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Question 1
ప్రశ్న 1
What is the principal purpose of the Hardy Cross method when applied to a looped water-distribution network?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A looped network must satisfy conservation of flow at its junctions. It must also satisfy conservation of energy around every independent closed loop. The Hardy Cross method begins with a trial flow distribution satisfying continuity and then adjusts those flows iteratively. The corrections reduce the algebraic imbalance of pipe head losses around the loops. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which two fundamental hydraulic conditions must be satisfied by the final solution of an ordinary looped water-distribution network?
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Explanation:
The continuity condition requires inflow at a junction to equal outflow plus any external demand. The energy condition requires the algebraic sum of head changes around a closed circuit to be zero. Both requirements must be satisfied simultaneously in a physically consistent network solution. Hardy Cross uses successive corrections to move an assumed flow distribution toward this condition. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
What requirement should normally be satisfied by the initial trial pipe flows before the balancing-head form of the Hardy Cross method is started?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The balancing-head Hardy Cross procedure starts with assumed pipe discharges that satisfy mass conservation. The trial values do not need to satisfy the loop energy equations initially. Head-loss imbalance is precisely what the subsequent loop corrections are intended to reduce. Because a correction circulates around a closed loop, properly applied corrections preserve the original nodal continuity. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
If clockwise traversal is chosen as positive for a Hardy Cross loop, how should a pipe whose assumed flow is anticlockwise relative to that traversal be treated?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The choice of clockwise as positive is an arbitrary but convenient sign convention. A flow in the same direction as the chosen loop traversal is assigned a positive sign. A flow opposite to that direction is assigned a negative sign. The corresponding friction head-loss term also carries the sign associated with the assumed flow direction. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
For a pipe obeying a resistance law with exponent n, which signed expression is convenient in Hardy Cross calculations?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The magnitude of friction head loss can be represented in the form R times the magnitude of flow raised to the appropriate exponent. A signed formulation is conveniently written as h = RQ|Q|^(n-1). This expression automatically gives positive or negative head loss according to the signed loop flow. For n = 2 it reduces to h = RQ|Q|. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
Which statement about the exponent n used in the Hardy Cross head-loss relation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The exponent depends on the friction relationship adopted for the network analysis. The current SI Hazen-Williams relation contains discharge raised to approximately 1.852, often rounded to 1.85 in Hardy Cross presentations. With Darcy-Weisbach and a friction factor treated as constant during an iteration, friction loss varies approximately with the square of velocity and discharge. The corresponding Hardy Cross exponent is therefore approximately 2. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
For h = RQ|Q|^(n-1), which expression gives the usual Hardy Cross loop-flow correction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hardy Cross correction is obtained by linearizing the nonlinear head-loss equation about the assumed flows. The numerator is the algebraic sum of signed head losses around the selected loop. The denominator contains the sum of the absolute derivative terms and is positive. For a constant exponent n, this becomes n times the sum of |h/Q|. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
For Questions 8–15, consider one loop with signed trial flows Q = (+4, +3, -2, -3) L/s and resistance coefficients R = (1, 2, 1, 1) in compatible units. Using h = RQ|Q|, what is the signed head loss in the second pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The second-pipe trial flow is +3 L/s relative to the selected positive loop direction. Its resistance coefficient is 2. Using h = RQ|Q| gives h = 2 × 3 × 3. The result is +18 m because the flow is in the positive loop direction. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
For the same loop, what is the algebraic sum of the four trial head losses?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The four signed head losses are +16, +18, -4 and -9 m. Their algebraic sum is 16 + 18 - 4 - 9. This equals +21 m. A nonzero loop sum shows that the initial trial flows do not yet satisfy the loop energy equation. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
For the same loop with n = 2, what is the denominator n Sigma|h/Q| used in the first Hardy Cross correction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The absolute values of h/Q are 16/4 = 4, 18/3 = 6, 4/2 = 2 and 9/3 = 3. Their sum is 4 + 6 + 2 + 3 = 15. Because n equals 2, the correction denominator is 2 × 15. This gives 30 in the corresponding correction units. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
For the same loop, what is the first Hardy Cross flow correction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The algebraic head-loss sum is +21 m. The correction denominator is 30. Therefore Delta Q = -21/30 = -0.70 L/s. The negative sign means the correction acts opposite to the selected positive clockwise loop direction. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
After applying the first correction of -0.70 L/s to the signed trial flows (+4, +3, -2, -3) L/s, what are the corrected signed flows?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The same signed loop correction is added algebraically to every signed flow in the selected loop. Thus +4 becomes +3.3 and +3 becomes +2.3 L/s. The negative flows become -2.7 and -3.7 L/s because adding -0.70 makes them more negative. This is consistent with a correction acting in the anticlockwise direction. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
Using the corrected flows (+3.3, +2.3, -2.7, -3.7) L/s and the same resistance coefficients, what is the new algebraic head-loss imbalance after the first correction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The corrected signed head losses are 3.3^2 = +10.89 m and 2(2.3^2) = +10.58 m. The two opposing terms are -(2.7^2) = -7.29 m and -(3.7^2) = -13.69 m. Their algebraic sum is 10.89 + 10.58 - 7.29 - 13.69 = +0.49 m. The large original imbalance has therefore been reduced substantially after one iteration. Therefore Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
For the same loop after the first correction, the new value of 2 Sigma|h/Q| is 28.6. What second Hardy Cross correction is required for the remaining +0.49 m imbalance?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The second correction is calculated from Delta Q = -Sigma h divided by the correction denominator. The remaining imbalance is +0.49 m and the stated denominator is 28.6. Thus Delta Q = -0.49/28.6 = approximately -0.0171 L/s. The much smaller second correction indicates that the trial solution is approaching convergence. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
After the second correction of approximately -0.0171 L/s, the signed flows become approximately (+3.2829, +2.2829, -2.7171, -3.7171) L/s. The resulting loop head-loss sum is approximately +0.00029 m. What is the correct interpretation under the CPHEEO ±0.01 m loop-imbalance criterion?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The calculated remaining imbalance is about 0.00029 m. Its magnitude is far smaller than 0.01 m. Current CPHEEO guidance identifies approximately ±0.01 m as an allowable loop head-loss imbalance for this balancing procedure. Additional iterations would change the flows by only a negligible amount for the stated accuracy. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
If the calculated Hardy Cross correction for a clockwise-positive loop is negative, what does the sign mean?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The correction sign is interpreted relative to the arbitrarily selected positive loop direction. A negative correction means that the required circulating adjustment is opposite to the clockwise-positive convention. Positive signed pipe flows are reduced by that correction. Negative signed pipe flows may become larger in magnitude in their existing anticlockwise direction. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
A pipe has an assumed signed flow of +0.30 L/s in a loop and the calculated correction is -0.50 L/s. What does the corrected flow of -0.20 L/s indicate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The corrected signed flow is +0.30 - 0.50 = -0.20 L/s. A negative result means that the final direction is opposite to the initially assumed positive direction. The physical discharge magnitude is 0.20 L/s. Hardy Cross calculations permit flow directions to reverse when the iterative solution demonstrates that the original assumed direction was incorrect. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
At a network junction, two pipes deliver 30 L/s and 20 L/s. The nodal demand is 12 L/s, and one outgoing pipe carries 15 L/s. What flow must the other outgoing pipe carry if continuity is satisfied?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The total flow entering the junction is 30 + 20 = 50 L/s. The external nodal withdrawal consumes 12 L/s of this quantity. One outgoing pipe carries a further 15 L/s. The other outgoing pipe must therefore carry 50 - 12 - 15 = 23 L/s. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
Why does adding the same circulating Hardy Cross correction around a closed loop preserve an initial trial solution that already satisfies nodal continuity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A loop correction represents an additional circulating flow around the closed circuit. At a junction on that loop, one connected loop pipe brings the correction toward the node while the next carries the same correction away. The net correction to the nodal mass balance is therefore zero. External demand at the node is not changed by the loop correction. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
A pipe is common to two adjacent loops. Refer its positive direction as A to B. Loop 1 produces a +0.40 L/s correction in the A-to-B direction, while Loop 2 produces a +0.15 L/s correction in its own positive traversal, which crosses the common pipe from B to A. What is the net algebraic correction to the shared pipe referred to A-to-B direction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Loop 1 correction contributes +0.40 L/s when the common pipe is referenced from A to B. Loop 2 traverses the shared pipe in the opposite B-to-A direction, so its +0.15 L/s loop correction contributes -0.15 L/s on the A-to-B reference. The net correction is therefore +0.40 - 0.15. This equals +0.25 L/s from A to B. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
Why must the Hardy Cross balancing procedure normally be repeated over all loops of a multi-loop network rather than stopping immediately after each loop has been corrected once?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Adjacent loops commonly share one or more pipes. Correcting one loop changes the discharge and hence the friction head loss in a shared pipe. That change can disturb the energy balance previously obtained in the neighbouring loop. The analyst therefore cycles successively through the independent loops until all loop imbalances become sufficiently small. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
If the algebraic sum of signed head losses around a loop is already exactly zero for the current trial flows, what Hardy Cross correction is obtained from the usual correction formula?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The numerator of the Hardy Cross correction formula is the algebraic loop head-loss sum. If this numerator is zero, no energy imbalance exists around that loop for the present flows. Dividing zero by the positive correction denominator gives a zero flow correction. The current loop flow distribution therefore needs no further adjustment from that loop equation. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
For a Hazen-Williams-based Hardy Cross iteration, take n = 1.852. If Sigma h = +4.0 m and Sigma|h/Q| = 3.6 m per (L/s), what is the approximate loop-flow correction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hardy Cross correction is Delta Q = -Sigma h/[n Sigma|h/Q|]. The denominator is 1.852 × 3.6 = 6.6672. Dividing -4.0 by 6.6672 gives approximately -0.600 L/s. The negative sign means the adjustment is opposite to the selected positive loop direction. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
When choosing initial trial flows for a Hardy Cross calculation, which statement is most appropriate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The essential initial requirement for the balancing-head approach is satisfaction of nodal continuity. The trial values do not need to be the final hydraulically correct flows. However, assigning plausible directions and allowing relatively larger flows through lower-resistance paths usually reduces the amount of iteration required. Poor initial guesses may still converge but can require more corrections or may initially reverse direction in some pipes. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
After Hardy Cross loop corrections have converged satisfactorily, what additional engineering check is still necessary before accepting a water-distribution design?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hardy Cross balancing determines a flow distribution consistent with continuity and loop energy equations. Those balanced flows can then be used to calculate head losses and hydraulic grades through the network. Actual nodal pressure also depends on source head, elevation and accumulated losses along the relevant paths. A hydraulically balanced network can still have inadequate pressure at a high or remote node if the design is deficient. Therefore Option A is correct.
వివరణ:
Question 26
ప్రశ్న 26
Consider the following statements about the Hardy Cross method: I. Initial trial flows for the balancing-head method should satisfy junction continuity. II. A balanced closed loop has an algebraic sum of friction head losses approximately equal to zero. III. A pipe shared by two loops may receive algebraic corrections from both loops. IV. Successive iterations are continued until the selected convergence criterion is met. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because continuity is established before the loop-energy corrections are applied. Statement II is correct because conservation of energy requires the algebraic head-loss sum around a closed loop to vanish within the adopted tolerance. Statement III is correct because a common pipe participates in both circuit equations and receives appropriately signed corrections from each. Statement IV is correct because Hardy Cross is a successive-approximation method rather than a single-pass calculation. Therefore all four statements are correct and Option D is the valid answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. To determine a flow distribution that satisfies both nodal continuity and energy balance around the loops సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. Continuity at junctions and zero algebraic head-loss sum around each closed loop సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: B. Continuity should be satisfied at all ordinary junctions సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: A. Its flow and associated loop head loss are assigned negative algebraic signs సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. h = RQ|Q|^(n-1) సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. n is approximately 1.852 for the Hazen-Williams relation and approximately 2 when Darcy-Weisbach is used with a constant friction factor సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: A. Delta Q = -Sigma h/[n Sigma |h/Q|] సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: B. +18 m సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: A. +21 m సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: C. 30 సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. -0.70 L/s సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. (+3.3, +2.3, -2.7, -3.7) L/s సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. Approximately +0.49 m సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. Approximately -0.0171 L/s సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: A. The loop is adequately balanced to the stated criterion సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: A. The assumed clockwise component is too large and the correction acts in the anticlockwise direction సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: A. The pipe carries 0.20 L/s in the direction opposite to the original assumed direction సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: C. 23 L/s సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. At each loop junction the correction entering through one loop pipe is balanced by the correction leaving through the next loop pipe సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. +0.25 L/s సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. Corrections to pipes shared between loops can alter the head-loss balance of adjacent loops సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. Zero సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: B. -0.600 L/s సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. Any trial distribution satisfying nodal continuity can be used, but a physically reasonable guess generally improves convergence సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: A. Calculate node heads and verify that residual pressures and other design criteria are satisfactory సమాధానం: A.
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Question 26 ప్రశ్న 26Answer: D. I, II, III and IV సమాధానం: D.