PROBLEM 1: Seventy kilojoules of work is done by each kilogram of fluid passing through an apparatus under steady-flow conditions. In the inlet pipe, which is located 30 m above the floor, the specific volume is 3 m³/kg, the pressure is 300 kPa, and the velocity is 50 m/s. In the discharge pipe, which is 15 m below the floor, the specific volume is 9 m³/kg, the pressure is 60 kPa, and the velocity is 150 m/s. Heat loss from the fluid is 3 kJ/kg. Determine the change in internal energy of the fluid passing through the apparatus.
PROBLEM 1: Seventy kilojoules of work is done by each kilogram of fluid passing through an apparatus under steady-flow conditions. In the inlet pipe, which is located 30 m above the floor, the specific volume is 3 m³/kg, the pressure is 300 kPa, and the velocity is 50 m/s. In the discharge pipe, which is 15 m below the floor, the specific volume is 9 m³/kg, the pressure is 60 kPa, and the velocity is 150 m/s. Heat loss from the fluid is 3 kJ/kg. Determine the change in internal energy of the fluid passing through the apparatus.
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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Please show sholutios completely use those formula that i give. thank youuu....
Please answer it in 2hrs.
![This is the most frequently used form because many steady-flow systems have only one inlet and
one exit.
Q= Em h+2+gz - Em
+gz +W
exits
inlets
Q=mg(z, - z, )+m(v -v,*)+m(h, –h, )+ W
(9-4)
Q-ΔΡE+ΔΚΕ + ΔΗ + W
For a Unit Mass
(9-5)
q=g(z, -z,)+ (v-v.?)+(h, -h, ) + w
PROBLEM 1: Seventy kilojoules of work is done by each kilogram of fluid
passing through an apparatus under steady-flow conditions. In the inlet pipe,
which is located 30 m above the floor, the specific volume is 3 m³/kg, the
pressure is 300 kPa, and the velocity is 50 m/s. In the discharge pipe, which is
15 m below the floor, the specific volume is 9 m³/kg, the pressure is 60 kPa,
and the velocity is 150 m/s. Heat loss from the fluid is 3 kJ/kg. Determine the
change in internal energy of the fluid passing through the apparatus.
Ans: 277.4 kJ/kg](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef4abb29-973a-44d3-9b2d-c4032542ce74%2Ffd814e88-a933-4ac3-b6a8-d663abc1bc94%2Fy3s0g6d_processed.png&w=3840&q=75)
Transcribed Image Text:This is the most frequently used form because many steady-flow systems have only one inlet and
one exit.
Q= Em h+2+gz - Em
+gz +W
exits
inlets
Q=mg(z, - z, )+m(v -v,*)+m(h, –h, )+ W
(9-4)
Q-ΔΡE+ΔΚΕ + ΔΗ + W
For a Unit Mass
(9-5)
q=g(z, -z,)+ (v-v.?)+(h, -h, ) + w
PROBLEM 1: Seventy kilojoules of work is done by each kilogram of fluid
passing through an apparatus under steady-flow conditions. In the inlet pipe,
which is located 30 m above the floor, the specific volume is 3 m³/kg, the
pressure is 300 kPa, and the velocity is 50 m/s. In the discharge pipe, which is
15 m below the floor, the specific volume is 9 m³/kg, the pressure is 60 kPa,
and the velocity is 150 m/s. Heat loss from the fluid is 3 kJ/kg. Determine the
change in internal energy of the fluid passing through the apparatus.
Ans: 277.4 kJ/kg
![First Law of Thermodynamics Conservation of Energy-Open systems
aiready determined.
In mathematical form,
ZEnergy = Energy
Inlet
exit
Considering all forms of energy flow into and out from the control volume with the mass, the
+pu ++gz
2
w+÷+gz +W
2
exits
inlets
h=u+po
but
Q=Em h++gz - Emh+-
+gz +W
2
exits
inlets
For a Unit Mass
q-Σh++gz-Σ| h+ .
+gz +w
exits
Inlets
where
w = specific work
Q= heat
p= pressure
v = specific volume
v = velocity
g= acceleration due to gravity
z = datum
W = work](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef4abb29-973a-44d3-9b2d-c4032542ce74%2Ffd814e88-a933-4ac3-b6a8-d663abc1bc94%2F5y37ild_processed.png&w=3840&q=75)
Transcribed Image Text:First Law of Thermodynamics Conservation of Energy-Open systems
aiready determined.
In mathematical form,
ZEnergy = Energy
Inlet
exit
Considering all forms of energy flow into and out from the control volume with the mass, the
+pu ++gz
2
w+÷+gz +W
2
exits
inlets
h=u+po
but
Q=Em h++gz - Emh+-
+gz +W
2
exits
inlets
For a Unit Mass
q-Σh++gz-Σ| h+ .
+gz +w
exits
Inlets
where
w = specific work
Q= heat
p= pressure
v = specific volume
v = velocity
g= acceleration due to gravity
z = datum
W = work
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