Respuesta :
Answer:
The heat is transferred is at the rate of 752.33 kW
Solution:
As per the question:
Temperature at inlet, [tex]T_{i} = 20^{\circ}C[/tex] = 273 + 20 = 293 K
Temperature at the outlet, [tex]T_{o} = 200{\circ}C[/tex] = 273 + 200 = 473 K
Pressure at inlet, [tex]P_{i} = 80 kPa = 80\times 10^{3} Pa[/tex]
Pressure at outlet, [tex]P_{o} = 800 kPa = 800\times 10^{3} Pa[/tex]
Speed at the outlet, [tex]v_{o} = 20 m/s[/tex]
Diameter of the tube, [tex]D = 10 cm = 10\times 10^{- 2} m = 0.1 m[/tex]
Input power, [tex]P_{i} = 400 kW = 400\times 10^{3} W[/tex]
Now,
To calculate the heat transfer, [tex]Q[/tex], we make use of the steady flow eqn:
[tex]h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}[/tex]
where
[tex]h_{i}[/tex] = specific enthalpy at inlet
[tex]h_{o}[/tex] = specific enthalpy at outlet
[tex]v_{i}[/tex] = air speed at inlet
[tex]p_{s}[/tex] = specific power input
H and H' = Elevation of inlet and outlet
Now, if
[tex]v_{i} = 0[/tex] and H = H'
Then the above eqn reduces to:
[tex]h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}[/tex]
[tex]Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}[/tex] (1)
Also,
[tex]p_{s} = \frac{P_{i}}{ mass, m}[/tex]
Area of cross-section, A = [tex]\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}[/tex]
Specific Volume at outlet, [tex]V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s[/tex]
From the eqn:
[tex]P_{o}V_{o} = mRT_{o}[/tex]
[tex]m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s[/tex]
Now,
[tex]p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg[/tex]
Also,
[tex]\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg[/tex]
Now, using these values in eqn (1):
[tex]Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW[/tex]
Now, rate of heat transfer, q:
q = mQ = [tex]0.925\times 813.33 = 752.33 kW[/tex]
