Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
I am Providing Answer in C Language Program.
Explanation:
Please find attachment regarding code of taking two numbers input and adding them.
I would like to recommend you please use software which supports C language.
#include <stdio.h>
int main () {
int a, b, sum;
printf ("\ nEnter two no:");
scanf ("% d% d", & d, & e);
sum1 = d + e;
printf ("Sum:% d", sum1);
return (0);
}
Answer:
Explanation:
Given that : -
The desirable limit is 500 mg / l , but
allowable upto 2000 mg / l.
The take volume is V = 160.000 m3
V = 160 , 000 x 103 l
The crainage gives 150 mg / l and lake has initialy 100 mg / l
Code of tpr frpm drawn = 150 x 60, 000 x 1000
Ci = 9000 kg / gr
Cl = 100 x 160,000 x 1000
Cl = 16, 000 kg
Since allowable limit = 2000 mg / l
Cn = ( 2000 x 160, 00 x 1000 )
= 320, 000 kg
so, each year the rate increases, by 9000 kg / yr
Read level = ( 320, 000 - 16,000 )
Li = 304, 000 kg
Tr=<u>304,000</u>
900
=33.77