Question

9-1 The above office requires 2,000,000 BTUH heating for exterior zones. A hot water heating system with a circulating pump is used for water circulation. Assume Heating coils have inlet temp 180 F outlet temp 140", and pump head is 30 FT. Answer the following questions: (3 points) (a) The required total hot water flow rate in GPM (b) BHP if the pump selected has an efficiency of 50% (c) what is the water main pipe size

Answer 1

Amount of heat to be supplied to the exterior zones is,

Q = 2,000,000 BTUH

This heat is supplied by flow of hot water through the heating system

So, $\dot{m}$$c_{p}$_{$\Delta T$ = Q}

Given, Q = 2,000,000 BTUH = 586.14207732 kW

$c_{p}$ = 4.186 kJ/kgK

_{$\Delta T$ = 180 - 140 = 40}

$\dot{m}$ = 586.14207732 kW/4.186 kJ/kgK $\times$ 40

$\dot{m}$ = 3.5 kg/s

Volume flow rate = $\dot{m}$/$\rho$ = 3.5/1000 = 0.0035 m³/s

Use conversion, 1 cubic meter/second = 13198.1549 gallon (UK)/minute

(a) Total hot water flow rate in GPM (UK gallons per minute) = 0.0035 $\times$ 13198.1549 = 46.2 GPM

(b) BHP = TDH $\times$ GPM/Efficiency $\times$ 3960

where TDH is the total discharge head measured in feet (30ft in this case)

GPM flow rate in gallons per minute (46.2 GPM in this case)

Efficiency of the pump is given as 50%(0.5 in this case)

and, 3960 is the conversion number we get by dividing 8.333 (the weight, in pounds, of one gallon of water) into 33,000 ( foot pounds in one horsepower).

Therefore, BHP = 30 $\times$ 46.2/0.5 $\times$ 3960 = 0.7BHP

(c) Let D be the diameter of the main pipe.

Area, A = = 0.7857D²

T D2/4

Using formula, $\dot{m}$ = $\rho$AV, where V is velocity through the main pipe.

Some information is missing as there are 2 unknowns in the above equation.