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Anarel [89]
3 years ago
5

You are to assess the biomechanics of a male’s arm using his bicep to hold a 20 kg object in his hand. The upper arm is perpendi

cular to the ground while the forearm is parallel to the ground. The male’s forearm is 26.5 cm long and weighs 1.42 kg. The man’s hand weighs 0.54 kg and is 19 cm long. The center of masses for the forearm and hand are 43% and 50% of the extremity lengths from the proximal joint, respectively.
You may assume that only the bicep generates the force, the angle between the bicep and forearm is 75 degrees, and the bicep insertion is 5cm from the elbow joint.

a) Draw a complete free body diagram of the above described activity. (5 pts)

b) Calculate the elbow moment due to external load and body segments weight. (5 pts)

c) Calculate the elbow contact force and force exerted by the bicep. (5 pts)

Engineering
1 answer:
cestrela7 [59]3 years ago
5 0

Answer:

Explanation:

The detailed analysis, with free body diagram and step by step calculations with appropriate substitution is as shown in the attached files.

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Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/
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Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

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Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

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\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

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