Answer:
Workdone, w = 68.935 kJ
Explanation:
m1h1 + m2h2 + Q = m3h3 - Wc
Final mass of mixture,
m3 = m1 + m2 = 5kg
p1V1 = m1R1T1
p2V2 = m2R2T2
Since they are at 1:1,
V1 = V2 and R1 = R2
Comparing equations,
p1/p2 = (m1/m2) x (T1/T2)
m1/m2 = (100/200) x ((50 + 273)/(25+273))
m1/m2 = 0.542
m1 = 0.542m2
m3 = m1 + m2
5 = m2 + 0.542m2
m2 = 3.243kg
m1 = 1.757 kg
Workdone is given as,
Wc = m3h3 - m1h1 - m2h2
h = Cp x T, since air is an ideal gas
Cp = 1.005kJ/kGK
Wc = (5 x 1.005 x (55+273)) - (1.757 x 1.005 x 298) -(3.243 x 1.005 x 323)
Wc = 68.935 kJ
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
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DescripciónAlimentos funcionales son aquellos alimentos que son elaborados no solo por sus características nutricionales sino también para cumplir una función específica como puede ser el mejorar la salud y reducir el riesgo de contraer enfermedades