Answer:
A pitot tube is used to measure fluid flow in engineering
Answer:
a) 
b) 
c) 
d) 
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature, 
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,

Rate of heat transfer,

a) To calculate the convection coefficient relationship for heat transfer by convection:

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe, 

b) Heat loss from the pipe to the environment:

d) The required fan control power is 25.125 W as calculated earlier above
Answer:
σ =5.39Mpa
Explanation:
step one:
The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces
Flexural strength test Flexural strength is calculated using the equation:
σ = FL/ (bd^2 )----------1
Where
σ = Flexural strength of concrete in Mpa
F= Failure load (in N).
L= Effective span of the beam
b= Breadth of the beam
step two:
Given data
F=40.45 kN= 40450N
b=0.15m
d=0.15m
L=0.45m
step three:
substituting into the expression we have
σ = 40450*0.45/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.0225 )
σ =18202.5/0.003375
σ =5393333.3
σ =5393333.3/1000000
σ =5.39Mpa
Therefore the flexure strength of the concrete is 5.39Mpa
Answer: A) Gradually decrease
Explanation:
The convection value of heat transfer rate are gradually decreasing with the flow of the heat. Flow in a circular pipe, flow direction does not change in the velocity path. The average of the coefficient of heat transfer and the number of pipes are needed and the effects are get neglected so that is why the flow are fully developed.
Answer:
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