Q6: Design a cast iron piston for a single acting four stroke engine for the following data: Maximum gas = 125 mm ;; Indicated mean effective pressure = 5 MPa; Cylinder bore (D) = 100 mm; Stroke (L) pressure (pm) = 0.75 MPa; Mechanical efficiency (nm) = 80%; Fuel consumption = 41.7× 10 kg/brake power/ Second; cast iron stress o 38 MPa; Higher calorific value of fuel (HCV) = 42 x 10³ kJ/kg ; Speed (N)= 2000 r.p.m; k= 46.6 W/m/°C; Tc - TE = 220 C; t1-3.4 mm; C is 0.05. th= 3pD2 Indicated power: IP = 16 % Pm LAn 60 R = P₂DI H th= 12.56 k (Tc-TE) H = C X HCV xm x BP BP = IP X1 R = μ TD² 4

Elements Of Electromagnetics
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CH
125 mm;; Indicated mean effective
Q6: Design a cast iron piston for a single acting four stroke engine for the following data: Maximum gas
pressure = 5 MPa; Cylinder bore (D) = 100 mm; Stroke (L)
pressure (pm) = 0.75 MPa; Mechanical efficiency (nm) = 80%; Fuel consumption = 41.7x 106 kg/brake
power/ Second; cast iron stress of 38 MPa; Higher calorific value of fuel (HCV) = 42 x 10³ kJ/kg ;
Speed (N)= 2000 r.p.m; k= 46.6 W/m/°C; Tc - TE = 220 C; t1=3.4 mm; C is 0.05.
=
Design of Clutch and spring
T = n.u.W.R
T = W.r
P x 60
Σπη
T =
C
3pD2 Indicated power: IP =
16 0
p.r = C(a constant) or p = C/r
TL
16
A = 2nRb
W
2n (1₁-1₂)
T =
τα3
= C/B
Pm L An
60
R = PbDl H = C X HCV xm x BP
H
12.56 k (Tc-TE)
4C - 1
4C-4
K =
+
0.615
C
8 WD³ n
P =
b=1₁-12
Do = D + d D₁ = D - d
Lend + 8 + 0.158
W = p. A
LF
ń - 1
R = (₁ + r₂)/2
Type of end
Plain ends
Ground ends
BP = IP X1 R = μ
T = KX
8 WD]
nd³
Total number
of turns n
n
n
n+?
C =
D
TD²
Solid
length
(n+1)d
nxd
(n+3)d
Transcribed Image Text:CH 125 mm;; Indicated mean effective Q6: Design a cast iron piston for a single acting four stroke engine for the following data: Maximum gas pressure = 5 MPa; Cylinder bore (D) = 100 mm; Stroke (L) pressure (pm) = 0.75 MPa; Mechanical efficiency (nm) = 80%; Fuel consumption = 41.7x 106 kg/brake power/ Second; cast iron stress of 38 MPa; Higher calorific value of fuel (HCV) = 42 x 10³ kJ/kg ; Speed (N)= 2000 r.p.m; k= 46.6 W/m/°C; Tc - TE = 220 C; t1=3.4 mm; C is 0.05. = Design of Clutch and spring T = n.u.W.R T = W.r P x 60 Σπη T = C 3pD2 Indicated power: IP = 16 0 p.r = C(a constant) or p = C/r TL 16 A = 2nRb W 2n (1₁-1₂) T = τα3 = C/B Pm L An 60 R = PbDl H = C X HCV xm x BP H 12.56 k (Tc-TE) 4C - 1 4C-4 K = + 0.615 C 8 WD³ n P = b=1₁-12 Do = D + d D₁ = D - d Lend + 8 + 0.158 W = p. A LF ń - 1 R = (₁ + r₂)/2 Type of end Plain ends Ground ends BP = IP X1 R = μ T = KX 8 WD] nd³ Total number of turns n n n n+? C = D TD² Solid length (n+1)d nxd (n+3)d
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