Answer:
Total saving would be of 36.917 $\yr
Explanation:
Given Data:
Degree F
Degree F
A = 30 ft^2
%
Heat loss before previous final value 

= 360 Btu/hr
Heat loss after new value
saving would be 
= 616.1782 kw hr/yr
$
= 36.917 $\yr
This question is incomplete, the complete question is;
(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2
. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.
D₃ = _____D.
{ the tolerance is +/-3% }
Answer:
the diameter of the second pipe D₃ is 1.13D
Explanation:
Given the data in the question;
Length = 2
pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.
Now, we know that for Laminar Flow;
V' = πD⁴ΔP / 128μL
where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃
Hence,
V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL
so
D₃ = D
ΔP₁₋₂ / ΔP₂₋₃ 
we substitute
D₃ = D
1.657 
D₃ = D( 1.134568 )
D₃ = 1.13D
Therefore, the diameter of the second pipe D₃ is 1.13D
Answer:
11.6 mm
Explanation:
With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:
σadm = strength / fos
σadm = 900 / 5 = 180 MPa
The stress is the load divided by the section:
σ = P / A
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)


Answer:
ηa=0.349
ηb=0.345
Explanation:
The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

The quality at state 4 is determined from the condition
and the entropies of the components at the condenser pressure taken from table:

The enthalpy at state 4 then is:

Part A
In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:

The enthalpy at state 2 is determined from an energy balance on the pump:

=346.67 kJ/kg
The thermal efficiency is then determined from the heat input and output in the cycle:

Part B
In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:

The enthalpy at state 2 is then determined from an energy balance on the pump:

=299.79 kJ/kg
The thermal efficiency in this case then is:

Answer
given,
Speed of vehicle = 65 mi/hr
= 65 x 1.4667 = 95.33 ft/s
e = 0.07 ft/ft
f is the lateral friction, f = 0.11
central angle,Δ = 38°
The PI station is
PI = 250 + 50
= 25050 ft
using super elevation formula



r = 1568 ft
As the road is two lane with width 12 ft
R = 1568 + 12/2
R = 1574 ft
Length of the curve


L = 1044 ft
Tangent of the curve calculation


T = 542 ft
The station PC and PT are
PC = PI - T
PC = 25050 - 542
= 24508 ft
= 245 + 8 ft
PT = PC + L
= 24508 + 1044
=25552
= 255 + 52 ft
the middle ordinate calculation


MO = 85.75 ft
degree of the curvature


D = 3.64°