A garden hose sits out in the sun on a hot surface for the p

A garden hose sits out in the sun, on a hot surface, for the purpose of warming water for an outdoor shower. The inner diameter of the garden hose is 3/8in. The surface temperature is constant at 36C . The inlet temperature to the hose is 12C, and the flow rate is 2 gallons per minute.

(a) Find the mean velocity in the hose.

(b) Calculate the Nusselt number.

(c) Solve for the required length of the hose, if the desired outlet temperature is 35C.

(d) Solve for the quantity Ts(x) Tm(x).

(e) Plot the mean and surface temperature distributions along the length found in (c).

Assume fully developed flow with constant properties evaluated at the inlet temperature. Also note, this will require some conversion to metric.

Solution

1

a)

Flow rate Q = 2 gal/min = 10.092*10^-5 m³/s

Cross-section area A = pi/4 * d² = 3.14 / 4 * (3/8)² = 0.1104 in² = 0.7123*10^-4 m²

Velocity V = Q / A

= (10.092*10^-5) / (0.7123*10^-4)

= 1.417 m/s

b)

For water, kinematic viscosity at 12 deg C, neu = 1.3*10^-6 m²/s and Prandtl number Pr = 8.86

Dia of pipe D = 3/8\" = 0.009525 m

Reynolds number Re = V*D / neu

= 1.417 * 0.009525 / (1.3*10^-6)

= 10382

Since Re is greater than the critical value of 2300, flow is turbulent.

For turbulent pipe flow with uniform heat flux, Nusselt number Nu = 0.023*Re^0.8Pr^0.4

Nu = 0.023*10382^0.8 * 8.86^0.4

Nu = 89.89

c)

Nu = hD/k

For water, thermal conductivity k = 0.6 W/m-K

89.89 = h*0.009525 / 0.6

h = 5662.6 W/m²-K

Surface area S = pi*D*L = 3.14*0.009525 * L

For constant wall temperature, h = -(m*Cp / S) ln [(Tsurf - Tout) / (Tsurf - Tin)].........where m = mass flow rate = rho*Q......where rho is density of water = 1000 kg/m3 and Cp for water = 4184 J/kg-K

5662.6 = -[1000*10.092*10^-5 * 4184 / (3.14*0.009525 * L)] ln [(36-35) / (36-12)]

L = 7.92 m = 25.98 feet

d)

Tsurf - Tm(x) = (Tsurf - T_in)*ln [-h*(3.14*D*x) / (m*Cp)]

= (36-12)* ln[-5662.6*3.14*0.009525*x / (1000*10.092*10-5 * 4184)]

= 24 * ln (-0.4*x)