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) 24 kg
b) 32 kg
Explanation:
The gauge pressure is of the gas is equal to the weight of the piston divided by its area:
p = P / A
p = m * g / (π/4 * d^2)
Rearranging
p * (π/4 * d^2) = m * g
m = p * (π/4 * d^2) / g
m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg
After the weight is added the gauge pressure is 2.8kPa
The mass of piston plus addded weight is
m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg
56 - 24 = 32 kg
The mass of the added weight is 32 kg.
Answer:
cpct gvxjjxjhdfjokjdzfjiyddzzsjhxf
Answer:
sum2 = 0
counter = 0
lst = [65, 78, 21, 33]
while counter < len(lst):
sum2 = sum2 + lst[counter]
counter += 1
Explanation:
The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>
While there are numbers in the <em>lst</em>, loop through <em>lst</em>
Add the numbers in <em>lst</em> to the sum2
Increment <em>counter</em> by 1 after each iteration
