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
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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
Transcribed Image Text: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
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)
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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