Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
Answer:
The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.
Explanation:
Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:
Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:
Using the volume rate:
Assuming no losses, the energy equation for fluids can be written as:
Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:
Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:
Answer:
point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))
Explanation:
The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted. For the general case, it appears you simply need to change how you have written the code.
point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))
Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.
You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.
Cheers.
