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frosja888 [35]
3 years ago
11

If link AB of the four-bar linkage has a constant counterclockwise angular velocity of 58 rad/s during an interval which include

s the instant represented, determine the angular acceleration of AO (positive if counterclockwise, negative if clockwise) and the acceleration of point D.
Engineering
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:Vb=-6i-(-0.1ωab+8)j m/s

Explanation:

Va=V0+Va0

Va=V0+(ra0 x ωao)

ω=Angular velocity of link A0

Using r0a=0.1m;

Va=V0+(0.1i x ω0a K)

Va=0

ixk=j

Va=0+0.1ω0aj

Calculating te velocity of using te equation below

Vb=Va+Vba

Vb=Va+ωab x rba

ωab=40rad/s

rab=-0.21i+0.15j

Va=0.1ω0aj

Vb=Va+ωabxrba

Vb=0.1ω0aj+40k x -(0.21i+0.15j)

Vb=0.1ω0aj-8j-6i

Vb=-6i-(-0.1ωab+8)j m/s

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A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
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flow(m) = 7.941 lbm/s

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Explanation:

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- T_6 = 940 = 1400 R

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- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

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3 years ago
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Trava [24]

Answer:

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Explanation:

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