Question #269224

Carbon dioxide flows steadily at 1 kg/s through a device where the pressure, specific volume, and velocity of the gas are tripled according pVn = C. The inlet conditions are p1 = 200 kPa, v1 = 0.4 m3/kg, and v1 = 100 m/s. Determine (a) n; (b) the initial and final temperatures; (c) the change of enthalpy; (d) the power; (e) the heat.


1
Expert's answer
2021-11-21T17:28:29-0500

Solution;

Given;

m˙=1kg/s\dot{m}=1kg/s

p1=200kPap_1=200kPa

V=0.4m3/kgV=0.4m^3/kg

v1=100m/sv_1=100m/s

p2=3p1=600kPap_2=3p_1=600kPa

V2=3V1=1.2m3/kgV_2=3V_1=1.2m^3/kg

v2=3v1=300m/sv_2=3v_1=300m/s

(a)

p1V1n=p2V2np_1V_1^n=p_2V_2^n

200600=(1.20.4)n\frac{200}{600}=(\frac{1.2}{0.4})^n

13=3n\frac13=3^n

31=3n3^{-1}=3^n

Therefore;

n=1n=-1

(b)

From;

PV=mRTPV=mRT

Mco2=44M_{co_2}=44

Hence;

R=8.31444=0.18895R=\frac{8.314}{44}=0.18895

Therefore;

T1=pVmR=200×0.41×0.18895=423.38KT_1=\frac{pV}{mR}=\frac{200×0.4}{1×0.18895}=423.38K

T2=p2V2mR=600×1.21×0.18895=3810.53KT_2=\frac{p_2V_2}{mR}=\frac{600×1.2}{1×0.18895}=3810.53K

(c)

Δh=cpΔT\Delta h=c_p\Delta T

cp=0.756kJ/kgKc_p=0.756kJ/kgK

Δh=0.756×(3810.53423.38)\Delta h=0.756×(3810.53-423.38)

Δh=2560.69kJ/kg\Delta h=2560.69kJ/kg

(d)

Power = Change in Kinetic Energy

P=12m(v22v12)P=\frac12m(v_2^2-v_1^2)

P=12×1(30021002)P=\frac12×1(300^2-100^2)

P=40000WP=40000W

P=40kWP=40kW

(e)

Heat

Assuming the device is horizontal;

z1=z2z_1=z_2

From steady flow energy equation;

Q=Δh+m˙(v12v222000)+WQ=\Delta h+\dot{m}(\frac{v_1^2-v_2^2}{2000})+W

By direct substitution;

Q=2560+40+40=2640.69kWQ=2560+40+40=2640.69kW


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS