It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of
AB. Determine the moment of this force about C for the two casesshown.
1 answer:
Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
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Answer:
a) the amount of energy produced in kJ/K is 0.73145 kJ/K
b) the amount of energy produced in kJ/K is 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.
Explanation:
Draw the T-s diagram.
a)
= 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280
R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar
Δs =
Substitute all parameters in the equation
Δs =
Δs = 5 kg × 0.14629 kJ/kg.K
= 0.73145 kJ/K
b)
Δs =
Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K
T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K
R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar
Δs =
= 5 kg (0.13795 kJ/kg.K)
= 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.
Answer:
Q=0.000604 m³/s
Explanation:
Given that
d₁=5 cm
d₂=1 cm
P= 30 KPa
Density of water ,ρ=1000 kg/m³
As we know that volume flow rate Q given as
A₁=0.0019 m²
A₂=0.000078 m²
Q=0.000604 m³/s