Answer:
Power required to overcome aerodynamic drag is 50.971 KW
Explanation:
For explanation see the picture attached
Answer:
Stat PVC = Stat(82+98.5)
Stat PVT = Stat(59+71.5)
Explanation
PVI = 71 + 35
Let G1 = Grade 1; G2 = Grade 2
G1 = +2.1% ; G2 = -3.4%
Highest point of curve at station = 74 + 10
General equation of a curve:
![y = ax^{2} +bx+c\\dy/dx=2ax+b\\](https://tex.z-dn.net/?f=y%20%3D%20ax%5E%7B2%7D%20%2Bbx%2Bc%5C%5Cdy%2Fdx%3D2ax%2Bb%5C%5C)
At highest point of the curve ![dy/dx=o](https://tex.z-dn.net/?f=dy%2Fdx%3Do)
![2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275](https://tex.z-dn.net/?f=2ax%2Bb%3D0%5C%5Cx%3D-b%2F2a%5C%5Cx%3DG1L%2F%28G2-G1%29%5C%5Cx%3DL%2F2%20%2B%28stat%2074%2B10%29-%28stat%2071%2B35%29%5C%5Cx%3DL%2F2%20%2B%20275)
![-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\](https://tex.z-dn.net/?f=-G1L%2F%28G2-G1%29%20%3D%20%28L%2F2%20%2B%20275%29%2F100%5C%5CL%20%3D%20-2327%20ft%5C%5CStation%20PVC%20%3D%20Stat%2871%2B35%29%2B%28-2327%2F2%29%5C%5C%5C%5CStat%20PVC%20%3D%207135-1163.5%5C%5CStat%20PVC%20%3D%20Stat%2882%2B98.5%29%5C%5C)
Station PVT
![Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)](https://tex.z-dn.net/?f=Station%20PVT%20%3D%20Stat%20PVI%20%2B%20%28L%2F2%29%5C%5CStation%20PVT%20%3D%20Stat%2871%2B35%29%2B%28-2327%2F2%29%5C%5CStation%20PVT%20%3D%207135-1163.5%5C%5CStat%20PVT%20%3D%20Stat%2859%2B71.5%29)
Answer:
COP of the heat pump is 3.013
OP of the cycle is 1.124
Explanation:
W = Q₂ - Q₁
Given
a)
Q₂ = Q₁ + W
= 15 + 7.45
= 22.45 kw
COP = Q₂ / W = 22.45 / 7.45 = 3.013
b)
Q₂ = 15 x 1.055 = 15.825 kw
therefore,
Q₁ = Q₂ - W
Q₁ = 15.825 - 7.45 = 8.375
∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124