Answer:
Explanation:
The image attached to the question is shown in the first diagram below.
From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.
IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :
From the diagram; we can determine the length of BC by using pyhtagoras theorem;
SO;
The cross -sectional of the cable is calculated by the formula :
where d = 4mm
A = 1.26 × 10⁻⁵ m²
However, looking at the maximum deflection in length 
; we can calculate for the force 
by using the formula:
where ;
E = modulus elasticity
= length of the cable
Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for and 0.006 m for ; we have:
---- (1)
Similarly; we can determine the force 
using the allowable maximum stress; we have the following relation,
where;
maximum allowable stress
Replacing 190 × 10⁶ Pa for 
; we have :
------ (2)
Comparing (1) and (2)
The magnitude of the force 
since the elongation of the cable should not exceed 6mm
Finally applying the moment equilibrium condition about point A
P = 1.9937 kN
Hence; the maximum load P that can be applied is 1.9937 kN
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
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Answer:
Explanation:
Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)
fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)
vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)
BFFS = 50+5, BFFS =55 (given) fLW= 6.6
TLC=6+3=9 fLC= 0.65
fM= 0.0
fA= 1.0
FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)
Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)
After: fA= 3.0
FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)
Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln
