Answer:
Explanation:
Given:
The two rods could be approximated as a fins of infinite length.
TA = 75 0C θA = (TA - T∞) = 75 - 25 = 50 0C
TB = 55 0C θB = (TB - T∞) = 55 - 25 = 30 0C
Tb = 100 0C θb = (Tb - T∞) = (100 - 25) = 75 0C
KA = 200 W/m · K
T∞ = 25 0C
Solution:
The temperature distribution for the infinite fins are given by
θ/θb=e⁻mx
θA/θb= e-√(hp/A.kA) x1 ....................(1)
θB/θb = e-√(hp/A.kB) x1.......................(2)
Taking natural log on both sides we get,
Ln(θA/θb) = -√(hp/A.kA) x1 ...................(3)
Ln(θB/θb) = -√(hp/A.kB) x1 .....................(4)
Dicving (3) and (4) we get
[ Ln(θA/θb) /Ln(θB/θb)] = √(KB/KA)
[ Ln(50/75) /Ln(30/75)] = √(KB/200)
Answer:
Option B
1025 psi
Explanation:
In a single shear, the shear area is
The shear strength=
and in this case
Shear strength=
hence making load the subject then
Load=Shear area X Shear strength
Load=
Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg