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:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature
degree celcius

Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8


Q = 125.538 W
Answer:
The pressure inside the cooker is 1.0804 atm.
Explanation:
Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.
<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>
So, Standard conditions:
T₁ = 373.15 K
P₁ = 1 atm
Given ,
The water boils at Temperature = 130 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature, T₂ = (130 + 273.15) K = 403.15 K
To find pressure inside the cooker (P₂) :
<u>Applying Amontons's Law as:</u>

So,



<u>Thus, The pressure inside the cooker is 1.0804 atm.</u>
My best guess is b but I honestly don’t know
Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N
(b) 17.1 m
Explanation: The length of wall under the surface can be given by

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.
![F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N](https://tex.z-dn.net/?f=F%28resultant%29%20%3D%20Pavg%20%28%20A%29%20%3D%20%28Patm%20%2B%20%20%5Crho%20g%20h%20c%29%2AA%20%5C%5C%3D%20%5B100000%20N%2Fm%5E2%20%2B%20%281000%20kg%2Fm%5E3%20%2A%209.81%20m%2Fs%5E2%20%2A%2025m%2F2%29%5D%2A%20%28140%2A25m%2Fsin60%29%5C%5C%3D%208.997%2A10%5E8%20N%20%5C%5C%3D%209.0%2A10%5E8%20N)
Noting from the Bernoulli equation that

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:
Substituting the values gives us the the distance of the surface to be equal to = 17.1 m