Answer:
1 The weight of the foundation block should be enough to withstand vibrations
2 The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.
3.The base should be sufficiently rigid to have the requisite rigidity
4 a distance should be created all across machine foundation to separate it from the adjacent parts of the building
Explanation:
1. The weight of the foundation block should be enough to withstand vibrations and to avoid friction between device and the surrounding soil as well. This can be done by increasing the base block weight in supporting with engine power
2.The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.
3.The base should be sufficiently rigid to have the requisite rigidity as the slightest misdirection of foundation may cause significant bearing disorders.
4.To avoid propagation of vibration from a device to the adjacent parts of the building, a distance should be created all across machine foundation to separate it from the adjacent parts of the building.
Answer:
R= 53.7 Ω
Explanation:
fc= 1/(2πRC)
3000=1/(2π× 1 × 10^-6 × R)
R= 53.7 Ω
Answer:
I think it is 10 percent because
Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
Answer:
POWER INPUT = 82.989 KW
Explanation:
For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg
For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg
Heat loss Q = 0.05w
Inlet diameter = 3 cm
exit diamter = 1.5 cm
volume of tank will be v = area * velocity
velocity at inlet
velocity at outlet
steady flow energy equation
solving wc = 1830.64 kJ/kg
wc in KWH
we know that