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
Ray Bans
Arrow Shirts
Gillette
Answer:
1121.7 × 10³⁰ photons per second
Explanation:
Data provided in the question:
Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W
Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz
Now,
P = 
here,
N is the number of photons
t is the time
E = energy = hf
h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s
Thus,
P = = 
[t = 1 s for per second]
or
550 × 10³ =
or
550 = N × 4903.24 × 10⁻³⁴
or
N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second
Answer:
2.455 W
Explanation:
The power dissipated in each branch is ...
P = V^2/R
So, the branch powers are ...
branch 1: 18^2/220 ≈ 1.473 W
branch 2: 18^2/330 ≈ 0.982 W
Total power is ...
1.473 W + 0.982 W = 2.455 W
