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Illusion [34]
3 years ago
14

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

fer rate. The air exits the compressor at 20 m/s through a tube of 10 cm diameter.
Engineering
1 answer:
aksik [14]3 years ago
4 0

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, T_{i} = 20^{\circ}C = 273 + 20 = 293 K

Temperature at the outlet, T_{o} = 200{\circ}C = 273 + 200 = 473 K

Pressure at inlet, P_{i} = 80 kPa = 80\times 10^{3} Pa

Pressure at outlet, P_{o} = 800 kPa = 800\times 10^{3} Pa

Speed at the outlet, v_{o} = 20 m/s

Diameter of the tube, D = 10 cm = 10\times 10^{- 2} m = 0.1 m

Input power, P_{i} = 400 kW = 400\times 10^{3} W

Now,

To calculate the heat transfer, Q, 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}

where

h_{i} = specific enthalpy at inlet

h_{o} = specific enthalpy at outlet

v_{i} = air speed at inlet

p_{s} = specific power input

H and H' = Elevation of inlet and outlet

Now, if

v_{i} = 0 and H = H'

Then the above eqn reduces to:

h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}

Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}                (1)

Also,

p_{s} = \frac{P_{i}}{ mass, m}

Area of cross-section, A = \frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}

Specific Volume at outlet, V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s

From the eqn:

P_{o}V_{o} = mRT_{o}

m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s

Now,

p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg

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

Now, using these values in eqn (1):

Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW

Now, rate of heat transfer, q:

q = mQ = 0.925\times 813.33 = 752.33 kW

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The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
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Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

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