Please help with heat Transfer homework A garden hose sits o

Please help with heat Transfer homework

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 5/8in and it receives heat flux from the sun at a rate of 1000W/m^2. The inlet temperature to the hose is 12 degree C, and the flow rate is 0.4 gallons per minute. (a) Find the mean velocity in the hose. Calculate the Nusselt number. Solve for the required length of the hose, if the desired outlet temperature is 35 degree C. Solve for the quantity Ts(x) - Tm(x). 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. Repeat problem 2 with the following changes: The flow rate is 2 gpm The hose inner diameter is 3/8in Instead of constant surface heat flux, assume the surface temperature is constant at 36 degree C

Solution

1

a)

Flow rate Q = 0.4 gal/min = 2.523*10^-5 m³/s

Cross-section area A = pi/4 * d² = 3.14 / 4 * (5/8)² = 0.30664 in² = 1.978*10^-4 m²

Velocity V = Q / A

= (2.523*10^-5) / (1.978*10^-4)

= 0.1276 m/s

b)

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

Dia of pipe D = 5/8\" = 0.015875 m

Reynolds number Re = V*D / neu

= 0.1276 * 0.015875 / (1.3*10^-6)

= 1558

Since Re is less than the critical value of 2300, flow is laminar.

For laminar pipe flow with uniform heat flux, Nusselt number Nu = 4.36

c)

Surface area S = pi*D*L = 3.14*0.015875 * L

Heat input = Heat flux * surface area

= 1000 * (3.14*0.015875 * L)

Also, Heat input = m*Cp*(Tout - 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

1000 * (3.14*0.015875 * L) = 1000*(2.523*10^-5)*4184*(35 - 12)

L = 48.71 m = 159.8 feet

d)

Nu = hD/k

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

4.36 = h*0.015875 / 0.6

h = 164.79 W/m²-K

Heat flux q = h*(Ts - Tm)

1000 = 164.79*(Ts - Tm)

Ts - Tm = 6.068 deg C