Question

A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p = a + bV, where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m3 and 1.20 m3. The specific internal energy of the gas is given by the relation
u = l.5 pv – 85 kJ/kg
Where p is the kPa and v is in m3/kg. Calculate the net heat transfer and the maximum internal energy of the gas attained during expansion.

Answer 1

Answer:

Net heat transfer IS =660KJ

Maximum internal energy of the gas attained during expansion=503.25KJ

Explanation:

A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p = a + bV, where a and b are constants.

The initial pressure is (p1)= 1000 kPa and  

The final pressure is (p2) =200 kPa

volumes (v1)=0.20 m3 and

volumes (v2)= 1.20 m3.

The specific internal energy of the gas is given by the relation

u = l.5 pv – 85 kJ/kg,

Where p is the kPa and v is in m3/kg.

 

ΔU=$U_{2} -U_{1}$

Δ$U=(1.5)*(P_{2} V_{2}- P_{1} V_{1} )$

ΔU=$1.5*(200*1.2-1000*0.2)$

ΔU=60KJ

  $W_{1-2}$=600KJ

    The net heat transfer IS

            $Q_{1-2} =W_{1-2}$ +ΔU

                  $Q_{1-2} =600+60\\ =660KJ$

u = l.5 pv – 85 kJ/kg

      $U=mu$

     $U=m(l.5 pv-85 kJ/kg)$

     $U=1.5PV-85*1.5$$U=1.PV-127.5$

     $U=1.5(a+bv)v-127.5$

     $\frac{d_{u} }{d_{v} } =1.5a+3vb-0\\\frac{d_{u} }{d_{v} } =0\\0=1.5a+3bv \\v=\frac{-a}{2b} \\U=1.5(a+bv)v-127.5\\U=1.5(a+b*(\frac{-a}{2b} )(\frac{-a}{2b})-127.5\\U=1.5(\frac{a}{2} )*(\frac{-a}{2b} )-127.5\\U=-\frac{a^{2}*1.5 }{4b} -127.5\\U_{max} =503.25 KJ$

$P=a+bv$

$1000=a+b(0.2)$

$200=a+b(1..2)$

$a=1160$

$b=-800$

     

Net heat transfer IS =660KJ

Maximum internal energy of the gas attained during expansion=503.25KJ

The project code is #SPJ2