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Fed [463]
3 years ago
14

Anne works with other engineers studying physical processes that involve the flow of particles. Which field of engineering would

her work fall under?
A.
chemical engineering
B.
mechanical engineering
C.
software engineering
D.
industrial engineering
E.
materials engineering
Engineering
1 answer:
slega [8]3 years ago
3 0

Answer:

Option A

Chemical engineering

Explanation:

Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.

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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
aksik [14]

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

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\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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Answer:

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