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Please Plot with solutionAn air-standard cycle with constant specific heats is executed in a closed system and is composed of the following four proce

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Answer #1

o first lets plot pou and T-s diagrum pm 2 T PVC pacy 9. Given! Pi= 100 kPa T, = 22°C = 295k Pe=600K Pu. T3 = 1500k. P4 = 10Now, to find net work output, heat input and thermal efficiency of cycle we will first find pressure and temperatures of difffor isentropic process (3-4), Tu * Th= 73 ( - Th= T / ) C: REA) Net work output, Whet q, de Waet: culto - T. 67,3%)-411() 44-54.5E 54 - 53.5E Efficiency 51.5F 51 E 50.5E 50 1500 1600 1700 1800 1900 2100 2200 2300 2400 2500 2000 T3105 16 Heat input Work output 1500 1600 1700 1800 1900 2200 2300 2400 2500 2000 2100 T3

Matlab code:

clear all
clc
cv=718; % Specific heat at constant volume for air (J/kgK)
cp=1005; % Specific heat at constant pressure for air (J/kgK)
gamma=1.4 % Specific heat ratio
rp=600/100; % rp=p2/p1
T1=295;
T3=1500:100:2500; % T3 from 1500 K to 2500 K in increment of 100K
rt=T3./T1;
T2=T1*rp^((gamma-1)/gamma);
Q1=cv.*(T3-T2);
figure,plot(T3,Q1)
hold on
T4=T3.*(1./(rt.*rp^(1/gamma))).^((gamma-1)/gamma)
Q2=cp.*(T4-T1);
Wnet=Q1-Q2;
plot(T3,Wnet)
xlabel('T3');
legend('Heat input','Work output')
eta=(Wnet./Q1).*100;
figure, plot(T3,eta)
xlabel('T3');
ylabel('Efficiency')

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In case of query feel free to comment i will be happy to help you

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Answer #1

o first lets plot pou and T-s diagrum pm 2 T PVC pacy 9. Given! Pi= 100 kPa T, = 22°C = 295k Pe=600K Pu. T3 = 1500k. P4 = 10Now, to find net work output, heat input and thermal efficiency of cycle we will first find pressure and temperatures of difffor isentropic process (3-4), Tu * Th= 73 ( - Th= T / ) C: REA) Net work output, Whet q, de Waet: culto - T. 67,3%)-411() 44-54.5E 54 - 53.5E Efficiency 51.5F 51 E 50.5E 50 1500 1600 1700 1800 1900 2100 2200 2300 2400 2500 2000 T3105 16 Heat input Work output 1500 1600 1700 1800 1900 2200 2300 2400 2500 2000 2100 T3

Matlab code:

clear all
clc
cv=718; % Specific heat at constant volume for air (J/kgK)
cp=1005; % Specific heat at constant pressure for air (J/kgK)
gamma=1.4 % Specific heat ratio
rp=600/100; % rp=p2/p1
T1=295;
T3=1500:100:2500; % T3 from 1500 K to 2500 K in increment of 100K
rt=T3./T1;
T2=T1*rp^((gamma-1)/gamma);
Q1=cv.*(T3-T2);
figure,plot(T3,Q1)
hold on
T4=T3.*(1./(rt.*rp^(1/gamma))).^((gamma-1)/gamma)
Q2=cp.*(T4-T1);
Wnet=Q1-Q2;
plot(T3,Wnet)
xlabel('T3');
legend('Heat input','Work output')
eta=(Wnet./Q1).*100;
figure, plot(T3,eta)
xlabel('T3');
ylabel('Efficiency')

Please do upvote if you like the solution

In case of query feel free to comment i will be happy to help you

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Add a comment
Answer #1

o first lets plot pou and T-s diagrum pm 2 T PVC pacy 9. Given! Pi= 100 kPa T, = 22°C = 295k Pe=600K Pu. T3 = 1500k. P4 = 10Now, to find net work output, heat input and thermal efficiency of cycle we will first find pressure and temperatures of difffor isentropic process (3-4), Tu * Th= 73 ( - Th= T / ) C: REA) Net work output, Whet q, de Waet: culto - T. 67,3%)-411() 44-54.5E 54 - 53.5E Efficiency 51.5F 51 E 50.5E 50 1500 1600 1700 1800 1900 2100 2200 2300 2400 2500 2000 T3105 16 Heat input Work output 1500 1600 1700 1800 1900 2200 2300 2400 2500 2000 2100 T3

Matlab code:

clear all
clc
cv=718; % Specific heat at constant volume for air (J/kgK)
cp=1005; % Specific heat at constant pressure for air (J/kgK)
gamma=1.4 % Specific heat ratio
rp=600/100; % rp=p2/p1
T1=295;
T3=1500:100:2500; % T3 from 1500 K to 2500 K in increment of 100K
rt=T3./T1;
T2=T1*rp^((gamma-1)/gamma);
Q1=cv.*(T3-T2);
figure,plot(T3,Q1)
hold on
T4=T3.*(1./(rt.*rp^(1/gamma))).^((gamma-1)/gamma)
Q2=cp.*(T4-T1);
Wnet=Q1-Q2;
plot(T3,Wnet)
xlabel('T3');
legend('Heat input','Work output')
eta=(Wnet./Q1).*100;
figure, plot(T3,eta)
xlabel('T3');
ylabel('Efficiency')

Please do upvote if you like the solution

In case of query feel free to comment i will be happy to help you

0 0
Add a comment
Answer #1

o first lets plot pou and T-s diagrum pm 2 T PVC pacy 9. Given! Pi= 100 kPa T, = 22°C = 295k Pe=600K Pu. T3 = 1500k. P4 = 10Now, to find net work output, heat input and thermal efficiency of cycle we will first find pressure and temperatures of difffor isentropic process (3-4), Tu * Th= 73 ( - Th= T / ) C: REA) Net work output, Whet q, de Waet: culto - T. 67,3%)-411() 44-54.5E 54 - 53.5E Efficiency 51.5F 51 E 50.5E 50 1500 1600 1700 1800 1900 2100 2200 2300 2400 2500 2000 T3105 16 Heat input Work output 1500 1600 1700 1800 1900 2200 2300 2400 2500 2000 2100 T3

Matlab code:

clear all
clc
cv=718; % Specific heat at constant volume for air (J/kgK)
cp=1005; % Specific heat at constant pressure for air (J/kgK)
gamma=1.4 % Specific heat ratio
rp=600/100; % rp=p2/p1
T1=295;
T3=1500:100:2500; % T3 from 1500 K to 2500 K in increment of 100K
rt=T3./T1;
T2=T1*rp^((gamma-1)/gamma);
Q1=cv.*(T3-T2);
figure,plot(T3,Q1)
hold on
T4=T3.*(1./(rt.*rp^(1/gamma))).^((gamma-1)/gamma)
Q2=cp.*(T4-T1);
Wnet=Q1-Q2;
plot(T3,Wnet)
xlabel('T3');
legend('Heat input','Work output')
eta=(Wnet./Q1).*100;
figure, plot(T3,eta)
xlabel('T3');
ylabel('Efficiency')

Please do upvote if you like the solution

In case of query feel free to comment i will be happy to help you

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