STUDENT’S SOLUTION MANUAL

(PARTIAL)

THERMAL ENVIRONMENTAL ENGINEERING

Second Edition

JAMES L. THRELKELD

St. paul, Minnesota

Includes: 5.4, 5.5, 5.6, 5.7

Submitted by: Jaycob O. Clavel

5.4 An aqua-ammonia system similar to that in Fig. 5.18 operates as follows: high-side pressure =

200 psia; t3 = 190F; t7 = 140F; t4 = 210F; m7 = 100 lb per min. Assume equilibrium conditions for

States 3, 4, and 7. Find (a) the lb per min of strong solution leaving the absorber, and (b) the lb

per min of cooling water required for the dephlegmator if the water temperature rise is 15 F.

You may make the same assumptions with regard to the rectifying column as made in Part(d) of

Example 5.4.

State Point

Pressure P

psia

Temperature t F

Concentration X

lb NH3/lb mix

Enthalpy h

Btu/lb mix

Flow Rate

lb mix/min

1 0.438

2 200 0.438

3-e 200 190 0.438 98

4-e 200 210 0.383 119

5 200 0.383

6 0.383

7-e 200 140 0.995 657 100

8 200 0.995 100

9 200 0.995 100

10 0.995 100

11 0.995 100

12 0.995 100

a.

CV at absorber

For the absorber we have

m6 + m12 = m1 m6 – m1 = -100

m6x6 + m12x12 = m1x1 0.383m6 – 0.483m1 = -100(0.995)

2 equations and 2 unknowns

m6 = 1012.73 lb per min

m1 = 1112.73 lb per min

b.

Let us assume that the vapor state inside the rectifying column is saturated with a temperature

10F higher than that of the strong solution which is entering.

qD/m7= he – h7 = 745-657 = 88 Btu/lb

qD = (qD/m7)(m7) = 88(100) = 8800 Btu per min

qD = mwcp(Te– Ti) ; Te– Ti = 15F, cp = 1Btu/lbm R

mw = 586.67 lb per min

5.5An aqua-ammonia system similar to that in Fig. 5.18 operates as follows: high-side pressure =

220 psia; low-side pressure = 20 psia;t4 = 210F; t8 = 80F; t12 = 40F;m4 = 1000 lb per min;m12 = 100

lb per min. Assume equilibrium states at1, 3, 4, and 12. Determine (a) the concentration of the

strong solution leaving the heat exchanger, (b) the heat removed in the absorber in Btu per min,

and (c) the tons of refrigeration produced.

State Point

Pressure P

psia

Temperature t F

Concentration X

lb NH3/lb mix

Enthalpy h

Btu/lb mix

Flow Rate

lb mix/min

1-e 20 52 -55 1100

2 220 1100

3-e 220 101 1100

4-e 220 210 0.387 119 1000

5 220 0.387 1000

6 20 0.387 1000

7 220 0.999 100

8 220 80 0.999 131 100

9 220 0.999 100

10 20 0.999 100

11 20 0.999 100

12-e 20 40 0.999 628 100

a.

x3 = ?

CV at the absorber

For the absorber we have,

m6x6 + m12x12 =m1x1 ;x1 =x3

0.387(1000)+ 0.999(100) = 1100(x3)

X3 = 0.44264lb NH3/lb mix

b.

h2 = h1+ ((P2-P1)(v1))/(J);P2 = 220 psia, P1 = 20 psia, J = 778 ft-lb per Btu, h1 = -55 Btu/lb

v1 = (1-x1)vH20+ 0.85x1vNH3; vH20 = vg@52F= 0.01602cu ft/lb, vNH3 = vg@52F= 1/38.9 cu ft/lb

v1 = 0.0186 cu ft per lb

h2 = -54.995 Btu/lb

CV at the heat exchanger below the generator

For the heat exchanger we have,

m4h4 + m2h2 =m3h3+ m5h5

1000h5 = 1000(119) + 1100(-54.995) – 1100(101)

h5 = -52.5945 Btu/lb

h6 =h5

Using the CV at the absorber we have,

m12h12 + m6h6 =m1h1+ qA

100(628) + 1000(-52.5945) = 1100(-55) + qA

qA = 70705.5 Btu per min

CV at the figure below,

m8h8 + qE =m12h12

100(131) + qE= 100(628)

qE= 49700 Btu

tons = qE/200

tons = 248.5

5.6 A lithium bromide-water system of the type shown in Fig.5.16 operates with a condensing

temperature of 110 F, evaporating temperature of 38 F, temperature of solution leaving

absorber of 100F, temperature of solution entering generator of 180F, and temperature of

solution leaving generator of 210F. Assume saturated conditions for States 3, 4, 8 and 10.

Neglect pressure drops in components and lines. Warm water from the load returns to the

machine at a temperature of 52F and at a rate of flow of 600 GPM. Chilled water leaves the

machine at a temperature of 44F. Saturated steam at 25psia enters the generator and leaves as

saturated water. Calculate the required rate of flow of steam in lb per hr.

State Point

Pressure P

mm Hg

Temperature t F

Concentration X

lbLi Br/lb mix

Enthalpy h

Btu/lb mix

Flow Rate m

lb mix/minton

1 5.82 100 0.57 2.1

2 65.96 0.57 2.1

3-e 65.96 180 0.57 -37 2.1

4-e 65.96 0.63 -24 1.9

5 65.96 0.63 1.9

6 5.82 0.63 1.9

7 65.96 210 0 155.5 0.2

8-e 65.96 110 0 78 0.2

9 5.82 0 78 0.2

10-e 5.82 38 0 1077.8 0.2

Convertions,

Condensing pressure = 2.5968 in Hg (2.54cm/1in)(10mm/1cm) = 65.96 mm Hg

Evaporating pressure = 0.22904 in Hg (2.54cm/1in)(10mm/1cm) = 5.82 mm Hg

600 gal/min (3.79 L/1gal)(1kg/1L)(2.2lbs/1kg)(60mins/1hr) = 299803.68 lb/hr

200 Btu/min(60min/1hr) = 12000 Btu/hr

q = mcp(Ti–Te); Ti= 52F, Te= 44F ,m = 600 GPM or 299803.68 lb/hr,cp = 1Btu/lbm R

q = (299803.68)(1)(52-44)

q = 2398429.44 Btu/hr or 199.87 tons

CV at the evaporator

qe= q

q is the amount of energy that was released by the water running through the machine

m9 = tons/(h10 – h9)

m9 = 199.87/(1077.8-78)

m9 = 0.2 lb/min ton

CV at the absorber

For the absorber we have,

m6 + m10 = m1 m6 – m1 = -0.2

m6x6 + m10x10 = m1x1 0.63m6 – 0.57m1 = -0(0.2)

2 equations and 2 unknowns

m6 = 1.9lb/min ton

m1 = 2.1lb/min ton

For steams, we can use the equation below;

h7 = 1061 +0.45t; t in F

h7 = 1155.5 Btu/lb

CV at the generator

qg= m4h4 + m7h7 – m3h3

qg= 1.9(-24) + 0.2(1155.5) -2.1(-37)

qg= 263.2

m = 60qg(ton)/hfg; hfg= hfg @ 25 psis for steam = 952.2, ton = 199.87

m = 3314.79lb per hr

5.7 Compare the cost in dollars per (hr)(ton) for generator steam in Example5.5 with cost of

electricity in dollars per (hr)(ton) for a refrigerant 12 compressor system. You may assume that

steam costs 40 cents per 1000 lb and that electricity cost is 2 cents per kwhr. Assume that the

Refrigerant 12 system is operated at 40F evaporating temperature and 100F condensing

temperature. You may obtain an estimate for the Hp/ton requirement requirement of the

Refrigerant 12 system by extrapolation of Fig 3.15. Assume an electric motor efficiency of 80

percent.

costfor generator steam[dollars per (hr ton)] = mass flow rate of steam consumption for the

generator[lb/ (hr ton)] x steam costs(cents per

lb)

cost for generator steam=15.78[lb/(hr ton)] x (.4 dollars/1000lb)

cost for generator steam =.006312 dollars per (hr ton)

instead of using solution, the problem wants us to use R12 as refrigerant for a single stage cycle

P1 =Psat@ 100F = 131.86 psia

h2 = hf @ 131.86 psia= 31.100 Btu/lb

P3 =Psat@ 40F = 51.667psia, h3 = hg@ 40F = 81.436 Btu/lb

v3 = .77357 cu ft/lb

s3 =s4 = sg@ 40F = 0.16586 Btu/lb R

h4 = h @ 131.86 psia and 0.16586 Btu/lb R = 88.55 Btu/lb

Hp/ton = 0.00606(nP3v3/(n-1) Ƞm(h3-h2))(( P4/P3)(n-1)/n-1) ; n=y=1.13, Ƞm = .8

Hp/ton =0.856837

Hp/ton(.7457kW/1hp)(.02dollars/kWhr) = $0.013 per(hr)(ton) for electricity

costfor electricity = $0.013 per(hr)(ton)