Answer:
The heat is transferred is at the rate of 752.33 kW
Solution:
As per the question:
Temperature at inlet,
= 273 + 20 = 293 K
Temperature at the outlet,
= 273 + 200 = 473 K
Pressure at inlet, ![P_{i} = 80 kPa = 80\times 10^{3} Pa](https://tex.z-dn.net/?f=P_%7Bi%7D%20%3D%2080%20kPa%20%3D%2080%5Ctimes%2010%5E%7B3%7D%20Pa)
Pressure at outlet, ![P_{o} = 800 kPa = 800\times 10^{3} Pa](https://tex.z-dn.net/?f=P_%7Bo%7D%20%3D%20800%20kPa%20%3D%20800%5Ctimes%2010%5E%7B3%7D%20Pa)
Speed at the outlet, ![v_{o} = 20 m/s](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D%2020%20m%2Fs)
Diameter of the tube, ![D = 10 cm = 10\times 10^{- 2} m = 0.1 m](https://tex.z-dn.net/?f=D%20%3D%2010%20cm%20%3D%2010%5Ctimes%2010%5E%7B-%202%7D%20m%20%3D%200.1%20m)
Input power, ![P_{i} = 400 kW = 400\times 10^{3} W](https://tex.z-dn.net/?f=P_%7Bi%7D%20%3D%20400%20kW%20%3D%20400%5Ctimes%2010%5E%7B3%7D%20W)
Now,
To calculate the heat transfer,
, we make use of the steady flow eqn:
![h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}](https://tex.z-dn.net/?f=h_%7Bi%7D%20%2B%20%5Cfrac%7Bv_%7Bi%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20gH%20%20%2B%20Q%20%3D%20h_%7Bo%7D%20%2B%20%5Cfrac%7Bv_%7Bo%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20gH%27%20%2B%20p_%7Bs%7D)
where
= specific enthalpy at inlet
= specific enthalpy at outlet
= air speed at inlet
= specific power input
H and H' = Elevation of inlet and outlet
Now, if
and H = H'
Then the above eqn reduces to:
![h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}](https://tex.z-dn.net/?f=h_%7Bi%7D%20%2B%20gH%20%2B%20Q%20%3D%20h_%7Bo%7D%20%2B%20%5Cfrac%7Bv_%7Bo%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20gH%20%2B%20p_%7Bs%7D)
(1)
Also,
![p_{s} = \frac{P_{i}}{ mass, m}](https://tex.z-dn.net/?f=p_%7Bs%7D%20%3D%20%5Cfrac%7BP_%7Bi%7D%7D%7B%20mass%2C%20m%7D)
Area of cross-section, A = ![\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20D%5E%7B2%7D%7D%7B4%7D%20%3D%5Cfrac%7B%5Cpi%200.1%5E%7B2%7D%7D%7B4%7D%20%3D%207.85%5Ctimes%2010%5E%7B-%203%7D%20m%5E%7B2%7D)
Specific Volume at outlet,
From the eqn:
![P_{o}V_{o} = mRT_{o}](https://tex.z-dn.net/?f=P_%7Bo%7DV_%7Bo%7D%20%3D%20mRT_%7Bo%7D)
![m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B800%5Ctimes%2010%5E%7B3%7D%5Ctimes%200.157%7D%7B287%5Ctimes%20473%7D%20%3D%200.925%20kg%2Fs)
Now,
![p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg](https://tex.z-dn.net/?f=p_%7Bs%7D%20%3D%20%5Cfrac%7B400%5Ctimes%2010%5E%7B3%7D%7D%7B0.925%7D%20%3D%20432.432%20kJ%2Fkg)
Also,
![\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%20h_%7Bo%7D%20-%20h_%7Bi%7D%20%3D%20c_%7Bp%7D%5CDelta%20T%20%3Dc_%7Bp%7D%28T_%7Bo%7D%20-%20T_%7Bi%7D%29%20%3D%201.005%28200%20-%2020%29%20%3D%20180.9%20kJ%2Fkg)
Now, using these values in eqn (1):
Now, rate of heat transfer, q:
q = mQ = ![0.925\times 813.33 = 752.33 kW](https://tex.z-dn.net/?f=0.925%5Ctimes%20813.33%20%3D%20752.33%20kW)