Answer:
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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



Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
A) 5 MPa , 55 MPa
B) maximum stress = 55 MPa, maximum shear stress = 25 MPa
Explanation:
using the given Data
free surface of a solid body
α
= 50 MPa, α
= 10 MPa , t
= -15 MPa
attached below is the detailed solution to the question