Engineering Questions with Answers - Multiple Choice Questions

Civil Engineering Drawing MCQ’s – Principle for the Preparation of Water Supply

1 - Question

The distribution pipes are generally laid below the road pavements, and as such their layouts generally follow the layouts of roads.
a) True
b) False
View Answer Answer: a
Explanation: There are, in general, four different types of pipe networks; any one of which either singly or in combinations, can be used for a particular place. They are: Dead End System, Grid Iron System, Ring System and Radial System.

2 - Question

Quantity = Per capita sewage contributed per day x Population.
a) True
b) False
View Answer Answer: a
Explanation: Wastewater Quantity Estimation – The flow of sanitary sewage alone in the absence of storms in dry season is known as dry weather flow (DWF). Sanitary sewage is mostly the spent water of the community draining into the sewer system. It has been observed that a small portion of spent water is lost in evaporation, seepage in ground, leakage, etc. Usually 80% of the water supply may be expected to reach the sewers.

3 - Question

 The most widely used expression for the growth rate of micro-organisms is given by Monod:
Total rate of microbial growth, civil-engineering-drawing-questions-answers-campus-interviews-q3
a) True
b) False

View Answer

Answer: a
Explanation: Biomass Growth Rate
The most widely used expression for the growth rate of micro organisms is given by Monod:
Total rate of microbial growth,
mm = maximum specific growth rate
X = micro organism concentration
S = substrate concentration
Ks = substrate concentration at one half the maximum growth rate
Similarly, rate of substrate utilization,
dS = k X S
dt Ks+ S
k = maximum specific substrate utilization rate.

4 - Question

Which is not a function of Distribution Reservoirs?
a) To absorb the hourly variations in demand
b) To maintain constant pressure in the distribution mains
c) Water stored can be supplied during emergencies
d) Work as a coolant present at the area
View Answer Answer: d
Explanation: Distribution reservoirs, also called service reservoirs, are the storage reservoirs, which store the treated water for supplying water during emergencies (such as during fires, repairs, etc.) and also to help in absorbing the hourly fluctuations in the normal water demand.

5 - Question

Which is not a type of reservoir?
a) Small ground level reservoirs
b) Large ground level reservoirs
c) Underground reservoirs
d) Sub-surface reservoir
View Answer Answer: d
Explanation: Types of Reservoirs- i. Underground reservoirs. ii. Small ground level reservoirs. iii. Large ground level reservoirs. iv. Overhead tanks.

6 - Question

Design a low rate filter to treat 6.0 Mld of sewage of BOD of 210 mg/l. The final effluent should be 30 mg/l and organic loading rate is 320 g/m3/d. Find Hydraulic loading rate?
a) 326.18 g/d/m3
b) 926.18 g/d/m3
c) 126.18 g/d/m3
d) 526.18 g/d/m3
View Answer

Answer: a
Assume 30% of BOD load removed in primary sedimentation i.e., = 210 x 0.30 = 63 mg/l. Remaining BOD = 210 – 63 = 147 mg/l.
Percent of BOD removal required = (147-30) x 100/147 = 80%
BOD load applied to the filter = flow x conc. of sewage (kg/d) = 6 x 106 x 147/106 = 882 kg/d
To find out filter volume, using NRC equation
Depth of filter = 1.5 m, Fiter area = 2704/1.5 = 1802.66 m2, and Diameter = 48 m < 60 m
Hydraulic loading rate = 6 x 106/103 x 1/1802.66 = 3.33m3/d/m2 < 4 hence o.k.
Organic loading rate = 882 x 1000 / 2704 = 326.18 g/d/m3 which is approx. equal to 320.

7 - Question

Design a rectangular sedimentation tank to treat 2.4 million litres of raw water per day. The detention period may be assumed to be 3 hours. Find overflow rate?
a) 28,000 l/d/m2
b) 12,000 l/d/m2
c) 24,000 l/d/m2
d) 4,670 l/d/m2
View Answer

Answer: c
Explanation: Raw water flow per day is 2.4 x 106 l. Detention period is 3h.
Volume of tank = Flow x Detention period = 2.4 x 103 x 3/24 = 300 m3
Assume depth of tank = 3.0 m.
Surface area = 300/3 = 100 m2
L/B = 3 (assumed). L = 3B.
3B2 = 100 m2 i.e. B = 5.8 m
L = 3B = 5.8 X 3 = 17.4 m
Hence surface loading (Overflow rate) = 2.4 x 106 = 24,000 l/d/m2 < 40,000 l/d/m2.

8 - Question

Which one is not a type of Equalization tanks?
a) Flow through type
b) Intermittent flow type
c) Variable inflow/constant discharge type
d) Intercept type
View Answer Answer: d
Explanation: The simple flow through type equalization tank is mainly useful in assisting self neutralization or evening out of fluctuating concentrations, not for balancing of flows since a flow through type tank once filled, gives output equal to input. Flow balancing and self-neutralization are both achieved by using two tanks, intermittently one after another. One tank is allowed to fill up after which it is checked for pH (or any other parameter) and then allowed to empty out. The second tank goes through a similar routine. Intermittent flow type tanks are economic for small flows from industries. When flows are large an equalization tank of such a size may have to be provided that inflow can be variable while outflow is at a constant rate, generally by a pump.The capacity required is determined from a plot of the cumulative inflow and a plot of the constant rate outflow and measuring the gaps between the two plots. A factor of safety may be applied if desired.

9 - Question

These are open drains provided for conveying water from kitchens, bathroom and rain water to main sewers.
a) Drains
b) Sewers
c) Soak pit
d) Surface drains
View Answer Answer: d
Explanation: These are usually provided at the side of the road and along the boundary line of the building. As far as possible drains should not be laid under the buildings. For efficient draining the surface drains should have certain qualities, such as should be laid in such a gradient to develop self -cleaning velocity, should have a reasonable free board at the top, joint should be smooth finished, easy curves, inner surface should be smooth, cheap in construction and maintenance.

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