Threlkeld chap 3 Answers

SOLUTION MANUAL THERMAL ENVIRONMENTAL ENGINEERING

SECOND EDITION

JAMES L. THRELKELD

St. Paul Minnesota

CHAPTER 3

ME 161

Submitted by:

Jaycob O. Clavel

BSMEN-4

ME 161 Chapter 3 Problems with Solutions

Example 3.2 Liquid refrigerant R12 leaves the condenser of a vapor compression system under a

pressure of 153 psia and at a temperature of 70F. Evaporating pressure is 23.74 psia and the

vapor leaves the evaporator at 5 F. Otherwise, the system follows the theoretical single-stage

cycle. Find (a) the enthalpy of the liquid entering the expansion valve, (b) the refrigerating effect

in Btu per lb, and (c) the coefficient of performance.

a. the enthalpy of a subcooled liquid refrigerant may be taken as equal to the enthalpy of

the saturated liquid at the same temperature.

h1 = 24.050 Btu/lb

b. h2 = h1 = 24.050 Btu/lb and h3= h @ 5F & 23.74 psia = 77.98 Btu/lb

2Q3=h3-h2= 53.93 Btu/lb

c. P4 = 153 psia s4 = s3 = 0.1705 Btu/lb R ; h4=92.55 Btu/lb

C.O.P. = (h3-h2)/(h4-h3)

C.O.P. =53.93/(92.39-77.98) =3.7

3.1 Solve Example 3.1 for Refrigerant 12.

An ammonia refrigerating plant following the theoretical single-stage cycle operates with

a condensing temperature of 90 F and an evaporating temperature of 0 F. the system produces

15 tons of refrigeration. Determine (a) the coefficient of performance, (b) refrigerating

efficiency, (c) horsepower per ton of refrigeration, (d) rate of refrigerant flow in lb per min, (e)

theoretical horsepower input to compressor, and (f) theoretical piston displacement of

compressor in cu ft per min.

h1 = h2 = hf @ 90F = 28.713 Btu/lb

h3 = hg @ 0F = 77.271 Btu/lb

v3 = vg @ 0F = 1.6089 cu ft/lb

s3 = s4 = 0.16888 Btu/lb R, P4 = Psat @ 90F = 114.5psia

h4 = 89.121 Btu/lb

(a) C.O.P. = (h3 – h2) / (h4 - h3) = (77.271 – 28.713) / (89 – 77.271) = 4.14

(b) ȠR = [ C.O.P. (T1 – T2) ] / T2 = (4.14)(90) / 459.6 = 0.81 or 81 percent

(c) Hp / ton = 4.72 / 4.14 = 1.14 hp per ton

(d) m = (tons)(200) / (h3 – h2) = (15)(200) / (77.271 – 28.713) = 61.78 lb/min

(e) Hp = m(h4 - h3) / 42.4 = (61.78)( 89 – 77.271) / 42.4 = 17.1 hp

Or

Hp = (Hp/ton)(tons) = (1.14)(15) = 17.1

(f) P.D. = mv3 = (61.78)(1.6089) = 99.4 cu ft/min

3.2 In a theoretical single-stage ammonia vapor compression system, liquid leaves the condenser

at 250 psia and 104 F. Evaporator pressure is 18.30 psia. Vapor leaves the evaporator at 0 F. the

system produces 25 tons of refrigeration. Determine (a) the coefficient of performance, and (b)

the piston displacement in cu ft per min, assuming a volumetric efficiency of 100 percent.

h1 = h2 = hf @104F = 159.9 Btu/lb

h3 = h @ 0F & 18.30 psia = 616 Btu/lb

s3 = s4 = s @ 0F & 18.30 psia = 1.40324 Btu/lb R

v3 = v @ 0F & 18.30 psia = 15.7322 cu ft/lb

P4 = 250 psia

h4 = 804 Btu/lb

(a) C.O.P. = (h3 – h2) / (h4 - h3) = (616 – 159.9) / (804 – 616) = 2.43

(b) m = (tons)(200) / (h3 – h2) = (25)(200) / (616 – 159.9) = 10.96 lb/min

P.D. = mv3 = (10.96)(15.7322) = 172 cu ft/min

3.3 A Refrigerant 12 system is arranged as shown in Fig 3.26. compression is isentropic. Assume

frictionless flow.the known data are: t1 = 100 F, t2 = 80 F, and t4 = -10 F. Find the system

horsepower per ton.

h1 = hf @100F =31.1 Btu/lb

h2 = hf @80F =26.365 Btu/lb

h3 = h2 = 26.365 Btu/lb, P3 = Psat @-10F =19.189 psia

P4 = P3 = 19.189 psia h4 = hg @-10F =76.2 psia

We already know that there is equal mass flow rate for inlets and exits, so we have

Energy Balance- h1 + h4 = h2 + h5

h5=80.935 Btu/lb P5 = P4 = 19.189 psia s5 = 0.1801 Btu/lb R

s6 = s5 P6 = P1 = 131.86 psia h6 = 96.67 Btu/lb

C.O.P.=( h4- h3)/( h6- h5)=3.11

Hp/ton=4.72/C.O.P.=1.52

3.4. Solve Example 3.3 for ammonia.

A Refrigerant 12 theoretical single-stage cycle operates with a condensing temperature

of 90 F and an evaporating temperature of 0 F. As shown in Fig. 3.7 we will assume a Carnot

cycle operating between the same temperatures. Determine (a) the Carnot-cycle work of

compression in Btu per lb, (b) the Carnot cycle refrigerating effect in Btu per lb, (c) the excess

work of compression in Btu per lb for the single-stage cycle caused by the superheat horn, (d)

the excess work of compression in Btu per lb for the single-stage cycle caused by throttling (e)

the loss in refrigerating effect in Btu per lb for the single-stage cycle caused by throttling, and (f)

the refrigerating efficiency.

a. Wc = (T1 – T2)( s3 - s1) = 90(1.3352-0.2958)= 93.546 Btu per lb

b. Qc = T2 (s3 – s1) = 459.6(1.3352-0.2958)= 477.7 Btu per lb

s3 = s4 = sg @ 0F =1.3352 Btu/lb R ; P4 = Psat @ 90F = 180.6 psia h4 = 724.1 Btu/lb

c. A1 = (h4 – hd)-T1( s3 – sd)= (724.1-632)-549.6(1.3352-1.1846)= 9.33024 Btu per lb

d. A2 = h1 – hb ; h1 = 143.5 Btu/lb s1 = sb=0.2958 Btu/lb R Pb =30.24 psia

0.2958= 0.0975 + x(1.3352-0.0975) ; x=0.16

hb = 42.9+0.16(611.8-42.9)= 134.05 Btu/lb

A2 = 143.5-134.05=9.45 Btu per lb

e. Since A3 = A2 we have A3 = 9.45 Btu per lb

f. ηr =(h3 – h2)( T1 – T2)/(h4 – h3)T2 = 90(611.8-143.5)/459.6(724.1-611.8) = 0.817

3.5. An ammonia system is arranged as shown in Fig. 3.27. Assume isentropic compression and

frictionless flow. Known data are P1 = 160 psia, t3 = 20 F, and t6 = -30 F. The capacity of

Evaporator A is 5 tons and the capacity of Evaporator B is 10 tons. (a) Draw schematic P-h and T-

s diagrams for the cycle. (b) Calculate the percent decrease in theoretical power required if the

above system was replaced with one having a separate compressor for each evaporator. Assume

the same evaporating and condensing pressures and tons capacity.

a.

b.

h1 = hf @ 160 psia = 134.7 Btu/lb

T2 = T3 = 20F, h2 = 134.7 Btu/lb

T3 = 20F, h3 = hg @ 20F = 617.8 Btu/lb

P4 = 13.90 psia, h4 = h3 = 617.8 Btu/lb

h5 = h1 = 134.7 Btu/lb

P6 = Psat @ -30F = 13.90 psia, h6 = hg @ -30F = 601.4 Btu/lb

Hp = m1,8,7(h8-h7)/ 42.4

5(200) = m2,3,4 (h3-h2) ; m2,3,4 = 2.07 lb/min

10(200) = m5,6 (h6-h5) ; m5,6= 4.29 lb/min

m5,6 + m2,3,4 = m1,8,7 ; m1,8,7 = 6.36 lb/min

Energy Balance:

m2,3,4 h4 + m5,6 h6= m1,8,7 h7

h7 = 606.74 Btu/lb, P7 = P6 = 13.90 psia, s7 = 1.412 Btu/lb R

s8 = s7 =1.412 Btu/lb R, P8 = P1 = 160psia, h8 = 771.1 Btu/lb

Hp = 6.36(1364.36)/ 42.4 = 24.64

5(200) = m2,3 (h3-h2) ; m2,3 = 2.07 lb/min

10(200) = m4,5,6 (h6-h5) ; m4,5,6= 4.29 lb/min

m4,5,6 + m2,3 = m1,8,7 ; m1,8,7 = 6.36 lb/min

h3= hf @ 20F = 617.8 Btu/lb

h5 = hf @ -30F = 601.4 Btu/lb, s5 = sf @ -30F = 1.398 Btu/lb R

P6 = Psat @ 20F = 48.22 psia, s6 = s5 = 1.398 Btu/lb R, h6 = 671.1 Btu/lb

m2,3 h3+ m4,5,6 h6 = m1,8,7 h7

h7 = 653.414 Btu/lb, P7 = P6 = 48.22 psia s7 = 1.366 Btu/lb R

P8 = 160 psia s8 = s7 = 1.366 Btu/lb R h8 = 737.4 Btu/lb

Hp = [m4,5,6 (h6-h5)+ m1,8,7(h8-h7)]/ 42.4 = 19.69

Percent decrease in theoretical power = (24.64-19.69)/24.64 = 0.2

There are many reasons to have multiple compressors aside from less work input

requirement. Surely we don’t want the compressor to be exhausted.

3.6 A Refrigerant 12 plant operating at 0 F evaporating temperature and 70 F condensing

temperature uses a compressor having 4 percent clearance. Under these conditions the plant

produces 20 tons of refrigeration. (a) Calculate the capacity of the plant if the condensing

temperature were to be increased to 100 F. Assume that other conditions are the same and the

compressor is operated at the same RPM. Assume theoretical single-stage cycle but include

effect of clearance. (b) Assume above system is operated at 100 F condensing temperature.

What evaporating temperature would be required for refrigerating capacity to be zero?

Ƞcv = 1+C-C(P4 /P3)1/ƴ; C=.04; P4 = Psat @ 70F = 84.888 psia, P3= Psat @ 0F = 23.849 psia, ƴ=1.13

Ƞcv = 0.91697

Without polytropic constant n, we will assume that the clearance volumetric efficiency is

equal to the total volumetric efficiency.

Ƞcv = mv3/P.D.

m = 200tons/(h3-h2); tons=20,

h3= hg @ 0F =77.271 Btu/lb, h2= h1 = hf @ 70F = 24.05 Btu/lb v3 = vg @ 0F = 1.6089 cu.ft/lb

P.D.= 131.87 cu.ft/min

a. constant piston displacement with 100F condensing temperature; P4 = Psat @100F=131.86 psia

Ƞcv = 1+C-C(P4 /P3)1/ƴ; C=.04; P4=131.86 psia, P3=23.849 psia, ƴ=1.13

Ƞcv = 0.85834

Ƞcv = mv3/P.D. ; Ƞcv = 0.85834, v3 = 1.6089 cu.ft/lb, P.D.=131.87 cu.ft/min

m = 70.35 lb/min

tons = m(h3-h2)/200; m = 70.35 lb/min h3=77.271 Btu/lb, h2=31.1 Btu/lb

tons =16.2

b. For refrigerating capacity to be zero we need to have Ƞcv equal to zero.

Ƞcv = 1+C-C(P4 /P3)1/ƴ; Ƞcv =0, C=.04; P4= Psat @100F =131.86 psia, P3=unknown , ƴ=1.13

P3 = 3.32 psia

T3 = 76 F

3.7. An ammonia refrigeration plant utilizes a water-jacketed compressor. Saturated vapor at 10

F enters the compressor, and the vapor leaves the compressor at 180.6 psia and 170 F. Flow

through the condenser and evaporator is at constant pressure. The liquid leaving the condenser

is saturated. The system produces 25 tons of refrigeration and the power input (power delivered

to the refrigerant passing through compressor) is 23.2 horsepower. Find the lb per min of

cooling which must flow through the compressor jacket, if the water temperature rise is 10 F.

State 1(Saturated Liquid):

P1 = 180.6 psia h1 = 143.5 Btu/lb

State 2(Saturated Liquid and Vapor Mixture):

P2 = 38.508 psia h2 = 143.5 Btu/lb

State 3(Saturated Vapor):

T3 = 10F h3 = 613.97 Btu/lb

State 4(Superheated):

P4 = 180.6 psia h4 = 686.20844 Btu/lb

The process in the compressor doesn’t have a heat transfer outside the sorroundings but

transfers heat from the ammonia to the cooling water to overcome extreme heat. If that is our

control volume in the figure above right away we have our first law equation:

3Q4 + Wc = ma(h4 – h3) + mw(h4w – h3w) ; 3Q4 = 0

2Q3 = ma(h3 – h2) ; 2Q3 = 25(200) Btu/min

ma = 10.63 lb/min

The enthalpy change of water (h4w – h3w) is equal to cp(T4w – T3w) ; cp = 1 Btu/lb R

23.2(42.4) = 10.63(686.20844-613.97) + mw(1)(10)

mw = 21.6 lb/min

3.8. An industrial plant has available a 4-cylinder, 3-in bore by 4-in stroke, 800 RPM, single-acting

compressor for use with Refrigerant 12. Proposed operating conditions for the compressor are

100 F condensing temperature and 40 F evaporating temperature. It is estimated that the

refrigerant will enter the expansion valve as a saturated liquid, that vapor will leave the

evaporator at a temperature of 45 F, and that vapor will enter the compressor at a temperature

of 55 F. Assume a compressor volumetric efficiency of 70 percent. Assume frictionless flow.

Calculate the refrigerating capacity in tons for a system equipped with this compressor.

Given: Compressor: number of cylinders = 4 D = 3 in. RPM = 800 Refrigerant: R-12 Condensing Temperature = 100F Evaporating Temperature = 40F Where T3 = 45F and T4 = 55F ȠV = 70% Required: Refrigerating Capacity (in tons) Solutions: P.D = 4π (3)2(4) (800)/4(12) (144) = 52.36 cu ft. /min.

State 1: T1 = 100F P1 = 131.6 psia h1 = hf @ P1 = 31.16 Btu/lb

State 2: T2 = 40F P2 = P3 = 51.68 psia h2 = h1 = 31.16 Btu/lb State 3: T3 = 45F h3 = 83.445 Btu/lb State 4: T4 = 55F v4 = 0.82572 ft3/lb Thus, m = ȠV (P.D/v4) = (0.7) (52.36)/0.82572 = 44.388 lb/min

Then,

2Q3 = m (h3-h2)/200 = (44.388) (83.445-31.16)/200 = 11.604 tons

3.9. The following are the conditions for a single-stage ammonia refrigeration plant: condensing

temperature 90 F with no liquid subcooling, evaporating temperature 0 F with saturated vapor at

evaporator outlet, polytropic compression with n = 1.24, compressor clearance = 5 percent.

Assume no pressure drop in piping and compressor valves and no temperature changes either in

piping or on intake stroke of compressor. (a) Calculate the coefficient of performance and

compare with coefficient of performance calculated for isentropic compression. (b) Assume that

the original system is operated with one additional change. The vapor picks up heat in the

suction line of the compressor so that the temperature of the vapor entering the compressor is

60 F. Calculate the percent decrease in the coefficient of performance. (c) Assume that the

original system is operated with the only change being a 4 psi drop in the suction valves and 6 psi

drop in the discharge valves of the compressor. Calculate the percent decrease in the coefficient

of performance.

(a)

Hp/ton = 0.00606(nPbvb/(n-1) Ƞm(h3-h2))(( Pc/Pb)(n-1)/n-1) ; n=1.24, Ƞm = 1

information obtained from the given:

Pb = Psat @ 0F = 30.42 psia or 4378.82 psfa, Pc = Psat @ 90F = 180.6 psia

vb = v3 = vg @ 0F = 9.116 cu ft/lb

h3 = hg @ 0F = 611.8 Btu/lb, h2 = hf @ 90F = 143.5 Btu/lb

Hp/ton = 1.099

C.O.P. = (200/42.4)/( Hp/ton) = 4.3

for isentropic compression we have :


h3 = hg @ 0F = 611.8 Btu/lb, s3 = sg @ 0F = 1.3352 Btu/lb R

h2 = hf @ 90F = 143.5 Btu/lb

s4 = s3, P4 = 180.6 psia, h4 = 725 Btu/lb

C.O.P. = (h3-h2)/ (h4-h3) = 4.16

(b)




Tb = 60F vb = 10.65 cu ft/lb


Hp/ton = 1.2834

C.O.P. = (200/42.4)/( Hp/ton) = 3.7

percent decrease in C.O.P. = 14 %

(c)



Pb = Psat @ 0F - 4 = 26.42 psia or 3804.48 psfa, Pc = Psat @ 90F + 6 = 186.6 psia

Tb = 0F vb = 10.48 cu ft/lb


Hp/ton = 1.225

C.O.P. = (200/42.4)/( Hp/ton) = 3.849

percent decrease in C.O.P. = 10.5 %

3.10 Work parts (b) and (c) of Problem 3.9 but calculate percent decrease in tons capacity.

Assume a constant displacement compressor.

W = {n Pb vb /[(n-1)778]}[(Pc/Pb)(n-1)/n] ; n = 1.24

Using previous information problem 3.9 a:


vb = v3 = vg @ 0F = 9.116 cu ft/lb


W =109.1602 Btu/lb

C.O.P = Q/W

Q = W (C.O.P); C.O.P. = 4.3

Q = 468.5286 Btu/lb

Q = m(h3-h2)

m = 1.000488 lb/min

Ƞv = [1+C-C(Pc/Pb)1/n] (v3 /vb); C = .05, v3 =vb=9.116 cu ft/lb, (v3 /vb) = 1

Ƞv = .83972

m = ȠV (P.D/v3)

P.D = 10.8613581 cu ft/min

For problem 3.9b we have P.D = 10.86 cu ft/min, v3 = 9.116 cu ft/lb

Ƞv = [1+C-C(Pc/Pb)1/n] (v3 /vb); v3 = 9.116 cu ft/lb, vb = 10.65 cu ft/lb

Ƞv = .71876

m = ȠV (P.D/v3)

m = 0.85638 lb/min

tonsold = m(h3-h2)/200; m = 1.000488 lb/min

tonsold = 2.3426

tonsnew= m(h3-h2)/200; m = 0.85638 lb/min

tonsnew= 2.00521428

percent decrease in tons capacity = 14 %

For problem 3.9c we have

P.D = 10.86 cu ft/min, v3 = 9.116 cu ft/lb

Pb = Psat @ 0F - 4 = 26.42 psia or 3804.48 psfa, Pc = Psat @ 90F + 6 = 186.6 psia

Ƞv = [1+C-C(Pc/Pb)1/n] (v3 /vb); v3 = 9.116 cu ft/lb, vb = 10.48 cu ft/lb

Ƞv = .70293

m = ȠV (P.D/v3)

m = 0.837509 lb/min

tonsold = m(h3-h2)/200; m = 1.000488 lb/min

tonsold = 2.3426

tonsnew= m(h3-h2)/200; m = 0.837509 lb/min

tonsnew= 1.961028

percent decrease in tons capacity = 16.3 %

3.11. Assume that you are the chief engineer of a creamery. You are in need of an ammonia

compressor for an addition to your plant. You have been informed that another creamery

operated by your company has a surplus compressor which will be made available to you if it is

of adequate capacity for your installation. It is estimated that the proposed installation would

operate under the following conditions: capacity = 18.5 tons, condensing pressure = 195 psia,

ammonia liquid leaves the condenser at saturated conditions, ammonia liquid enters the

expansion valve at 76 F, evaporating temperature = 10 F, vapor leaves evaporator at saturated

conditions, vapor is superheated 20 F in compressor suction line. The data supplied to you about

the surplus compressor are only the following: It is a 4-cylinder, vertical, reciprocating, single-

acting, single-stage, water-jacketed compressor, with a maximum RPM of 600. Cylinder diameter

is 4 in and the stroke is 5 in. Based on your past experience you make the following supplemental

assumptions to allow further calculations: clearance = 4 percent, pressure drop in suction valves

= 4 psi, vapor in superheated 15 F in cylinder on the intake stroke after passing the suction

valves, polytropic compression with n = 1.27, pressure drop in discharge valves = 6 psi,

compressor mechanical efficiency = 80 percent. (a) Determine whether the surplus compressor

may be used and, if so, what RPM it should operate at. (b) Estimate the horsepower required to

drive the compressor.

Given:

Compressor:

Number of cylinders = 4 RPM = 600 D = 4 in. L = 5 in.

Ammonia

Capacity = 18.5 tons Condensing Pressure = 195 psia

T1 = Tsat @ 195 psia T2 = 76F

Evaporating Temperature = 10F

T4 = Tsat = 10°F T5 = 20°F (superheated)

C = 4% ∆Ps = 4 psi, superheated by 15°F on intake ∆Pd = 6 psi n = 1.27 ηm = 80%

Solutions: For surplus compressor: P.D = 4π (4)2(5) (600)/4(12) (144) = 87.266 cu ft/min. Note: P.Dsurplus ≥ P.Dold unit Then, State 1: P1 = 195 psia T1 = 94.654F h1 = hf @ T1 = 148.709 Btu/lb State 2: P1 = P2 = 195 psia T2 = 76F h2 = hf @ T2 = 127.092 Btu/lb State 3: P3 = 38.508 psia T3 = 10F h3 = h2 = 127.092 Btu/lb State 4: P4= P3 = Psat @ 10F = 38.508 psia T4 = 10F h4 = hg @ 10F h4 = 613.97 Btu/lb State 5: P5 = P4 = 38.508 psia T5 = 10 + 20 = 30°F h5 = 625.589 Btu/lb v5 = 7.711 cu ft/lb So, Pa = P5 - ∆Ps = 38.508 – 4 = 34.508 psia (approx. 35 psia) ha = h5 = 625.589 Btu/lb Ta = 27.86F Pb = Pa = 34.508 psia Tb = Ta + 15 = 42.86F

vb = 8.891 cu ft/lb Pc = 195 + 6 = 201 psia Then, ηv = * 1 + C – C (Pc/ Pb)1/n](v5/ vb) = 0.7636 Then, m = tons (200)/ (h4 – h3) m = 18.5 (200)/ (613.97-127.092) m = 7.599 lb/min So, P.Dold unit = mv5/ηv = 7.599(7.711)/0.7636 = 76.7363 cu ft/min. P.Dold unit < P.Dsurplus Answers: a.) Compressor’s RPM: 76.7363 cu ft/min = 4π (4)2(5) (RPM)/4(12) (144) RPM = 527.622

The surplus compressor may be used and to be operated at 527.622 RPM. b.) 5W6 = nPbvb / (n-1)(778)[ (Pc/Pb)(n-1)/n – 1 ] = 0.8429 x 144 = 121.3776 Btu/lb So,

HP = 7.599(0.8429)/42.4(0.8) x 144 HP = 27.1918

3.12. An ammonia vapor-compression system is arranged as shown in Fig. 3.28. Assume

isentropic compression and frictionless flow. Given data are t1 = 90 F, t6 = -10 F, and t8 = -20 F. (a)

Compare the coefficient of performance for the cycle of Fig. 3.28 with that obtained if the liquid

cooler were omitted. (b) What conclusion can be drawn from theoretical reasoning on the use of

liquid intercoolers in single-stage systems? (c) Using practical reasoning, discuss the conditions

under which the arrangement of Fig. 3.28 might be advantageous.

With Intercooler

h1 = h2 = h4 = h3 = hf @90F = 143.5 Btu/lb

h6 = h7 = hf @ -10F = 32.1 Btu/lb

h5 = h8 = h9 = hf @ -20F = 605 Btu/lb s5 = sg @ -20F = 1.376 Btu/lb R

s10 = s8, h10 = h @ 180.7 psia & 1.376 Btu/lb R = 753.9 Btu/lb

C.O.P. = (h8 – h7) / (h10 – h9) = (605 – 32.1) / (753.9 – 605) = 3.85

Without Intercooler

h1 = h2 = 143.5 Btu/lb

h3 = 605 Btu/lb

h4 = 753.9 Btu/lb

C.O.P. = (h3 – h2) / (h4 - h3) = (605 – 143.5) / (753.9 – 605) = 3.10

(b) Intercoolers increases the C.O.P of the vapor-compression system.

(c) Intercoolers lessens the amount of work input in the compressor which means lesser bills for

electric consumption.

3.13. A two stage ammonia system is arranged as shown in Fig. 3.22 except that the water

intercooler is omitted. Condensing temperature is 90 F, intermediate saturation temperature is

15 F, and evaporating temperature is -40 F. Liquid leaves the intercooler at a temperature of 25

F. Saturated liquid leaves the condenser. Saturated vapor leaves the evaporator. Saturated vapor

leaves the intercooler. It is known that the volumetric efficiency for each compressor is the

same. The piston displacement of the second-stage (high-pressure) compressor is 100 cu ft per

min. Calculate the piston displacement of the first-stage compressor.

h1= hf @ 90F =143.5 Btu/lb

h2= h1= 143.5 Btu/lb

h3= hf @ 25F =70.2 Btu/lb

v5= vg @ -40F =24.86 cu ft/lb, s5= sg @ -40F =1.4242 Btu/lb R

s6= s5, P6= Psat @ 15F =43.14psia h6= h @ 43.14psia & 1.4242 Btu/lb R = 678.9 Btu/lb

h7= hg @ 15F = 616.3 Btu/lb, v7= vg @ 15F = 6.556 cu ft/lb

100 = m1,7,8 v7

m1,7,8 = 15.24 lb/min

P.D. = m3,4,5,6 v5

CV @ shell and coil intercooler

m3,4,5,6 h1 + m2 h2 + m3,4,5,6 h6 = m3,4,5,6 h3 + m1,7,8 h7

CV @ these pipes

m1,7,8 = m2 + m3,4,5,6

Using substitution or elimination for these 2 equations we will get:

m3,4,5,6 = 11.837 lb/min

P.D. = m3,4,5,6 v5

P.D. = 294.3 cu ft per min

3.14. A Refrigerant 12 system is arranged as shwn in Fig.3.23. Condensing pressure is 120 psia,

intermediate pressure is 30 psia, and evaporating pressure is 7 psia. The following temperatures

are known: t1 = 80 F, t3 = 20 F, t5 = -40 F, and t7 = 20 F. Assume isentropic compression and

frictionless flow. Calculate the coefficient of performance.

h1 = hf @ 80F = 26.36 Btu/lb (subcooled)

h2 = h1 =26.36 Btu/lb

h3 = hf @ 20F = 12.863 Btu/lb

h4 = h3 = 12.863 Btu/lb

P5 =7 psia T5 =-40F h5 = h @ 7psia & -40F = 73.12 Btu/lb s5 = s @ 7psia & -40F = 0.1788 Btu/lb R

P6 =30 psia s6 = s5 = 0.1788 Btu/lb R h6 = h @ 30psia & 0.1788 Btu/lb R = 83.78 Btu/lb

P7=30 psia T7 =20F h7 = h @ 30psia & 20F = 79.77 Btu/lb

C.V at these pipes

mass balance for this figure is;

m1,8,9=m3,4,5,6 + m2 Eqn 1

C.V at these pipes

Mass and Energy balance for this figure is;

m7h7 + m3,4,5,6 h6 = m1,8,9h8 Eqn 2

m1,8,9=m3,4,5,6 + m7 Eqn 3

Adding Eqn 3 and Eqn 1 will give us m2=m7

C.V at the shell and coil liquid intercooler

Energy balance for this figure is;

m3,4,5,6 h1 + m2h2 = m3,4,5,6 h3 + m7h7; m2=m7

m3,4,5,6(h1-h3) = m2(h7-h2)

m2=0.253 m3,4,5,6 Eqn 3

m1,8,9=(103.94/ h8) m3,4,5,6- from Eqn 2 combined with Eqn 3

m1,8,9=(0.253+1)m3,4,5,6- from Eqn 1 combined with Eqn 3

Dividing previous equations we get :

h8= 82.97 Btu/lb

P8 = 30 psia s8 = s @ 30psia & 82.97 Btu/lb = 0.1772 Btu/lb R

s9= s8 =0.1772 Btu/lb R P9 = 120 psia h9 = s @ 120psia & 0.1772 Btu/lb R = 94.39 Btu/lb

C.O.P.= m3,4,5,6(h5 – h4)/ [m3,4,5,6(h6 – h5)+ m1,8,9 (h9 – h8)]; m1,8,9=1.253 m3,4,5,6

C.O.P.= m3,4,5,6(60.257)/ m3,4,5,6[10.66+1.253(11.42)]=2.414

3.15. An ammonia system is arranged as shown in Fig. 3.24. Condensing temperature is 100 F,

intermediate saturation temperature is 12 F, low-side evaporating temperature is -50 F, t3 = 20 F,

and t7 = 100 F. Saturation states occur as shown in the P-h diagram of Fig. 3.24. Assume

isentropic compression and frictionless flow. Tons capacity of the high-temperature evaporator

is four times that of the low-temperature evaporator. Assuming equal volumetric efficiencies,

calculate the ratio of the piston displacement of the low-pressure compressor to that of the

high-pressure compressor.

State 1: T1 = 100 F P1 = Psat @ 100F = 211.9 psia h1 = hf @ 100F = 155.2 Btu/lb

State 2: P2 = Psat @ 12F = 40.31 psia h2 = h1 = 155.2 Btu/lb

State 3: T3 = 20 F P3 = Psat @ 20F = 211.9 psia h3 = hf @ 20F = 64.7 Btu/lb

State 4: P4 = Psat @ -50F = 7.67 psia h4 = h3 = 64.7 Btu/lb State 5: P5 = P4 = 7.67 psia h5 = hg @ -50F = 593.2 Btu/lb

s5 = sg @ -50F = 1.4787 Btu/lb R v5 = vg @ -50F = 33.08 cu.ft/lb

State 7: T7 = 100F P7 = P2 = 40.31psia

h7 = h @ 40.31psia & 100F =667.9 Btu/lb State 8: P8 = P2 =40.31psia

h8 = hg @ 12F = 615.5 Btu/lb

State 9: h9 = h2 =155.2 Btu/lb State 10: h10 = h8 =615.5 Btu/lb State 11: v11 = vg @ 12F = 6.996 cu.ft/lb

CV at these pipes

m1,12,11=m8 + m9,10-Eqn 1

CV at these pipes

m1,12,11= m9,10 +m2 + m34567-Eqn 2

Adding Eqn 2 and Eqn 1 will result to an equation which is:

m34567=m8-m2-Eqn 3

And because tons capacity of the high-temperature evaporator is four times that of the low-

temperature evaporator we have;

m9,10(h10-h9)=4 m34567(h5-h4) m9,10 =4.593 m34567

CV at the shell and coil intercooler

m34567 h1+m2 h2+ m34567 h7 = m34567 h3+m8 h8Eqn 4

Using Eqn 4 combined with Eqn 3 we get m8 =4.22m2,

Transforming all mass flow rate in terms of m2 using m8 =4.22m2, m34567=m8-m2 and

m1,12,11= m9,10 +m2 + m34567 to help us solve for the ratio of piston displacement.

Ratio of P.D.= m34567v5/ m1,12,11V11

Ratio of P.D.= (m8-m2)v5/( m9,10 +m2 + m34567)v11

Ratio of P.D.= (4.22m2-m2)v5/[(4.593+1)( m8-m2)+ m2]v11

Ratio of P.D.= 3.22 m2 v5/19.015 m2 v11

Ratio of P.D.= 0.801

3.16. A Refrigerant 12 system is arranged as shown in Fig. 3.29. Assume isentropic compression

and frictionless flow. The following data are given: condensing temperature = 90 F; Evaporator A

has a capacity of 5 tons and an evaporating temperature of -20 F; Evaporator B has a capacity of

10 tons and an evaporating temperature of -70 F; vapor leaves each evaporator in dry and

saturated condition; t1 = 84 F, t6 = -5 F, t11 = 20 F, t16 = -5 F, and t17 = -55 F; each compressor has

5 percent clearance.

Draw schematic P-h and T-s diagrams for the cycle and determine (a) the C.O.P. for the system,

(b) the piston displacement, cu ft per min, for each compressor, and (c) the theoretical

horsepower input for each compressor.

Given: Condensing Temp. = 90F Evaporator A = 5 tons of capacity Evaporating temp. A = -20F Evaporator B = 10 tons of capacity Evaporating temp. = -70F Note: Saturated conditions @ evaporator exits T1 = 84F T6 = -5F T11 = 20F T17 = -55F C = 5% for each compressor Required: a.) C.O.P of the system

b.) P.D of each compressors

c.) HP theoritcal for compressor Solutions: State 1: Subcooled P1 = 114. 3 psia T1 = 84F h1 = hf @ 84F = 27.24 Btu/lb State 2 and 8 are equal to state 1: State 5: P5 = Psat @ -20F T5 = 15.28F h5 = hg @ -20F = 75.87 Btu/lb State 6: Superheated P6 = P5 T5 = -5F h6 = 77.8 Btu/lb

C.V Superheater A:

Energy balance: h2 + h5 = h3 + h6 h3 = h2 + h5 – h6 h3 = 25.31 Btu/lb Then, at P3 = 114.3 psia of h3 = 25.31 Btu/lb, T3 = 76F Then, state 4 would be: P4 = 15.28 psia h4 = h3 = 25.31 Btu/lb T4 = -20F Then, m2 = m3 = m4 = m5 = m6

C.V Evaporator A:

QA = 5 tons So, m4 = 5(200)/ (75.87-25.31)

m4 = 19.78 lb/min

C.V Superheater B:

State 13: P13 = 3.98 psia h13 = hg @ 3.98 psia = 69.95 Btu/lbm State 17: P17 = 3.98 psia T17 = -55F (superheated) h17 = 71. 774 Btu/lb v17 = 8. 273 cu ft/lb CV of a proposed system

Mass Balance: m8 = m16 + m17 Energy Balance: m8h8 = m16h16 + m17h17 - QB

Liquid Cooler:

State 11: Subcooled P11 = 114.3 psia T11 = 20F h11 = hf @ 20F = 12.55 Btu/lb

C.V Evaporator B:

Qb = 10 tons m12 = m13 = 10(200)/ (69.95-12.55) = 34. 843 lbm/ min. Note: m11 = m12 = m13 = m17 = m18 Energy Balance: m8h8 + Qb = m16h16 + m17h17 27.24 m8 – 77.8 m16 = 500.82 >> Equation 1: Mass Balance: m8 = m16 + m17

m8 – m16 = 34.843 >>Equation 2: Using calculator: m8 = 43.7097 lb/min m16 = 8.87 lb/min

For State 7: m6h6 + m16h16 = m7h7 m7 = m6 + m16 = 28.65 lb/min 19.78(77.8) + 77.8(8.87) = m7h7 h7 = 77.8 Btu/lb State 18: P18 = 15. 28 psia s18 = 17 0.1846 Btu/lb R h18 = 81.3 Btu/lbm t18 ≈ 20F State 19: P19 = P18 = 15.28 psia

C.V for these pipes

Energy Balance:

m7h7 + m18h18 = m19h19 h19 = 79.72 Btu/lb Remark: Superheated Then, v19 = 2.6544 cu ft/lb s19 = 0. 18115 Btu/lb

State 20: P20 = 114.3 psia s20 = s19 = 0. 18115 Btu/lb R T20 =142. 97F h20 = 96. 355 Btu/lb Answer for the questions required: a.) C.V. Condenser Qcond = m1 (h20 – h1) = (34.843 + 28.65) (96.355 – 27.24) Qcond = 4388.32 Btu/min C.O.P = ql/qh-ql = (10 + 5) tons (200 Btu/min.ton)/ (4388.32 -3000) C.O.P = 2.16 b.) P.D =? Compressor A: m19 = 63.493 lb/min v19 = 2.6544 ft3/lb ηcva = *1.05 – 0.05 (114.3/15.28)1/1.13] ηcva = 0.7533 Then, P.DA = (63.493) (2.6544)/0.7533 P.DA = 224 cu ft/min Compressor B: m17 = 34.843 lb/min v17 = 8.273 cu ft/min ηcvb = *1.05 – 0.05 (15.28/3.98)1/1.13] ηcvb = 0.89 Then, P.DB = (34.843) (8.273)/0.89 P.DB = 324 cu ft/min

For c.) H.P

19W20 = (63.493) (96.355 – 79.72)/42.4 Compressor A = 24.9 hp

17W18 = (34.843) (81.3 – 71.774)/42. 4 Compressor B = 7.83 hp

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Threlkeld chap 3 Answers