System sketch:

State 1 State 2

DLLL

Assumptions:

- Water vapor and ice are in equilibrium at state 1

- Pure saturated liquid or pure saturated vapor at state 2

Solution:

(a)

Vtank = 𝑚1𝑣1 = 𝑚1,𝑖𝑐𝑒𝑣1,𝑖𝑐𝑒 + 𝑚1,𝑔𝑣1,𝑔

1 kg ice

0.05 kg water vapor

P1= 0.0381 kPa

T1= -30oC

1.05 kg of

Saturated liquid

or

Saturated vapor Q

From Table A-6,

𝑣1,𝑖𝑐𝑒 = 1.0858 × 10−3 𝑚3

𝑘𝑔

𝑣1,𝑔 = 2943 𝑚3

𝑘𝑔

Then, the equation we wrote can be,

Vtank = (1 𝑘𝑔)(1.0858 × 10−3 𝑚3

𝑘𝑔) + (0.05 𝑘𝑔) (2943

𝑚3

𝑘𝑔) = 147.15 𝑚3

The volume of the tank is 𝟏𝟒𝟕. 𝟏𝟓 𝒎𝟑

(b)

P1 = (0.0381 kPa)(1 𝑏𝑎𝑟

105𝑃𝑎) (

1000 𝑃𝑎

1 𝑘𝑃𝑎) = 0.000381 𝑏𝑎𝑟

(c)

𝑣2 = 𝑣1 =𝑉𝑡𝑎𝑛𝑘

𝑚1=

𝑉𝑡𝑎𝑛𝑘

𝑚2=

147.15 𝑚3

1.05 𝑘𝑔= 140.14

𝑚3

𝑘𝑔

which is greater than 𝑣𝑐𝑟𝑖𝑡 , 0.0031𝑚3

𝑘𝑔. Then, the state is a saturated vapor.

(d)

From Table A-2,

𝑣𝑔 = 147.120 𝑚3

𝑘𝑔 at 5 oC

𝑣𝑔 = 137.734 𝑚3

𝑘𝑔 at 6 oC

Interpolate to know temperature at 𝑣𝑔 is 140.14 𝑚3

𝑘𝑔.

This provides T2 = 5.744 oC.

(e)

(SP 2)

Given:

Two-phase R-134a

State 1: T1= 50oF

State 2: superheated vapor at P2= 4 psia

Find:

(a) min and max of T2

(b) x1,min

(c) P-h diagram of x1=0.9

(d) plot x1 vs T2

System sketch:

Assumptions:

1. Steady-state, Steady-flow

2. ∆𝐾𝐸 = ∆𝑃𝐸 = 0

3. Adiabatic

4. No work

Basic equations:

Solution:

Apply our assumptions into the energy balance equation,

The equation above is now,

��1ℎ1 = ��2ℎ2

Because of ��1 = ��2,

ℎ1 = ℎ2

(a)

Using the properties we know, establish an equation with the above equation. 𝑥1 𝑎𝑛𝑑 𝑇2 are

independent value and dependent value.

ℎ1(𝑇1, 𝑥1) = ℎ2(𝑇2, 𝑃2)

Use EES to solve (EES code is attached).

T2,max= 29.5 oF for x1=1 and T2,min= -60 oF for x2=1

Maximum T2 when x1 occurs at 1.0 (sat. vapor at the throttle inlet). Minimum T2 when x2

occurs at 1.0 (sat. vapor at the throttle outlet).

∵ 1 ∵ 3 ∵ 4 ∵ 2 ∵ 2

(b)

Minimum x1 occurs for T2= T2,min. See the attached EES code.

x1,min= 0.804

(c) and (d)

EES code is attached.

"Specified inlet temperature and outlet pressure" T[1] = 50 [F] P[2] = 4 [psia] "Maximum outlet temperature where device works is when X[1] = 1 (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) h_1_max = enthalpy(R134a,T=T[1],x=1.0) h_2_max = h_1_max T_2_max = temperature(R134a,P=P[2],h=h_2_max) "Minimum outlet temperature where device works is when X[2] = 1 (saturated vapor)" T_2_min = temperature(R134a,P=P[2],x=1.0) h_2_min = enthalpy(R134a,P=P[2],x=1.0) h_1_min = h_2_min X_1_min = quality(R134a,P=P[1],h=h_1_min) "Data for process line on P-h diagram with X[1] = 0.9" x[1] = 0.9 h[1] = enthalpy(R134a,T=T[1],x=x[1]) h[2] = h[1] T[2] = temperature(R134a,P=P[2],h=h[2]) "Data for plot of x_1 versus T_2" h_2 = enthalpy(R134a,T=T_2,P=P[2]) h_1 = h_2 x_1 = quality(R134a,T=T[1],h=h_1)

(SP 3)

Given:

4L of water inside rigid body cooker from State 1 to State 2

State 1: 2L of saturated liquid water @ 𝑃1 = 200 𝑘𝑃𝑎

2L of saturated water vapor @ 𝑃1 = 200 𝑘𝑃𝑎

State 2: 4L of saturated water vapor @ 𝑃2 = 200 𝑘𝑃𝑎

𝑇𝑠𝑡𝑜𝑣𝑒 = 1000 ℃, 𝑇0 = 20℃, ��𝑙𝑜𝑠𝑠 = 300 𝑊, ��𝑖𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, ∆𝑡 = 1 ℎ𝑜𝑢𝑟

Find:

(a) The total mass of the steam (kg) entering the environment.

(b) The rate of heat transfer (kW) from the source.

(c) The total entropy production (kJ/K) in one hour.

(d) The total exergy destroyed (kJ) in one hour.

System: open system with saturated water inside 4L pressure cooker

Assumptions:

1. Steady flow (No steady state!)

2. No change in PE and KE

3. No work

4. Temperature and pressure are uniform throughout the cooker

State1 2L saturated liquid water + 2L saturated water vapor at P= 200 kPa

State 2

4L saturated water vapor

at P= 200 kPa

5. Constant properties

6. A half of the volume is saturated liquid water and a half of the volume is saturated water

vapor inside the cooker at state 1

7. Water vapor inside the cooker at state 2

8. Water vapor leaves the cooker at state 2

Basic Equations:

𝑑𝑚𝑐𝑣

𝑑𝑡= ∑ ��𝑖𝑛 − ∑ ��𝑜𝑢𝑡

𝑑𝐸𝑐𝑣

𝑑𝑡= 𝑄𝑐𝑣

− 𝑊𝑐𝑣 + ∑ ��𝑖𝑛(ℎ + 𝑃𝐸 + 𝐾𝐸)𝑖𝑛 − ∑ ��𝑜𝑢𝑡(ℎ + 𝑃𝐸 + 𝐾𝐸)𝑜𝑢𝑡

𝑑𝑆𝑐𝑣

𝑑𝑡= ∑

��𝑐𝑣

𝑇𝑏+ ∑ ��𝑖𝑛𝑠𝑖𝑛 − ∑ ��𝑜𝑢𝑡𝑠𝑜𝑢𝑡 + ��𝐶𝑉

𝐸�� = 𝑇0𝜎𝑐𝑣

Solution:

(a)

Because pressure 200 kPa is given for the saturated water, we can find properties of

saturated water from table A-3.

Tsat = 120.2℃

vf = 0.0010605 m3/kg vg = 0.8857 m3/kg

uf = 504.49 kJ/kg ug = 2529.5 kJ/kg

hf = 504.7 kJ/kg hg = 2706.7 kJ/kg

sf = 1.5301 kJ/kgK sg = 7.1271 kJ/kgK

First, mass balance equation for open system is

𝑑𝑚𝑐𝑣

𝑑𝑡= ∑ ��𝑖𝑛 − ∑ ��𝑜𝑢𝑡

Integrate the equation w.r.t. time and then the equation can be written as

∆𝑚 = 𝑚2 − 𝑚1 = − 𝑚𝑜𝑢𝑡

The mass at state 1 is

𝑚1 = 𝑚𝑓 + 𝑚𝑔 = 𝑉𝑓

𝑣𝑓⁄ +

𝑉𝑔𝑣𝑔

⁄ = 2𝐿0.0010605 𝑚3/𝑘𝑔⁄ + 2𝐿

0.8857 𝑚3/𝑘𝑔⁄

∵ no inlet

= 0.002𝑚3

0.0010605𝑚3

𝑘𝑔⁄

+ 0.002𝑚3

0.8857𝑚3

𝑘𝑔⁄

= 1.886 𝑘𝑔 + 0.002 𝑘𝑔

= 1.888 𝑘𝑔

The mass at state 2 is

𝑚2 = 𝑉𝑣𝑔

⁄ = 4𝐿0.8857 𝑚3/𝑘𝑔⁄

= 0.004𝑚3

0.8857𝑚3

𝑘𝑔⁄

= 0.005𝑘𝑔

Thus,

∆𝑚 = 𝑚2 − 𝑚1 = 0.005 𝑘𝑔 − 1.888 𝑘𝑔 = −1.883 𝑘𝑔

− 𝑚𝑜𝑢𝑡 = −1.883 𝑘𝑔

𝑚𝑜𝑢𝑡 = 1.883 𝑘𝑔

Therefore, 1.883 kg of steam enters into the environment

(b)

Now, the energy balance equation for open system is

𝑑𝐸𝑐𝑣

𝑑𝑡= 𝑄𝑐𝑣

− 𝑊𝑐𝑣 + ∑ ��𝑖𝑛(ℎ + 𝑃𝐸 + 𝐾𝐸)𝑖𝑛 − ∑ ��𝑜𝑢𝑡(ℎ + 𝑃𝐸 + 𝐾𝐸)𝑜𝑢𝑡

Apply no work, no KE, no PE, and one outflow, then, the equation above can be

𝑑𝐸𝑐𝑣

𝑑𝑡= 𝑄𝑐𝑣

− 𝑊𝑐𝑣 + ∑ ��𝑖𝑛(ℎ + 𝑃𝐸 + 𝐾𝐸)𝑖𝑛 − ∑ ��𝑜𝑢𝑡(ℎ + 𝑃𝐸 + 𝐾𝐸)𝑜𝑢𝑡

Rewrite the equation,

𝑑𝐸𝑐𝑣

𝑑𝑡= 𝑄𝑐𝑣

− ��𝑜𝑢𝑡ℎ𝑜𝑢𝑡

Integrate the equation w.r.t. time,

∆𝐸𝑐𝑣 = 𝑄𝑐𝑣 − 𝑚𝑜𝑢𝑡ℎ𝑜𝑢𝑡

Because no KE and no PE, the equation above can be written as

𝑚2𝑢2 − 𝑚1𝑢1 = 𝑄𝑖𝑛 − 𝑄𝑙𝑜𝑠𝑠 − 𝑚𝑜𝑢𝑡ℎ𝑜𝑢𝑡

0.005𝑘𝑔 ∗ 2529.5𝑘𝐽/𝑘𝑔 − (1.886𝑘𝑔 ∗ 504.49𝑘𝐽/𝑘𝑔 + 0.002𝑘𝑔 ∗ 2529.6𝑘𝐽/𝑘𝑔)

= 𝑄𝑖𝑛 − 0.3𝑘𝐽/𝑠 ∗ 3600𝑠 − 1.883𝑘𝑔 ∗ 2706.7𝑘𝐽/𝑘𝑔

𝑄𝑖𝑛 = 5232.8 kJ

The rate of heat transfer is

��𝑖𝑛 =𝑄𝑖𝑛

∆𝑡⁄ =5232.8 𝑘𝐽

3600 𝑠 = 1.454 kW

Therefore, 𝟏. 𝟒𝟓𝟒 𝒌𝑾 of the rate of heat transfer is entering into the cooker from

the stove.

(c)

The entropy balance equation for open system is

𝑑𝑆𝑐𝑣

𝑑𝑡= ∑

��𝑐𝑣

𝑇𝑏+ ∑ ��𝑖𝑛𝑠𝑖𝑛 − ∑ ��𝑜𝑢𝑡𝑠𝑜𝑢𝑡 + ��𝐶𝑉

Apply assumptions and integrate the equation w.r.t. time,

∆𝑆𝑐𝑣 = ∑𝑄𝑐𝑣

𝑇𝑏− 𝑚𝑜𝑢𝑡𝑠𝑜𝑢𝑡 + 𝜎𝑐𝑣

Rewrite,

𝑚2𝑠2 − 𝑚1𝑠1 =𝑄𝑖𝑛

𝑇𝑠𝑡𝑜𝑣𝑒−

𝑄𝑙𝑜𝑠𝑠

𝑇0− 𝑚𝑜𝑢𝑡𝑠𝑜𝑢𝑡 + 𝜎𝑐𝑣

𝑤ℎ𝑒𝑟𝑒 𝑚1𝑠1 = 𝑚𝑓𝑠𝑓 + 𝑚𝑔𝑠𝑔

Once you substitute all known values, you will get one unknown,

0.005 𝑘𝑔 ∗ 7.1271𝑘𝐽

𝑘𝑔𝐾– (1.886 𝑘𝑔 ∗ 1.5301

𝑘𝐽

𝑘𝑔𝐾+ 0.002 𝑘𝑔 ∗ 7.1271

𝑘𝐽

𝑘𝑔𝐾)

=5232.8 𝑘𝐽

1273 𝐾−

0.3 𝑘𝑊 ∗ 3600 𝑠

293 𝐾− 1.883 𝑘𝑔 ∗ 7.1271

𝑘𝐽

𝑘𝑔𝐾+ 𝜎𝑐𝑣

𝜎𝑐𝑣 = 0.0356 − 2.9 − 4.111 + 3.686 + 13.420 = 10.131 kJ/K

Therefore, 𝟏𝟎. 𝟏𝟑𝟏 𝐤𝐉/𝐊 of the total entropy production is generated in one hour.

(d)

The total exergy destruction can be calculated by

𝐸�� = 𝑇0𝜎𝑐𝑣 = 293 𝐾 ∗ 10.131𝑘𝐽

𝐾= 2968.3 𝑘𝐽

Therefore, 𝟐𝟗𝟔𝟖. 𝟑 𝐤𝐉 of the total exergy is destroyed during one hour.