To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K,
D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius?
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Answer:
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Explanation:
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To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K, D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius
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Answer:
a) The change in Kinetic energy, KE = -1.95 kJ
b) Power output, W = 10221.72 kW
c) Turbine inlet area,
Explanation:
a) Change in Kinetic Energy
For an adiabatic steady state flow of steam:
.........(1)
Where Inlet velocity, V₁ = 80 m/s
Outlet velocity, V₂ = 50 m/s
Substitute these values into equation (1)
KE = -1950 m²/s²
To convert this to kJ/kg, divide by 1000
KE = -1950/1000
KE = -1.95 kJ/kg
b) The power output, w
The equation below is used to represent a steady state flow.
For an adiabatic process, the rate of heat transfer, q = 0
z₂ = z₁
The equation thus reduces to :
w = h₁ - h₂ - KE...........(2)
Where Power output, ..........(3)
Mass flow rate,
To get the specific enthalpy at the inlet, h₁
At P₁ = 10 MPa, T₁ = 450°C,
h₁ = 3242.4 kJ/kg,
Specific volume, v₁ = 0.029782 m³/kg
At P₂ = 10 kPa, , x₂ = 0.92
specific enthalpy at the outlet, h₂ =
h₂ = 3242.4 + 0.92(2392.1)
h₂ = 2392.54 kJ/kg
Substitute these values into equation (2)
w = 3242.4 - 2392.54 - (-1.95)
w = 851.81 kJ/kg
To get the power output, put the value of w into equation (3)
W = 12 * 851.81
W = 10221.72 kW
c) The turbine inlet area
Answer:
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Explanation:
solution is attached below
