3/63 A 2‐kg sphere S is being moved in a vertical plane by a robotic arm. When the angle θ is 30°, the angular velocity of the a
rm about a horizontal axis through O is 50 deg/s clockwise and its angular acceleration is 200 deg/s2 counterclockwise. In addition, the hydraulic element is being shortened at the constant rate of 500 mm/s. Determine the necessary minimum gripping force P if the coefficient of static friction between the sphere and the gripping surfaces is 0.50. Compare P with the minimum gripping force Ps required to hold the sphere in static equilibrium in the 30° position.
1 answer:
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Answer:
Explanation:
Please check the below file for the attached file
Answer:
A) condenser pressure = 9.73 kPa,
B) 10242 kw
C) 36.9%
Explanation:
given data
entrance pressure of steam = 12.5 MPa
temperature of steam = 550⁰c
flow rate of steam = 7.7 kg/s
outer pressure = 2 MPa
reheated steam temperature = 450⁰c
isentropic efficiency of turbine( nt ) = 85% = 0.85
isentropic efficiency of pump = 90% = 0.90
From steam tables
at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg, S3 = 6.6317 kJ/kgK
also for an Isentropic expansion
S4s = S3 .
therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa
h4s = 2948.1 kJ/kg
nt = 0.85
nt (0.85) = =
making h4 subject of the equation
h4 = 3476.5 - 0.85 (3476.5 - 2948.1)
h4 = 3027.3 kj/kg
at P5 = 2 MPa and T5 = 450⁰c
h5 = 3358.2 kj/kg, s5 = 7.2815 kj/kgk
at P6 , x6 = 0.95 and s5 = s6
using nt = 0.85 we can calculate for h6 and h6s
from the chart attached below we can see that
p6 = 9.73 kPa, h6 = 2463.3 kj/kg
B) the net power output
solution is attached below
c) thermal efficiency
thermal efficiency = 1 - = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%
