Answer:
0.0406 m/s
Explanation:
Given:
Diameter of the tube, D = 25 mm = 0.025 m
cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²
Mass flow rate = 0.01 kg/s
Now,
the mass flow rate is given as:
mass flow rate = ρAV
where,
ρ is the density of the water = 1000 kg/m³
A is the area of cross-section of the pipe
V is the average velocity through the pipe
thus,
0.01 = 1000 × 4.9 × 10⁻⁴ × V
or
V = 0.0203 m/s
also,
Reynold's number, Re = 
where,
ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s
thus,
Re = 
or
Re = 611.39 < 2000
thus,
the flow is laminar
hence,
the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s
or
maximum velocity = 0.0406 m/s
Answer:
A famous example of concurrent engineering is the development of the Boeing 777 commercial aircraft. The aircraft was designed and built by geographically distributed companies that worked entirely on a common product database of C A TIA without building physical mock-ups but with digital product definitions.
Answer:
work is 50 kj
Explanation:
Given data
heat (Q) = 50 kj
To find out
work input for the compression stroke per kilogram of air
Solution
we will apply here "first law of thermodynamics" i.e.
The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.
ΔU = Q – W ................1
here ΔU is change in internal energy, Q is heat and W is work done
here U = 0 because air compressor the compression takes place at a constant internal energy in question
so that by equation 1
Q = W
and Q = 50
so work will be 50 kj
Answer:
21.456 kJ/h
Explanation:
See the figure attached. In this case
Coefficient of performance in heat pump is defined by
Now it is necessary to change units, remember that Watt (W) is defined as J/s
