1. (8pts) Calculate the velocity at the exit of a diffuser when air at 100kPa an
ID: 1862664 • Letter: 1
Question
1. (8pts) Calculate the velocity at the exit of a diffuser when air at 100kPa and 20oC enters it with a velocity of 500m/s and exit state is 200kPa and 90oC.
2. (12pts) R-134a at 800kPa and 100oC enters an adiabatic nozzle steadily with a velocity of 20m/s and leaves as sat-V at 400kPa. Determine 1) temperature at the outlet; 2) the velocity at the outlet; 2) ratio of the inlet to outlet area (A1/A2)
3. (12pts) A compressor compresses helium from 100kPa and 20oC to 400kPa and 100oC. Helium enters this compressor through a 0.2m2 pipe at a velocity of 15m/s. A heat loss of 200kJ/kg occurs during the compression process. Neglecting the kinetic energy and potential energy, please determine 1) the mass flow rate at the outlet; and 2) the power input to the compressor.
4. (8pts) Saturated mixture of water at 1500kPa is throttled to 100kPa and 150oC in a steam line. Please calculate the temperature and quality of the water at the inlet.
5. (10pts) Air is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. Air enters the heat exchanger at 120kPa and 10oC at a rate of 0.5m3/s. The exhaust gases (cp = 1.2kJ/kg.K) enters at 200oC at a rate of 1.5kg/s and leaves at 80oC. Determine the rate of heat transfer to the air and air temperature at the outlet.
Explanation / Answer
1) h1 +V1^2/2 = h2 + V2^2/2
Cp*T1 + V1^2/2 = Cp*T2 + V2^2/2
T1 = 20 deg = 293 K
T2 = 90 deg = 363 K
V1 = 500 m/s
Cp = 1.004 KJ/kg.K
1.004*10^3*293 + 500^2/2 = 1.004*10^3*363 +V2^2/2
V2 = 330.82 m/s
2)h1 +V1^2/2 = h2 + V2^2/2 ----(1)
@ P1 = 800 kPa and T1 =100 deg
h1 = 485.50 KJ/kg and v1 = 0.03518 m^3/kg
V1 = 20 m/s
@ P2 = 400 kPa
T2 = 8.84 deg
h2 = 403.56 KJ/kg
v2 = 0.05136 m^3/kg
From 1
485.5*10^3 + 20^2/2 = 403.56*10^3 + V^2/2
V2 = 405.32 m/s
From mass Flow rate
m = A1*V1/v1 = A2*V2/v2
A1/A2 = (V2/V1)*(v1/v2) = (405.32/20)*(0.03518/0.05136)
A1/A2 = 13.88
3)P1 = 100 kPa and T1 = 20 deg = 293 K and R = 2078.5
P2 = 400 kPa and T2 = 100 deg = 373 K
Pv =RT
v1 = 2078.5*293/10^5 = 6.09 m^3/s
Mass flow is always constant
Mass flow at oulet = mass flow at inlet = Ai*Vi/v1 = 0.2*15/6.09
Mass flow at oulet = 0.493 kg/s
gamma(k) =1.66
P*v^k = const
v2 = v1*(P1/P2)^1/k = 6.09*(100/400)^1/1.66 = 2.64 m^3/kg
work = P2*v2 - P1*v1/(1-gamma) = (400*2.64 - 100*6.09)/(1-1.66) = -677 KJ/kg
Work done on compressor (w) = 677.27 KJ/kg
Heat Lost = 200 KJ/kg
Total = 877.27 KJ/kg
Power Input = m*Total = 0.493*877.27 = 432.49 KW
4)@ P2 = 100 kPa and T2 =150 deg - ---superheated condition
h2 = 2776.38 KJ/kg
In throttling h1 = h2 = 2776.38 KJ/kg
@ P1 = 1500 kPa
T1 = 198.32 deg
hf = 844.87 KJ/kg
hg = 2792.15 KJ/kg
h1 = (1-x)*hf + x*hg
2776.38 = (1-x)*844.87 +x*2792.15
x = 0.992
5) Density of air = 1.2 kg/m^3
Volumetric flow of air = 0.5 m^3/kg
Mass flow of air = 1.2*0.5 kg/s = 0.6 kg/s
Heat gain by air = Heat loss by hot exhaust gases
mass flow of air* Cp of air*dT = mass flow of gas*Cp of gas*dT
0.6*1.004*(T-10) = 1.5*1.2*(200-80)
T = 368.57 Deg
Heat transfer to air = 1.5*1.2*(200-80) = 216 KW
Air oulet temp = T = 368.57 deg
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