Theory of Machines Test 4

Result

Theory of Machines Test 4

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Question 1
1 / -0

In a horizontal reciprocating engine, 4 kg of mass is to be balanced which is located at radius of 10 cm. If 50% of reciprocating unbalance is to be balanced by mass rotating at radius of 8 cm. What is value of rotating mass?

A
5 kg

B
2.5 kg

C
10 kg

D
1.25 kg

Solution

Concept:
Bb = cmr
Where, B = rotating mass (kg), b = radius of rotating mass, c = fraction of reciprocating mass to be balanced
m = Reciprocating mass, r = radius of reciprocating mass
Calculation:
Bb = cmr
B × 8 = 0.5 × 4 × 10
∴ B = 2.5 kg
Question 2
1 / -0

A engine is running at 800 rpm. The magnitude of maximum primary forces is observed to be 500 N. If crank and connecting rod length are found to be 200 mm and 800 mm respectively. What is magnitude of maximum secondary force (N)?

A
125-125

B
1254-1251

C
1257-1256

D
1259-1252

Solution

Concept:
Primary force (F_p) = mrω² cos θ
\({\rm{Secondary\;force\;}}\left( {{F_s}} \right) = \frac{{mr{\omega ^2}\cos 2\;\theta }}{n}\)
Calculation:
\({\left( {{F_p}} \right)_{max}} = mr{\omega ^2} = 500\;N\)
\({\left( {{F_s}} \right)_{max}} = \frac{{mr{\omega ^2}}}{n}\)
\(n = \frac{\ell }{r} = \frac{{800}}{{200}} = 4\)
\(\therefore {\left( {{F_s}} \right)_{max}} = \frac{{500}}{4}\)
∴ (F_s)_max= 125 N
Question 3
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Consider a 2-cylinder locomotive engine having cranks perpendicular to each other. The plane gap between the cylinders is 0.7 m and radius of the crank is 0.3 m. The total mass of reciprocating part is 80 kg. If (2/3) of the reciprocating mass is to be balanced, then what is value of the swaying couple (in kN.m) if the crank angle is 60°. The engine speed is 300 rpm.

A
3.6-3.9

B
3.64-3.91

C
3.67-3.96

D
3.69-3.92
Solution
Concept:
- Let c be the fraction of the reciprocating mass is to be balanced.
- a be the plane gap between the cylinders
- θ be the crank angle, r be the crank radius
- m be the mass of the reciprocating parts, ω be the angular velocity of the crank
Then Swaying couple is given by ‘S’
\(s = \frac{a}{2}\left( {1 - c} \right)mr{\omega ^2}\left( {\cos \theta + \sin \theta } \right)\)
Calculation:
Given: a = 0.7 m, r = 0.3 m, m = 80 kg, c = 2/3, θ = 60°, N = 300 rpm
ω = 2π N/60 = 2π × 300/60 = 31.41 rad/sec
\(s = \frac{a}{2}\left( {1 - c} \right)mr{\omega ^2}\left( {\cos \theta + \sin \theta } \right)\)
\(s = \frac{{0.7}}{2}\left( {1 - \frac{2}{3}} \right) \times 80 \times 0.3 \times {\left( {31.41} \right)^2}\left( {\cos 60 + \sin 60} \right)\)
s = 3773.57 Nm = 3.77 kN.m
Question 4
1 / -0

A horizontal engine has a crank pin radius of 300 mm. The mass of reciprocating parts is 250 kg and the connecting rod length is 1.2 m. If the cylinder bore is 0.5 m and the difference between driving and back pressures is 0.45 N/mm²(when crank has travelled 63⁰ from IDC) and the engine runs at 260 rpm. Then the piston effort is __________ kN

A
70-72

B
704-721

C
707-726

D
709-722

Solution

Explanation:
Given:
r = 300 mm = 0.3, m_R = 250 kg, θ = 63°, P₁ – P₂ = 0.45 N/mm², L = 1.2 m, D = 0.5 m, N = 260 rpm
Now,
\(\omega = \frac{{2\pi N}}{{60}} = 27.23\;rad/s\)
Now,
Load on piston:
\({F_L} = \left( {{P_1} - {P_2}} \right)\frac{\pi }{4}{D^2}\)
\({F_L} = 0.45 \times \frac{\pi }{4} \times {\left( {500} \right)^2}\)
∴ F_L= 88357.29 N
Now,
\(n = \frac{L}{r} = \frac{{1.2}}{{0.3}} = 4\)
\({\rm{Inertia\;force\;}}\left( {{F_I}} \right) = m{\omega ^2}r\left( {\cos \theta + \cos \frac{{2\theta }}{n}} \right)\)
∴ Inertia force (F_I) = 17074.87 N
∴ Piston effort = 88357.29 – 17074.87
∴ F_P = 71.282 kN