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A tank is situated at top of a house. RL of ground was kept as 200 m while the RL of the


terrace was measured to be 206.5 m. The RL of top of tank was found to be 208 m. An orifice


is placed at the bottom of the tank (Cd = 0.65) having diameter of 40 mm. A pipe was fit to


the orifice having diameter of 40 mm and diameter of the pipe was suddenly increased to 100


mm. Bend of 90o was provided and a gate valve is provided to control the flow. Just before


the tap a bend of 90o was again provided and a sudden contraction was provided and pipe


diameter was reduced to 20 mm. Calculate the discharge coming out from the tap if the total


length of pipe is 120 m (10 m 40 mm pipe, 90 m 100 mm pipe and 20 m 20 mm pipe) and


pipe is made up of steel with f = 0.015 if


a) Tank is full filled


b) Tank is half filled

A mean draft 380000 m³/day is produced from a drainage area of 330 km². At the flow line, the reservoir is estimated to convert about 1,600 hectares. The annual rainfall is 96.5 cm, the mean annual runoff is 22.8 cm and the annual lake evaporation is 77.1 cm. Find the net gain or loss and storage and compute the volume of water evaporated. Calculate volume in m³

4. A mean draft of 100mgd is produced from a drainage area of 200mi². At the flow line the reservoir is estimated to cover about 4,000 acres. The annual rainfall is 37in., The mean annual runoff is 10in, and the mean annual lake evaporation is 30in. Find the net gain or lost a storage. Compute the volume of water evaporated. Hoe significant is this amount.

Compute the rainfall for gauge x in problem 1. If the storm readings at A, B and C were 3,7,4,1,and 4.8in, respectively

State a laplace transform function and explain the meaning



identify and discuss the factors that affect the adoption of new irrigation technologies.


A glulam beam is manufactured with a depth of 16 in. and a 8% moisture content. Assuming common


values for estimating dimensional changes, compute the new depth of the beam if the moisture content is increased to 40%. (The fiber saturation point is 39%, and it shrinks 6.53% in drying from FSP to oven dry in the tangential direction.)

A 370-mm wide plank is cut in such a way that the width is in the tangential direction of annual rings.


Compute the change in width as the moisture content changes from 18% to 42%. (The fiber saturation point is 35%, and it shrinks 8.21% in drying from FSP to oven dry in the tangential direction.)

A timber wall 75 mm wide, 75 mm deep and 3.50 m high is assumed to be fixed at the top and bottom.


It is subjected to uniformly distributed load of 388 N/m acting along the weaker axis of its entire height. The wall is made up of Apitong wood with 80% stress grade with properties as shown:


Bending and tension parallel to the grain = 16.5 MPa


Modulus of Elasticity in Bending = 7310 MPa


Compression parallel to the grain = 9.56 MPa


Compression perpendicular to the grain = 2.20 MPa


Compute for the following:


a) Allowable compressive stress


b) Allowable bending stress


c) Maximum axial load

A simple span member is 6.65 m in length is made up of Apitong 185 mm x 270 mm wooden section,


with an allowable stress based on 80% stress grade of 16.5 MPa in bending and tension parallel to grain. The beam carries a concentrated load of 18 kN at the center and neglecting its own weight.


Weight of wood = 7.5kN/m3. The beam carries an axial tensile load of 192 kN. Bending and tension parallel to the grain = 16.5 MPa Modulus of Elasticity in Bending = 7310 MPa


Compression parallel to the grain = 9.56 MPa


Compute for the following:


a) actual tensile stress if only tensile force is acting.


b) interaction value for both bending and tensile stress


c) ratio of the difference between its actual bending and tensile stress to the adjusted bending


stress for slenderness

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