Answer:
heat rate= 1281W
length = 15.8m
Explanation:
we have this data to answer this question with
Tmi = 85 degrees
Tmo = 35 degrees
Ts = 25 dgrees
flow rate = 25 degrees
using engine oil property from table a-5
Tm = Tmo - TMi/2 = 333k
u =0.522x10⁻²
k = 0.26
pr = 51.3
cp = 2562 J/kg.k
mcp(Tmo-Tmi) =
0.01 x 2562(35-85)
= 1281 W
we find the change in Tim
= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]
= -50/ln0.167
= -50/-1.78976
= 27.9°c
we finf the required reynold number
4x0.01/πx0.003x0.522x10⁻²
= 0.04/0.00004921
= 812.8
= 813
we find approximate correlation
NuD = hd/k
NuD = 3.66
3.66 = 0.003D/0.26
cross multiply
0.003D = 3.66x0.26
D = 3.66x0.26/0.003
= 317.2
As = 1281/317x27.9
= 0.145
As = πDL
L = As/πD
= 0.145/π0.003
= 0.145/0.009429
L = 15.378
Answer:
(a) The mass of an object remains the same always. It is independent of its location. In this case, an iron nail is ground into powder.
Explanation:
please mark this answer as brainliest
Answer:
691.6 torr
Explanation:
Given data:
Initial temperature = 273 K
Initial pressure = 1 atm
Final temperature = -25 °C (-25 + 273 = 248 k)
Final pressure = ?
Solution:
P₁/T₁ = P₂/T₂
P₁ = Initial pressure
T₁ = Initial temperature
P₂ = Final pressure
T₂ = Final temperature
Now we will put the values in formula.
P₁/T₁ = P₂/T₂
P₂ = P₁T₂/T₁
P₂ = 1 atm × 248 K / 273 k
P₂ =248 atm. K / 273 K
P₂ = 0.91 atm
In torr:
0.91 × 760 = 691.6 torr
A gas can be treated as ideal gas when it is at higher temperatures or low pressures relative to its critical temperature and pressure. Because, at high temperatures and low pressures, the density of gas decreases and at low densities real gases behave as ideal gases.
I would tell you... but there is a lot of information... just type in online what you asked, and you will get lots of info.
