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Chilled-water pump: head worksheet to annual power cost

Pump sizing · flow, head, power and operating cost

A 300 gpm chilled-water pump is being selected for a mid-rise retrofit. The head worksheet closes out at 40 ft of static head, 18 ft of pipe and fitting friction at design flow, and 2 ft carried for the velocity head at the discharge. The pump quoted runs at 70% efficiency at that duty point, the plant logs 4,000 hours a year of run time, and the utility bills 12 cents per kilowatt-hour.

Given
  • Q = 300 gpm — Design flow
  • h_s = 40 ft — Static head
  • h_f = 18 ft — Friction head at design flow
  • h_v = 2 ft — Velocity head allowance
  • η = 70 % — Pump efficiency at the duty point
  • t = 4000 h/yr — Annual run time
  • p_e = 0.12 $/kWh — Electricity rate
Determine
  1. (a)the total dynamic head the pump must develop
  2. (b)the hydraulic power delivered to the water
  3. (c)the shaft power the motor must supply
  4. (d)the electrical energy a year of running consumes
  5. (e)what that year of pumping costs
Step 1 of 5(a) · solve for Total dynamic head

Everything downstream is priced off this one number, so the head worksheet closes first: 40 + 18 + 2 = 60 ft. The classic mistake is quoting the pump on static head alone — at design flow the friction is almost a third of the duty, and a pump bought 18 ft short runs off the right edge of its curve.

Rearranged for TDH
TDH=hs+hf+hvTDH = h_s + h_f + h_v
Your values, in your units
TDH=(40 ft)+(18 ft)+(2 ft)TDH = \left( 40\ \text{ft} \right) + \left( 18\ \text{ft} \right) + \left( 2\ \text{ft} \right)
Converted to base units
TDH=(12.192 m)+(5.4864 m)+(0.6096 m)TDH = \left( 12.192\ \text{m} \right) + \left( 5.4864\ \text{m} \right) + \left( 0.6096\ \text{m} \right)
Answer
TDH=18.288 mTDH = 18.288\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Total Dynamic Head solver →

Step 2 of 5(b) · solve for Hydraulic power

P = ρgQh is the power the water actually receives — about 3.39 kW, or 4.55 hp, which is exactly what the trade shortcut WHP = Q·H/3960 = 300 × 60 / 3960 gives. It has to be computed before efficiency enters, because efficiency divides this number and nothing else.

Rearranged for P
P=ρgQhP = \rho g Q h
18.288 mcarried from step 1
Your values, in your units
P=(1,000 kg/m3) (9.80665 m/s2) (300 gpm) (18.288 m)P = \left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 300\ \text{gpm} \right) \, \left( 18.288\ \text{m} \right)
Converted to base units
P=(1,000 kg/m3) (9.80665 m/s2) (1,135.62 L/min) (18.288 m)P = \left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 1{,}135.62\ \text{L/min} \right) \, \left( 18.288\ \text{m} \right)
Answer
P=3.3945 kWP = 3.3945\ \text{kW}

Carried onward at full precision, not this rounded figure.

Open the Hydraulic Power (P = ρgQh) solver →

Step 3 of 5(c) · solve for Shaft power

The motor pays for the shaft power, not the water power: 3.394 kW / 0.70 = 4.849 kW. Dividing by efficiency — never multiplying — is the step people flip; multiplying gives 2.38 kW and a breaker that trips on day one.

Rearranged for P_shaft
Pshaft=PhydηP_{shaft} = \tfrac{P_{hyd}}{\eta}
3.3945 kWcarried from step 2
Your values, in your units
Pshaft=(3,394.45 W)(0.7)P_{shaft} = \tfrac{\left( 3{,}394.45\ \text{W} \right)}{\left( 0.7 \right)}
Answer
Pshaft=4.8492 kWP_{shaft} = 4.8492\ \text{kW}

Carried onward at full precision, not this rounded figure.

Open the Pump Efficiency from Hydraulic and Shaft Power solver →

Step 4 of 5(d) · solve for Energy

Power sustained over the run hours becomes energy — 4.849 kW × 4,000 h is about 19,400 kWh. The trap is billing the motor nameplate here: a 7.5 hp motor loaded to 4.85 kW draws 4.85 kW, not its rating.

Rearranged for E
E=PtE = P t
4.8492 kWcarried from step 3
Your values, in your units
E=(4,849.22 W) (4,000 h)E = \left( 4{,}849.22\ \text{W} \right) \, \left( 4{,}000\ \text{h} \right)
Converted to base units
E=(4,849.22 W) (14,400,000 s)E = \left( 4{,}849.22\ \text{W} \right) \, \left( 14{,}400{,}000\ \text{s} \right)
Answer
E=69.829 GJE = 69.829\ \text{GJ}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Step 5 of 5(e) · solve for Energy cost

Energy times the tariff: about $2,328 a year, every year, for one pump. Set against a $6,000 purchase price, the electricity overtakes the pump in under three years — which is why the efficiency point in step 3 is a money number, not a datasheet number.

Rearranged for C_e
Ce=E peC_e = E \, p_e
69.829 GJcarried from step 4
Your values, in your units
Ce=(69,828,800,000 J) (0.12 $/kWh)C_e = \left( 69{,}828{,}800{,}000\ \text{J} \right) \, \left( 0.12\ \text{\$/kWh} \right)
Converted to base units
Ce=(19,396.9 kWh) (0.12 $/kWh)C_e = \left( 19{,}396.9\ \text{kWh} \right) \, \left( 0.12\ \text{\$/kWh} \right)
Answer
Ce=2,327.6 $C_e = 2{,}327.6\ \text{\$}

Carried onward at full precision, not this rounded figure.

Open the Energy Cost from a Utility Rate solver →

The pump must develop 60 ft of head, delivering 3.39 kW to the water and drawing 4.85 kW at the shaft, so a 4,000-hour year consumes about 19,400 kWh and costs roughly $2,328 at 12 ¢/kWh.

Why this order

The order is the estimator's order, and it only runs one way. Head has to close first because both power numbers are linear in it — get the head 20% wrong and every dollar downstream is 20% wrong. Hydraulic power comes second because it is the physics floor: no pump, however good, delivers 300 gpm against 60 ft for less than 3.39 kW. Efficiency then divides that floor into the shaft power the motor must actually supply, and only at that point do hours and tariff turn watts into money. Running it in SI keeps the two trade constants honest at once: WHP = Q·H/3960 is nothing but ρg in gpm–ft–hp clothing, and the cross-check is that 300 × 60 / 3960 = 4.545 hp = 3.39 kW, the same answer by both routes.

Two mistakes cost real money here. The first is quoting head from the drawings' static lift and letting friction ride — friction is 30% of this duty, it grows with the square of flow, and a pump selected short runs out on its curve where efficiency collapses and NPSH margin vanishes. The second is treating efficiency as a paperwork number. The gap between a 70% pump and a 55% pump at this duty is 1.32 kW at the shaft, which at 4,000 hours and 12 cents is about $634 a year — on a pump that will run twenty years, that is the price of the pump again, spent on nothing. And when the bill needs checking, remember the meter reads energy at the motor terminals: motor efficiency (say 92%) sits between the shaft and the panel, so the real invoice runs a few percent above this chain's answer.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.