A circular pipe of 25-mm outside diameter is placed in an airstream at 25 °C and 1-atm pressure. The air moves in cross flow over the pipe at 15 m/s, while the outer surface of the pipe is maintained at 100 °C. a) What is the drag force exerted on the pipe per unit length of the pipe? (50 pts) b) What is the rate of heat transfer from the pipe per unit length of the pipe? (50 pts)

A) the drag force exerted on the pipe per unit length of the pipe is; f_d = 3.24 N/m

B) The rate of heat transfer from the pipe per unit length of the pipe is; Q' = 518.36 W/m

We are given;

Diameter of pipe; D = 25 mm = 0.025 m

Pressure; P = 1 atm

Velocity; V = 15 m/s

Surface temperature; T_s = 100°C

Airstream Temperature; T_∞ =  25 °C

Now, from table of properties of air;

Kinematic viscosity of air; ν = 19.31 × 10⁻⁶ m²/s

k = 0.0288 w/m.k

Density; ρ = 1.048 kg/m³

Formula for Reynold's Number is;

Re = VD/ν

Thus; Re = 15 × 0.025/(19.31 × 10⁻⁶)

Re = 1.941 × 10⁴

f_d = C_d * D * ρ * V²/2

where C_d is drag coefficient and since this is a smooth cylinder, from online tables, the drag coefficient is 1.1. Thus;

f_d = 1.1 * 0.025 * 1.048 * 15²/2

f_d = 3.24 N/m

Q' = h'(πD) * (T_s - T_∞)

Formula for h' the convection heat transfer coefficient is;

h' = (kC/d) * R_e * m * Pr^(¹/₃)

where;

The Prandtl number; Pr = 0.702

C = 0.193

m = 0.618

Thus;

h' = (0.0288 * 0.913/0.025) * 1.941 × 10⁴ * 0.702^(¹/₃)

h' = 88 w/m².kg

Thus;

Q' = 88(π * 0.025) * (100 - 25)

Q' = 518.36 W/m

Read more on drag force at; link

a) Fd = 3.24 N/m

b) Q = 520 w/m

Explanation: please find the attached files for the solution

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

answer:

yes that is right have a good day