Answer:
We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.
Explanation:
Answer:
The correct answer is "1341.288 W/m".
Explanation:
Given that:
T₁ = 300 K
T₂ = 500 K
Diameter,
d = 0.2 m
Length,
l = 1 m
As we know,
The shape factor will be:
⇒ ![SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}](https://tex.z-dn.net/?f=SF%3D%5Cfrac%7B2%20%5Cpi%20l%7D%7Bln%5B%5Cfrac%7B1.08%20b%20%7D%7Bd%7D%20%5D%7D)
By putting the value, we get
⇒ ![=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%20%5Cpi%20l%7D%7Bln%5B%5Cfrac%7B1.08%5Ctimes%201%7D%7B0.2%7D%20%5D%7D)
⇒ 
hence,
The heat loss will be:
⇒ 



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
Answer:
Outside temperature =88.03°C
Explanation:
Conductivity of air-soil from standard table
K=0.60 W/m-k
To find temperature we need to balance energy
Heat generation=Heat dissipation
Now find the value
We know that for sphere

Given that q=500 W
so

By solving that equation we get
=88.03°C
So outside temperature =88.03°C