with no mechanical work added to, or taken from an incompressible airflow in a n
ID: 1842257 • Letter: W
Question
with no mechanical work added to, or taken from an incompressible airflow in a non-leaky duct, between 2 locations in the duct: All of the responses below The total mechanical energy is constant The static pressure plus dynamic pressure plus elevation pressure is constant If the duct becomes large in the flow the kinetic energy of the flow will reduce and up as pressure energy if the constant If the duct is of uniform dross sectional area, and points the potential energy of the flow will reduce and the flow work of the flow will and the kinetic energy is constant.Explanation / Answer
solution:
1)answer to above question is 1)all of the responses below because
i)for flow of fluid between two points by bernoulli's principle total conservtaion of energy and mass is achieved for imcompressible flow.
ii)for option 2,states that total mechanical energy is constant,as mechanical energy is sum of internal energy,kinetic energy and potential energy.
by steady flow equation we get that
Q=heat supplied
W=work supplied
dH=enthalphy change
d(K.E)=kinetic energy change
d(P.E)=potential energy change
Q'-W'=m(dH+d(K.E)+d(P.E))
if Q'=W'=0
(dH+d(K.E)+d(P.E))=0
mechanical energy change=0
hence mechanical energy is constant between two points
iii)for option 3,bernoulli's principle state that algebric sum of static pressure,dynamic pressure and elevation pressure is constant.
p/(density*g)+V^2/(2*g)+Z=constant
hence option 3 is true
iv)option 4 is also true,as by bernoulli's law,if for constant elevation decrease in kinetic energy is maintained by increase in pressure energy.
by contuinity equation for flow in close conduit we have
Q=A1*V1=A2*V2
if A2>A1
for constant Q,
V2<V1
p/(density*g)+V1^2/(2*g)+Z1=p2/(density*g)+V2^2/(2*g)+Z2
for constant elevation,Z1=Z2
(p2/(density*g)-(p1/(density*g))=(V1^2/(2*g)-V2^2/(2*g))
hence option 4 is also true and it is proved as above
v)here option 5 is also true,by steady flow energy equation we can write that mechanical energy is constant and for constant kinetic energy as no variation in cross section of flow,enthalphy and potential energy can be related as stated as follows
Q'-W'=m(dH+d(K.E)+d(P.E))
(dH+d(K.E)+d(P.E))=0
for constant K.E,d(K.E)=0
dH=-d(P.E)
h=enthalphy=internal energy+flow work
internal energy=u=CvT
for constant temperature.u=constant
so we get that
d(flow work)=-d(P.E)
flow work2-flow work1=P.E1-P.E2
hence it is proved that decrease in potential energy for imcompressible flow is responsible for rise in flow work
2)in this way all responses satis fies bernoulli's principle and hence response with option 1 is correct
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