Answer:
526.5 KN
Explanation:
The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.
But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.
h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg
where ρ = density of the fluid and g = acceleration due to gravity
h = ΔP/ρg
ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa
Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with
Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa
Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²
Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN
Answer:
![h=1.99998\ W/m^2.C](https://tex.z-dn.net/?f=h%3D1.99998%5C%20W%2Fm%5E2.C)
![k=33.333\ W/m.C](https://tex.z-dn.net/?f=k%3D33.333%5C%20W%2Fm.C)
Explanation:
Considering the one dimensional and steady state:
From Heat Conduction equation considering the above assumption:
Eq (1)
Where:
k is thermal Conductivity
is uniform thermal generation
![T(x) = a(L^2-x^2)+b](https://tex.z-dn.net/?f=T%28x%29%20%3D%20a%28L%5E2-x%5E2%29%2Bb)
![\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%5C%20T%28x%29%7D%7B%5Cpartial%20x%7D%3D%5Cfrac%7B%5Cpartial%5C%20a%28L%5E2-x%5E2%29%2Bb%7D%7B%5Cpartial%20x%7D%3D-2ax%5C%5C%5Cfrac%7B%5Cpartial%5E2%5C%20T%28x%29%7D%7B%5Cpartial%20x%5E2%7D%3D%5Cfrac%7B%5Cpartial%5E2%5C%20-2ax%7D%7B%5Cpartial%20x%5E2%7D%3D-2a)
Putt in Eq (1):
![-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C](https://tex.z-dn.net/?f=-2a%2B%5Cfrac%7B%5Cdot%20e_%7Bgen%7D%7D%7Bk%7D%3D0%5C%5C%20k%3D%5Cfrac%7B%5Cdot%20e_%7Bgen%7D%7D%7B2a%7D%5C%5C%20k%3D%5Cfrac%7B1000%7D%7B2%2A15%7D%5C%5C%20k%3D33.333%5C%20W%2Fm.C)
Energy balance is given by:
![Q_{convection}=Q_{conduction}](https://tex.z-dn.net/?f=Q_%7Bconvection%7D%3DQ_%7Bconduction%7D)
Eq (2)
![T(x) = a(L^2-x^2)+b](https://tex.z-dn.net/?f=T%28x%29%20%3D%20a%28L%5E2-x%5E2%29%2Bb)
Putting x=L
![T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC](https://tex.z-dn.net/?f=T%28L%29%20%3D%20a%28L%5E2-L%5E2%29%2Bb%5C%5CT%28L%29%3Db%5C%5CT%28L%29%3D40%5EoC)
![\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdx%7D%3D%5Cfrac%7Bd%28a%28L%5E2-x%5E2%29%2Bb%7D%7Bdx%7D%3D-2ax%5C%5CPut%5C%20x%5C%20%3D%5C%20L%5C%5C%5Cfrac%7BdT%7D%7Bdx%7D%3D-2aL%5C%5C%28%5Cfrac%7BdT%7D%7Bdx%7D%29_L%3D-2%2A15%2A0.04%3D-1.2)
From Eq (2)
![h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B-k%2A-1.2%7D%7B%2840-20%29%7D%20%5C%5Ch%3D%5Cfrac%7B-33.333%2A-1.2%7D%7B%2840-20%29%7D%5C%5Ch%3D1.99998%5C%20W%2Fm%5E2.C)
Answer:
Explanation:
- a) Given C [ cal pro fat calc sod]
[140 27 3 13 64]
- P = [cal pro fat calc sod]
[180 4 11 24 662]
- B = [cal pro fat calc sod]
[50 5 1 82 20]
To find C+2P+3B = [140 27 3 13 64] + 2[180 4 11 24 662] + 3[50 5 1 82 20]
= [650 54 28 307 1448]
The entries represent skinless chicken breast , One-half cup of potato salad and One broccoli spear.