Question #251694

3. For a certain ideal gus R = 25.8 d k = 1.09 What is the value of c, in points) You sent b.What rass (in lb) of this gas would occupy a volume of 15 eu ft at 75 psia and 80 degrees Fahrenheit? points) f 30 BTU are transferred to thais gas at constant volume in (bj, what is the resulting temperature in degrees Fahrenheit? (1 BTU-778 ft h An adiabatic expansion of air occurs through a nozzle from 828 kPa and 71 degrees Celsius to 138 kPa. The initial kinetic energy is negligible. Far an isentropic expansion compile the speed the (5 points) exit section (1S peints)


1
Expert's answer
2021-10-15T12:42:00-0400

Given Data

  • The Gas constant is: R=25.8 ft. lb/lb*R
  • The constant is: k=1.09

The equations are

CpCv=RCpCv=kC_p−C_v=R\\ \frac{C_p}{C_v}=k

Take Cv common in first equation

Cv=Rk1=25.81.091=286.67Btu/lb°FC_v= \frac{R}{k-1}= \frac{25.8}{1.09-1}=286.67 Btu/lb⋅°F

Cp286.67=25.8    Cp=312.467Btu/°FlbC_p−286.67=25.8 \implies C_p=312.467 Btu/°F⋅lb

The mass of the gas is

m=PVRT=[(75×144)lb/ft2]×15ft325.8×(80+459.67°R)=11.63lbm=\frac{PV}{RT}=\frac{[(75×144) lb/ft^2]×15 ft^3}{25.8×(80+459.67°R)}=11.63 lb

Q=mCvdTQmCv+T1=T2T2=30286.67×11.635+80=80.009°FQ=mC_vdT\\ \frac{Q}{mC_v}+T_1=T_2\\ T_2=\frac{30}{286.67×11.635}+80=80.009°F


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