A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that
1 answer:
Answer:
The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )
Explanation:
width of channel = 3.0 m
Flow rate = 5 m^3/s
Normal depth = 0.50 m
Flow encounters a dam rise of 0.25 m
To know if the hydraulic jump will occur
we will Determine the new normal depth
Y2 = 3.77m
Yc ( critical depth )= 0.66m
Attached below is the detailed solution
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Answer:
25 mm = 0.984252 inches
Explanation:
Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:
<u>1 mm = 1/25.4 inches</u>
From the question, we have to convert 25 mm into inches
Thus,
<u>25 mm = (1/25.4)*25 inches</u>
So,
Thus, solving we get:
<u>25 mm = 0.984252 inches</u>
Answer:
engine B is more efficient.
Explanation:
We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.
We also kn ow that the efficiency of Carnot cycle given as follows
Here temperature should be in Kelvin.
For engine A
For engine B
So from above we can say that engine B is more efficient.
Answer:d
Explanation:
Given
Temperature
Also
R=287 J/kg
Flow will be In-compressible when Mach no.<0.32
Mach no.
(a)
Mach no.
Mach no.=0.63
(b)
Mach no.
Mach no.=0.31
(c)
Mach no.
Mach no.=1.27
(d)
Mach no.
Mach no.=0.127
From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.