PROBLEM 1: Assume 5 Ib/sec of fluid enter a steady-state, steady flow system with p1 = 100 psia, p2 =0.2 Ib/ft³, v1 =100 fps, u1=800 Btu/lb, and leave with p2 =20 psia, p2 =0.05 lb/ft, v2 = 500 fps, and u2=780 Btu/lb. During passage through the open system, each pound rejects 10 Btu of heat. Find the work in horsepower. Ans: 167.8 hp
PROBLEM 1: Assume 5 Ib/sec of fluid enter a steady-state, steady flow system with p1 = 100 psia, p2 =0.2 Ib/ft³, v1 =100 fps, u1=800 Btu/lb, and leave with p2 =20 psia, p2 =0.05 lb/ft, v2 = 500 fps, and u2=780 Btu/lb. During passage through the open system, each pound rejects 10 Btu of heat. Find the work in horsepower. Ans: 167.8 hp
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![PROBLEM 1:
Assume 5 Ib/sec of fluid enter a steady-state, steady flow
system with p1 = 100 psia, p2 =0.2 lb/ft³, v1 =100 fps, u1=800 Btu/lb, and leave
with p2 =20 psia, p2 =0.05 lb/ft, v2 = 500 fps, and u2=780 Btu/lb. During
passage through the open system, each pound rejects 10 Btu of heat. Find the
work in horsepower.
Ans: 167.8 hp](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef4abb29-973a-44d3-9b2d-c4032542ce74%2F2f66fac3-1014-4d00-81bb-c5c447118521%2F9sjg8ge_processed.png&w=3840&q=75)
Transcribed Image Text:PROBLEM 1:
Assume 5 Ib/sec of fluid enter a steady-state, steady flow
system with p1 = 100 psia, p2 =0.2 lb/ft³, v1 =100 fps, u1=800 Btu/lb, and leave
with p2 =20 psia, p2 =0.05 lb/ft, v2 = 500 fps, and u2=780 Btu/lb. During
passage through the open system, each pound rejects 10 Btu of heat. Find the
work in horsepower.
Ans: 167.8 hp
![Considering all forms of energy flow into and out from the control volume with the mass, the
- Em
+gz
pu++gz
M.
exits
inlets
h=u+pv
but
Q= Em
+gz
+gz +W
exits
inlets
For a Unit Mass
q=Eh++gz -Eh++gz +w
exits
inlets
w = specific work
Q= heat
p= pressure
v= specific volume
v = velocity
where
g= acceleration due to gravity
z = datum
W = work
This is the most frequently used form because many steady-flow systems have only one inlet and
one exit.
h+
+gz + W
2
exits
inlets
Q=mg(z, -z, )+÷m(v -v,?)+ ṁ(h, –h, ) + W
Q-ΔΡE+ΔΚΕ + ΔΗ + W
(9-4)
For a Unit Mass
q=g(z, -z, )+(v* -v;')+ (h, -h,. )+ w
(9-5)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef4abb29-973a-44d3-9b2d-c4032542ce74%2F2f66fac3-1014-4d00-81bb-c5c447118521%2Fy27wlzd_processed.png&w=3840&q=75)
Transcribed Image Text:Considering all forms of energy flow into and out from the control volume with the mass, the
- Em
+gz
pu++gz
M.
exits
inlets
h=u+pv
but
Q= Em
+gz
+gz +W
exits
inlets
For a Unit Mass
q=Eh++gz -Eh++gz +w
exits
inlets
w = specific work
Q= heat
p= pressure
v= specific volume
v = velocity
where
g= acceleration due to gravity
z = datum
W = work
This is the most frequently used form because many steady-flow systems have only one inlet and
one exit.
h+
+gz + W
2
exits
inlets
Q=mg(z, -z, )+÷m(v -v,?)+ ṁ(h, –h, ) + W
Q-ΔΡE+ΔΚΕ + ΔΗ + W
(9-4)
For a Unit Mass
q=g(z, -z, )+(v* -v;')+ (h, -h,. )+ w
(9-5)
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