Thermodynamics, Combustion, and Compressor Analysis Homework Solution

This document presents solutions to a mechanical engineering homework assignment. The first part addresses heat transfer calculations involving hot oil and cold water, determining the arithmetic mean temperature difference (AMTD), logarithmic temperature difference (LMTD), overall heat transfer rate, and mass flow rates. The second part focuses on a mass balance for the stoichiometric combustion of ethanol, calculating the mass of oxygen required, the mass of air and nitrogen, the air/fuel ratio, and the mass of carbon dioxide and water produced. The third part involves calculating the mass flow rate of ethanol. The final part analyzes a reciprocating compressor, including a pressure-volume graph, calculations for indicated power, volumetric efficiency, mass flow rate, isothermal efficiency, and the actual power required considering mechanical efficiency.

THERMODYNAMIC SOLUTION
Answer-1
Hot oil
th 1=200 ° C`
th 2=80 °C
Cold water
tc1 =18° C
tc2 =70° C
a) Arithmetic mean temperature difference (AMTD)
AMTD= th 1 +th 2
2 − tc1 +tc 2
2
AMTD= 200+80
2 − 70+18
2
AMTD=140−44
AMTD=96 ° C
b) Logarithmic temperature difference (LMTD or θm)
LMTD= θ1−¿ θ2
ln θ1
θ2
¿
θ1=th 1−tc 2=200−70=130 ° C
θ2=th 2−tc 1=80−18=62 °C
LMTD= 130−62
ln 130
62
LMTD=91.84 ° C
c) Overall heat transfer rate
200°C
18°C
80°C
70°C
Temperature Diagram

Q=nUA θm
U= Overall heat transfer coefficient
=75 W/m2K
A=πDL
= 0.141 m^2
Q=80∗75∗0.141∗91.84
Q=77.7 KW
d) Mass flow rate of oil
Q=mo co(th 1−th2 ¿
77.7=mo∗2.19∗(200−80)
mo=0.296 Kg
s
Mass flow rate of water
Q=mw cw(tc2 −tc 1 ¿
77.7=mw∗4.19∗(70−18)
mw=0.357 Kg
s
Answer -2

C2H5OH + 3O2 → 2CO2 + 3H2O
If 2.5kg of Ethanol is burnt in air stoichiometrically (i.e. to completion),
perform a mass balance to determine:
i. The mass of Oxygen(O2) required
Molecule wt of ethanol =46 grms
For 2.5 Kg of ethanol , No. of moles = 2500/46 = 55
For 1 mole of ethanol = 3 moles of O2
So No. of moles of O2 required = 55 * 3 =165
Molecule wt of O2 =32 grms
So For 2.5 Kg of ethanol , O2 required = 165 *32/1000
So For 2.5 Kg of ethanol , O2 required = 5.28 Kg
ii. The mass of Air (23% O2& 77% N2) required
As 23 % of O2 = 5.28 * 23 /100 = 1.214 Kg
As 77 % of N2 = 5.28 * 77 /100 = 4.06 Kg
iii. The mass of Nitrogen (N2) required
Molecule wt of N2 =28 grms
N2 required = 165 *28/1000
N2 required = 4.62 Kg
iv. The Air/Fuel ratio

A /F = 5.28 / 2.5 =2.112
v. The mass of Carbon Dioxide (CO2) produced by the combustion
Molecule wt of CO2 =44 grms
For 1 mole of ethanol = 2 moles of CO2
= 2 * 55 *44/1000
CO2 produced = 4.84 Kg
vi. The mass of Water (H2O) produced by the combustion
Molecule wt of H2O =18 grms
For 1 mole of ethanol = 3 moles of H2O
= 3 * 55 *18/1000
H2O produced = 2.97 Kg
b) mass flow rate in kg/h of Ethanol (m)
Q=mneff E
350 = m * 0.73 *29 .7 1000
m = 58.11 Kg /hr

Answer-4
(a) Sketch a Pressure (p) vs Volume (v) graph for the reciprocating compressor with full
labelling.
(a) the (theoretical) indicated power;
IP= pm LAN
60
pm= n P1
n−1∗
{( P2
P1 )n−1
n −1 }
n =1.3
P1 = 101.3 Kpa
P2 = 8.5bar
Put all values we get
pm= 1.3∗101.3
1.3−1 ∗
{( 850
101.3 ) 1.3−1
1.3 −1 }
pm=276.79 Kp a
L*A =Vs =2.5 L
N = 1200 rpm
1
23
4
stroke
P
V
Clearance
volume

Put all values we get
IP=276.79∗2.5∗1200
60∗1000
IP=13.8 KW
(b) the (theoretical) volumetric efficiency;
eff vol =1+ c−c {( P2
P1 )1
n
}
c= Vc
Vs =0.04
eff vol =1+ 0.4−0.4∗
{( 850
101.3 ) 1
1.3
}
eff vol =83.5 %
(c) the (theoretical) mass flow rate/ air delivery
pv=mRT
101.3∗2.5
1000 =m∗0.287∗(273+20)
M = 0.025 Kg /s
(d) the (theoretical) isothermal efficiency;

eff iso=
P1 V 1∗
{(ln P2
P1 ) }
n P1 V 1
n−1 ∗
{( P2
P1 )n−1
n −1 }
Put all values we get
eff iso=
101.3∗2.6
1000 ∗{(ln 850
101.3 ) }
1.3∗101.3∗2.6
( 1.3−1 )∗1000 ∗ {( 850
101.3 )1.3−1
1.3 −1 }
eff iso= 0.56
0.7196
eff iso=77.85 %
(e) the actual power required to drive the compressor, assuming that the actual indicated
power is 15% greater than the theoretical value, and the mechanical efficiency is 80%.
eff mech= BP
IP
0.8= BP
15.9
BP = 12.72 kW