SSC JE MAINS PAPER SOLUTIONS
CIVIL ENGINEERING CONVENTIONAL PAPER SOLUTIONS 2018
QUESTION NO--01
(a) For A Town with population of of 200000 a water supply scheme is to be designed.The maximum daily demand may be assumed as 200 litre per capita per daY. The storage Reservoir is situated 5 kilometre away from the town. Assuming loss of head from source to town as 10m and coefficient of friction for the pipe material as 0.012, recommend the size of supply main. 50% of the daily demand has to be pumped in 8 hours for the purposed scheme.
Solutions-
video solution's link for ssc je 2018 mains paper solutions
Solution:-
Maximum Daily Water Demand = 200 LPCD
Maximum Water Demand = 200000 x 200 = 40 MLD
Maximum Water Demand for which supply main is to be designed = 50% of daily demand
Q = (50 / 100) x 40 x 106
Liters per 8 hours
Q = (50 / 100) x (40 x 106) / (8 x 60 x 60)
Q = 0.694 m3/s
L = 5000 m HL
= 10 m f = 0.048
HL
= (f * L*v2) / (2gd)
10 = (0.048 x 5000 x v2) / (2 x 9.81 x d)
v = ( Q / (∏ /4)) x d2 v = ( 0.694 / (3.14/4)) x d2
v = 0.884 d2
10 = (0.048 x 5000 x (0.884 d2)2) / (2 x 9.81 x d)
d = 0.955 m
FOR
NEXT QUESTIONS- CLICK HERE
For A Town with population
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CIVIL ENGINEERING CONVENTIONAL PAPER SOLUTIONS 2018
QUESTION NO--01
(a) For A Town with population of of 200000 a water supply scheme is to be designed.The maximum daily demand may be assumed as 200 litre per capita per daY. The storage Reservoir is situated 5 kilometre away from the town. Assuming loss of head from source to town as 10m and coefficient of friction for the pipe material as 0.012, recommend the size of supply main. 50% of the daily demand has to be pumped in 8 hours for the purposed scheme.
Solutions-
video solution's link for ssc je 2018 mains paper solutions
Solution:-
Maximum Daily Water Demand = 200 LPCD
Maximum Water Demand = 200000 x 200 = 40 MLD
Maximum Water Demand for which supply main is to be designed = 50% of daily demand
Q = (50 / 100) x 40 x 106
Liters per 8 hours
Q = (50 / 100) x (40 x 106) / (8 x 60 x 60)
Q = 0.694 m3/s
L = 5000 m HL
= 10 m f = 0.048
HL
= (f * L*v2) / (2gd)
10 = (0.048 x 5000 x v2) / (2 x 9.81 x d)
v = ( Q / (∏ /4)) x d2 v = ( 0.694 / (3.14/4)) x d2
v = 0.884 d2
10 = (0.048 x 5000 x (0.884 d2)2) / (2 x 9.81 x d)
d = 0.955 m
FOR
NEXT QUESTIONS- CLICK HERE
For A Town with population
ssc je mains test series,
ssc je mains solved paper pdf,
ssc je mains paper,
ssc je civil conventional question papers pdf,
ssc je mains paper 2018,
ssc je mains paper pdf,
ssc je mains paper 2018,
ssc je mains book pdf
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