Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 1 (Q.No. 30)
30.
The maximum efficiency of transmission through a pipe is
Discussion:
11 comments Page 1 of 2.
PANDURANG KARMURE said:
6 years ago
Max Power delivered = Pactual = 2pQgh/3.
Max efficiency = (Pactual)max/Pideal.
= (2pQgh/3)/pQgh.
= 2/3 = 0.6667.
= 66.67%.
Max efficiency = (Pactual)max/Pideal.
= (2pQgh/3)/pQgh.
= 2/3 = 0.6667.
= 66.67%.
PANDURANG KARMURE said:
6 years ago
Max.Efficiency = (H+H/3)/H.
= (2H/3)/H.
= (2H/3)(1/H).
= 2/3.
= 0.6667x100.
= 66.67%.
= (2H/3)/H.
= (2H/3)(1/H).
= 2/3.
= 0.6667x100.
= 66.67%.
Sagar said:
6 years ago
66.67%.
Salman said:
7 years ago
(H-H/3)/H * 100 = 2/3 * 100 = 0.6667 * 100 = 66.67%.
Mahesh said:
7 years ago
Maximum loss due to friction is H/3.
(H-H/3)/H * 100 = 2/3 * 100 = 0.6667 * 100 = 66.67%.
(H-H/3)/H * 100 = 2/3 * 100 = 0.6667 * 100 = 66.67%.
Pradeep gk said:
8 years ago
Hf=h/3.
Suraj kakade said:
10 years ago
Maximum efficiency = (H-H/3)/H = 2/3 = 0.6667 = 66.67%.
Pratik chudasama said:
10 years ago
I think there should be use word minimum not maximum.
Because here use maximum loss due to friction is 33% not minimum.
Because here use maximum loss due to friction is 33% not minimum.
E KISHORE KUMAR said:
10 years ago
m.e = 9(H-H/3)/H = 2/3 = 0.6667 = 66.67%
M BALA said:
1 decade ago
So maximum efficiency = (h-h/3)h =2/3 = 0.6667 = 66.67.
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