CHEMICAL ENGINEERING

FLOW AND PRESSURE MEASUREMENT 87

PD 5. 74 ð 107 ð 1 / 1000 

D 5. 74 ð 104 N/m^2 or 57.4kN/m^2

Power requiredDhead loss (m)ðGðg
D 5. 74 ð 104 / 1000 ð 9. 81  5. 97 ð 9. 81 D343 W

For a pump efficiency of 60%, the actual power requirementD 343 / 0. 6 D571 W.

Water velocityD 5. 97 / 0. 00197 ð 1000 D 3 .03 m/s.

For this low-power requirement at a low head and comparatively low flowrate, a
centrifugal pump, electrically driven and made of stainless steel, would be suitable.

PROBLEM 6.12

A rotameter has a tube 0.3 m long which has an internal diameter of 25 mm at the top
and 20 mm at the bottom. The diameter of the float is 20 mm, its effective density is
4800 kg/m^3 , and its volume 6.6cm^3. If the coefficient of discharge is 0.72, at what
height will the float be when metering water at 100 cm^3 /s?

Solution

See Volume 1, Example 6.4.

PROBLEM 6.13

Explain why there is a critical pressure ratio across a nozzle at which, for a given upstream
pressure, the flowrate is a maximum. Obtain an expression for the maximum flow for a
given upstream pressure for isentropic flow through a horizontal nozzle. Show that for
air (ratio of specific heats,D 1 .4) the critical pressure ratio is 0.53 and calculate the
maximum flow through an orifice of area 30 mm^2 and coefficient of discharge 0.65 when
the upstream pressure is 1.5MN/m^2 and the upstream temperature 293 K.
Kilogram molecular volumeD 22 .4m^3.

Solution

The reasons for critical pressure ratios are discussed in Section 4.4.1. The maximum rate
of discharge is given by:

GmaxDCDA 0

√

kP 1 /v 1  2 /kC 1 kC^1 /k^1  (equation 6.29)

For an isentropic process,kDD 1 .4forair.

The critical pressure ratio, wcD 2 /kC 1 k/k^1  (equation 4. 43 

Substituting forkDD 1 .4,wcD 2 / 2. 4 ^1.^4 /^0.^4 D 0. 523