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VAAL UNIVERSITY OF TECHNOLOGY
FACULTY OF ENGINEERING
DEPARTEMENT: MECHANICAL
NATIONAL DIPLOMA: ENGINEERING: MECHANICAL
SUBJECT:
DATE:
DURATION:
EXAMINER:
THERMODYNAMICS III
NOV 2007
3 HOURS
DW SPIRET
EMTGA3-MAIN
MODERATOR: PROF JH WICHERS
REQUIREMENTS:
Steam Tables
INSTRUCTIONS:
Assume STP as 101.3 kPa and 15°C unless given
Answer all the questions as instructed
TOTAL: 107
FULL MARKS 100
THIS QUESTION PAPER CONSISTS OF: 1 Front Page
2 Typed pages
3 Formulae sheets
(Properties of Water and Steam)
DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED
VAAL UNIVERSITY OF TECHNOLOGY EMTGA3 2007 NOV (MAIN)
QUESTION 1
A single-stage, double-acting air compressor is required to deliver 15.6m^3 /min of air at 103 kPa and 20°C. The delivery pressure is 800 kPa and the compressor speed is 320 rev/min. Assume the clearance volume is 5.5% of the swept volume with the law of compression of PV1?32=C. Assume the induction pressure and temperature is 90 kPa and 32°C
Draw the P-V diagram for the cycle. (2)
Determine:
- The volumetric efficiency. (7)
- The swept volume of the cylinder. (5)
- The power required to drive the compressor. (2)
- The Isothermal efficiency. (2)
- The heat rejected by the cylinder. (4) [22]
QUESTION 2
The following measurements were taken from a refrigeration system.
Refrigerant type Condensing pressure Evaporating pressure Temperature after compression Temperature at condenser outlet Compressor speed Swept volume Volumetric efficiency Temperature rise of condenser water
R
9.607 bar 2.610 bar 50°C 30°C 1450 rev/min 60 cmVstroke 90% 5°C
Assume the refrigerant enters the compressor dry saturated.
Complete the following:
- Draw the line, T-S and P-H diagram for this system. (5)
- Find the enthalpy at entrance to each component. (5)
- Show that the cycle obeys the First Law of Thermodynamics. (4)
- Calculate the C.O.P. of the system. (1)
- Determine the mass flow of refrigerant. (4)
- Determine the power requirement of the compressor. (1)
- Determine the mass flow of cooling water through the condenser if the temperature rise of the water is restricted to 5°C. (2)
- If the refrigeration unit runs continuously for 24 hours determine how much ice can be manufactured per 24 hours if the latent heat of fusion for ice is 336 kJ/kg.K and water is supplied to the evaporator at 12°C. (4) [26]
QUESTION 3
A four-stroke, single-cylinder engine has a cylinder diameter of 180 mm and a stroke of 340 mm. During a trail 38 kg/min of cooling water was circulated through the jacket, with inlet and outlet temperatures being 17.5 and 59°C respectively. The indicated mean effective pressure was 555 kPa at a speed of 900 rev/min, while a nett brake load of 1900 N was measured at a radius of 0. m. The calorific value of the fuel was 43 MJ and fuel consumption was 0.6 kg/min. Assume an air/fuel ratio of 14:1 and the atmospheric conditions are 85 kPa and 20°C. Exhaust gas temperature is 350°C and cpg=1.15 kJ/kg.K
Determine:
- The mechanical efficiency. (8)
- The volumetric efficiency. (3)
- The energy rejected to radiation. (8)
- The brake thermal efficiency. (1)
- The specific fuel consumption. (1) [21]
THERMODYNAMICS
General formulae
Q+W =AU m [ Z,g + Vz C,^2 + h, ] +Q + W = m [ Z 2 g + '/ 2 C 22 + h 2 ] h = u + pv m = CA/v
Vapour Model
Sfg = Sg - Sf V = Vf + %Vfg » X- hfiuid« c T
(a 2 -a,)/(a 3 -a 1 ) = (b 2 -b,)/(b 3 -bi)
Ideal Gas Model h = cpT u = cvT Y = Cp/cv cv=R/(X-l) R = cp - cv
pV = mRT m = n.m 0
= 8.3145kJ.kmor'.K'^1
Expansion law V=C PV=C PVn=C P=C
p,/T,=p 2 /T L PlV!=P2V 2
P^fV/V,]" T,/T2=[V 2 A^ 1 ]"-^1 T 1 /T2=p,/p 2 /n
Process Isochoric Isobaric Isothermal Isentropic Polytropic
Law V=C p=C pV=C pV=C pVn=C
AU
w —
AH
Gas Process AS cv.ln (T 2 /T,) cp.ln (T 2 /T,) Q/T = R.ln (V 2 /V,) 0 R.ln (V 2 /V,)+ c.ln (T 2 /T,)
W
-P(V 2 -V,) -C.ln(p,/p 2 ) -C[V 2 l"Y-V 1 '"y]/(y-l) (P2V 2 -p,V,)/(n-l)
Q
AU
AH
-W
W[(n-Y)/(y-l)]
Vapour Process Process Isochoric Isobaric Isothermal Isentropic Polytropic
Other
Law V=C p=C T=C AS=Q- pVn=C pV=C None
AU AH
Tables Tables Tables Tables Tables
Tables
AS
Tables Tables Q/T 0 Tables
Tables
W
p(VrV 2 ) AU-Q AU (p 2 V 2 -p,V 1 )/(n-l) -pV.ln(pi/p 2 ) AU-Q
Q
AU
AH
TAS
-W + AU
-W + AU
Steam plant
oikr - msxAh / mfxC.V. msxAhtulb / mfxC.V. E.E. = ms/mf
E.E from & @ 100°C = E.E.xAh/ S.F.C = mf x3600 / (msxAh) S.S.C. = ms x3600 / (msxAh) Atbleed = (tb-tc)/(feed heaters + 1)
Refrigeration C.O.P= Q AV
Reciprocating Compressors
W = nmR(T 2 -T,)/(n-l)
rP = (Pi/P 2 )1/z Wiso = mRTln(P 2 /Pi) rivoi = Vi/vs= l+(V(/Vs)x(l-rp1/n
Standard cycles
Tlcamot^ 1—T1/T
p Efficiency Ratio = thermal r]/Air standard _r_
Internal combustion engines FP=IP-BP
BP/(mfxCV) Tjn- = IP/(mfxCV) BMEP = BP/vs IMEP = IP/vs
General •Hcycle = =^ W (^) t u r b / Q (^) i n Work Ratio = net work/gross work Tllsentropic =^ ( h l - h 2 ) / (hi~h 2 ') S.F.C. = m (^) f x 3 6 0 0 / B P rivoi = v;/vs V = cdAV(2000gh) vsup » 0.23(hsup - 1943)/px CpS«1.88kJ/kg.K
Turbines/Compressors W=£CbCw F=mAC
(steam) (steam)
mve=(n/p)x(psinpe-t)lCre
aj) = blade speed ratio =diagram power/ kinetic power
^REACTION ~ cal
2cosor; -
Cai (^) [C cos a,.
^IMPULSE ~ 4| ^ ' ai (^) cai
Maximum Efficiency Conditions: Cb/Cai =coscci (reaction) Ch/Cai =0.5cosaj (impulse) Cb/Cai = 0.25coscti (velocity, axial exit) Tldiag =^ 2(COS Cti)/(1+ COS (reaction) (impulse & velocity; axial exit)
work factor = actual power input/diagram power Slip factor = C'we/Cwe A=Ahbiades/Ahstage Aturb=Cf(tanPe-tanPi)/2Cb Acom=Cf(tanpe+tanPi)/2Cb ACw=Ci(tanpi-tanPe)
Vortex blading Cf = constant CbACw = constant C^r = constant
T + (y-)CbACJ
: + l
r-i
Nozzles ho = h + 0.5C^2 p 0 = p+0.5pC^2 pc/p 0 = Cc = a = VyRT = Vkpv Ma = C/a
Tc/T 0 =
p A. J
V|/=CO/(Og
h (^) r h 2 = {k/(k-l) x (piv,-p 2 v 2 )} Piv,k^ = p 2 v 2 k ka=1.
Psychrometry co =0.622ps/(p-ps) vj/=100x(p-pg)/(p-ps) h = ha + cohs = cpmat cpma =c
hs = hsup» (hg at ps) + cps(t-tg at ps) «C+cpst whg at t
C«2500 kJ/kg.K(steam) Room ratio line = sensible heat load / total heat load
