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Question 1 (a) For small flow rates, the volume flow rate Q through a small square-section tube of side h and length L is a function of dynamic viscosity ?, pressure drop per unit length ?p/L and h....

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Question 1
(a) For small flow rates, the volume flow rate Q through a small square-section tube of
side h and length L is a function of dynamic viscosity ?, pressure drop per unit length
?p/L and h. Using Buckingham pi theorem, express the relation between variables in a
non-dimensional form.
(b) Reynolds number is probably the most widely used non-dimensional number in
Fluid dynamics. Explain its physical meaning by reference to a creeping flow (a very
slow motion) and a high speed subsonic flow
Question 2
Two reversible Carnot engine operate in series between a source at 800oC and a sink
at 30oC. The amount of heat taken from the high temperature reservoir is 5 kW. If the
efficiency of the first engine operating with the source is 1.2 times the efficiency of the
second engine, calculate
(a) the temperature at which heat is supplied to the second engine
(b) the work done by each engine
(c) the heat rejected to sink
Question 3
(a) Calculate the heat and work transfers, cycle efficiency, work ratio and specific
steam consumption of a Carnot cycle, using steam between pressures of 66 and
0.04 bar. Recalculate the problem with introduction of isentropic efficiencies of 85%
for the compression and expansion processes of the ideal cycle.
(b) By reference to a line diagram of the above cycle, identify how irreversibility might
have occurred during heat input to the boiler. Also show how the first and second laws
of thermodynamics are not violated by considering the energy flow through the turbine.
Answered Same Day Dec 21, 2021

Solution

Robert answered on Dec 21 2021
125 Votes
Solution: (1)
(a)
Flow rate depends upon pressure drop per unit length, dynamic viscosity, and h.
Let


)
(


)


K is constant.
Make the dimension equal on both left and right side of the above equation:
(


) ) ) )
By equating the power of M we get;
0 = a + b
By equating the power of L we get;
3 = -a - 2b + c
By equating the power of T we get;
-1 = -a - 2b = -a – b – b = -(a+b) – b = 0 – b
b = 1
Then a = -1
And c = 4
Then (


)


Answer:
(b)
Inertial force, as the name implies is the force due to the momentum of the fluid. This is usually
expressed in the momentum equation by...
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