Answer:
Explanation:
To solve this problem we use the expression for the temperature film
Then, we have to compute the Reynolds number
Re<5*10^{5}, hence, this case if about a laminar flow.
Then, we compute the Nusselt number
but we also now that
but the average heat transfer coefficient is h=2hx
h=2(8.48)=16.97W/m^{2}K
Finally we have that the heat transfer is
In this solution we took values for water properties of
v=16.96*10^{-6}m^{2}s
Pr=0.699
k=26.56*10^{-3}W/mK
A=1*0.5m^{2}
I hope this is useful for you
regards
$2.
Both tickets cost $1.50
$1.50 x 2 = $3
$5 - $3 = $2
Answer:
The speed of shaft is 1891.62 RPM.
Explanation:
given that
Amplitude A= 0.15 mm
Acceleration = 0.6 g
So
we can say that acceleration= 0.6 x 9.81
We know that
So now by putting the values
We know that
ω= 2πN/60
198.0=2πN/60
N=1891.62 RPM
So the speed of shaft is 1891.62 RPM.
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V = 
V = Av = 
V = 
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = 
= 
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = 
v2 = 
v2 = 6.017 m/s
