A Brayton cycle with regeneration (heat- exchanger for heat recovery) has a pres

Question

A Brayton cycle with regeneration (heat- exchanger for heat recovery) has a pressure ratio of 9. the temperature of compressor inlet is 300 K and the temperature of turbine inlet is 1200k. The compress has an isentropic efficiency of 80% and the turbine has an efficiency of 85%. The temperature o t e compressed air entering the combustion chamber is 755K. Drawn T -s diagram of the cycle an determine. The temperature at the compressor exit The temperature at the turbine exit The back-work ratio The network output The thermal efficiency

Explanation / Answer

According to the given problem,

let the wire be x-axis, then the plastic stick has its ends at points (0, 2)cm and (0, 8)cm;

First consider a smaller problem,

wire and a point charge q at a distance y from it;

Elementary force on the charge q by elementary charge

dk=p*dx positioned at x on the wire is vector

df=dk*q*r/(4*0*|r|3), where p=1*10-7 C/m, vector r = (-x,y) is a vector connecting dk and q, |r|=(x2 +y2) is distance between dk and q;

total f=m*dx*(-x,y)/(x2 +y2)3 {x=- to +}, where m=p*q/4*0;

dx*x/(x2 +y2)3 =-1/(x2 +y2) {x=- to +} =0; as expected!

dx*y/(x2 +y2)3 = (x/y)/(x2 +y2) {x=- to +} =2/y;

vector f = m*(0, 2/y);

Elementary force on a piece of stick dy is vector

dF = dm*(0, 2/y), where dm= p*dq/4*0, dq=dy*Q/b,

Q/b is linear charge density on the stick,

Q=13*10-9 C, b = 6cm = 0.06m;

|F|= 2*(p*Q/4*0*b) dy/y {y=0.02m to 0.02+b};

dy/y =ln(0.08/0.02) =ln4;

|F|= 2*p*Q*ln4/4*0*b

|F|=2*1*10-7*13*10-9*ln4 /(4*8.854*10-12*0.06)

|F|=5.4*10-4 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.