Weak Shower, Chattering Pump and Slow Tank Filling
A ground-plus-two house is supplied from a 1000 L HDPE overhead tank on the roof slab. The tank outlet is 1.2 m above the slab, and the first-floor bathroom tap is 7.5 m below the tank outlet, while the second-floor tap is 4.5 m below it. A 0.5 hp monobloc booster pump with a pressure switch draws from a 1000 L underground sump and fills the tank through a 25 mm line. The owner complains that the first-floor shower is weak in the morning, that the second-floor tap is acceptable, that the pump switches on and off about every 40 seconds while filling, and that the tank now takes nearly 50 minutes to fill instead of the earlier 25. The gate valve between the sump and the pump is kept half open to protect the pump. The suction line is 3 m of 25 mm GI and the delivery is 20 m of 25 mm GI with eight elbows.
1. Calculate the static pressure at the first-floor tap in kPa and bar, and decide whether pressure alone explains the weak shower.
Read the worked answer
Head = 7.5 m. P = density × gravity × head = 1000 × 9.81 × 7.5 = 73,575 Pa, i.e. about 73.6 kPa or 0.74 bar (second floor: 4.5 × 9.81 = 44 kPa, about 0.44 bar). A shower will work at 0.74 bar although it is not strong, and the key evidence is comparative: the second-floor tap has only 0.44 bar and is judged acceptable, while the first-floor outlet with more head is weak. A lower-pressure outlet performing better rules out static pressure as the cause. The fault must be a local flow restriction on the first-floor branch, or a shared restriction that shows up where the demand is highest: scale or a choked flexible hose, a half-closed angle valve, a kinked or undersized branch, or air trapped at a high point. Check the flow rate at the tap in litres per minute, not just the pressure.
2. Explain the pump cycling and the doubling of the filling time, naming the checks you would make.
Read the worked answer
The tank needs 1000 L: 25 minutes means about 40 L/min (2.4 m³/h) and 50 minutes means about 20 L/min (1.2 m³/h), so the pump is delivering roughly half its earlier flow. Cycling every 40 seconds means the pump is being starved or the pressure switch is seeing rapid swings. Checks: (1) the half-open gate valve on the suction - throttling a suction line is the opposite of protection; it lowers pressure at the impeller eye, causes cavitation and starves the pump; (2) the suction strainer and foot valve - a choked strainer or a leaking foot valve gives air ingress and loss of prime; (3) air leaks at suction joints, which make the pump chatter; (4) scale in 20 m of 25 mm GI with eight elbows - GI scales internally and the bore may be down to 15 mm; (5) pressure switch differential set too narrow, so it cuts in and out; (6) worn impeller or low voltage. The correct protection is adequate suction diameter and a short, rising suction line, not a throttled valve.
3. Recommend the minimum changes to restore the flow, with reasons and a rough order of cost.
Read the worked answer
Cheapest and first: open the gate valve fully and clean the strainer; check and re-seat or replace the foot valve; set the pressure switch to a wider differential so cycling stops. Next: replace the 3 m suction with 32 mm pipe (suction never smaller than delivery, short, with a continuous rise to the pump) and re-make joints to eliminate air leaks; verify that the pump holds prime overnight. Then: descale or replace the 20 m delivery with 25 mm CPVC, which does not scale the way GI does and has smoother bends; reduce elbows by using two 45° bends instead of one 90° where possible. Check the float valve in the tank is the right size, since an undersized float valve can restrict filling and cause the tank to take longer. Order of cost: valve and strainer work is almost free; suction replacement is a few hundred rupees in material; the delivery line is the largest item. The suction and valve work should recover most of the lost flow, so do not quote the full replacement until the cheap checks are done.