The fuel pump in a classic car forces gasoline from the gas tank to the carburet

Question

The fuel pump in a classic car forces gasoline from the gas tank to the carburetor through a 10-foot-long, 1/4 - inch fuel line, as shown in the figure. The line has five 90 degree smooth bends with an R/D = 4 (K_L = 0.21). The gasoline (T = 60 degree F) discharges through a nozzle creating a 1/32 - inch jet in the carburetor to a pressure of 14 psia. The pressure in the tank is 14.7 psia. The pump is 80% efficient. What power must be supplied to the pump if the engine is consuming fuel at a rate of 0.12 gpm? Refer to gasoline properties data reported in Table 1.5 (MYO).

Explanation / Answer

1) Power gained by gasoline from pump can be expressed as

P= mw

P is power

M is mass flow rate (kg/s)

W is specific work (Nm/kg, j/kg)

2) specific work W is given by:

W=gh

h-head (mass)

g - gravity (9.81 m/s^2) (32.174ft/s^2)

3) Mass flow rate:

m = pQ

p = density (kg/m^3)

Q = volume flow rate (m^3/s)

Combining 1,2&3,

Power gained is given by,

P = pQgh

pg = Y = specific weight (N/m^3)

So P=YQh

head is given by,

h = p2-p1/Y

So, P = Q(p2-p1)

Now, here pressure in tank = 14.7 psia

Pump efficiency n = 80% = 0.8

Volume rate = 0.12gpm

T = 60°F

Pressure required to discharge = 14 psia

p density of gasoline = 42.5 lb/ft^3 / 32.174 ft/s^2 = 1.32094 lb/ft2

h = p2 -p1/y = 14.7-14/42.5 = 0.01647

P = YQh = 42.5×0.12×( 0.7/42.5)

= 0.0834 bhp

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.