Carbon dioxide enters an adiabatic compressor at 100 kPa and 300K at a rate of 0.5 kg/s and leaves at 600 kPa and 450K. Neglecting kinetic energy changes, determine a) the volume flow rate of the carbon dioxide at the compressor inlet (Ans. around 0.3 m3/s) and b) the power input to the compressor (Ans. around 70 kW).

The answers to the question are

a) The volume flow rate of the carbon dioxide at the compressor inlet is 0.2834 m³/s ≈ 0.3 m³/s

b) The power input to the compressor is 73.35 kW ≈ 70 kW

Explanation:

We note the following

Mass flow rate = 0.5 kg/s

Inlet pressure = 100 pKa

Outlet pressure = 600 kPa

Inlet temperature = 300 K

Outlet temperature  =  450 K

Molar mass of CO₂ = 44.01 g/mol

R Universal Gas Constant = 8.314 4621. J K−1 mol−1

a) Number of moles = = = 11.361 moles

P·V= n·R·T ∴ V = = = 0.2834 m³

Therefore the volume flow rate = 0.2834 m³/s ≈ 0.3 m³/s

b) Cp at 300 K = 0.846 kJ/(kg K)

Cp at 600 K = 0.978 kJ/(kg K)

Cv = 0.657

K = 1.289

While the power input to the compressor can be calculated by

m'×Cp×(T₂-T₁)

Where m' = mass flow rate = 0.5 kg/s

Therefore power = 0.5 kg/s×0.978 kJ/(kg K)×(450 K - 300 K)

= 73.35 kJ/s = 73.35 kW ≈ 70 kW

15.3 cos 65 degrees trig

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.