Lecture 22 Mechanical Energy Balance Contents Exercise 1 Exercise 2 Exercise 3 Key Words: Fluid flow, Macroscopic Balance, Frictional Losses, Turbulent Flow Exercise 1 It is proposed to install a fan to draw air at rest in a horizontal straight duct of 250m m 350mm cross section. The duct is 60 m long. Figure 1: Horizontal straight duct The air enters the duct at 30 m min measured at 298 K and 755 mm Hg. Calculate the horse power of the fan, if the fan discharges air at 755 mm Hg pressure. 1mmHg 133.2 N m. Friction loss due to contraction and expansion are 0.4 and 1 respectively. K Viscosity of air 1.78 10 SOLUTION Applying mechanical energy balance between plane 1 and 2 and noting that Z Z (duct is horizontal), P P (inlet pressure at 1 = exit pressure at plane 2 and V V (velocity at plane 1 = velocity at plane2), we get. F M 0 (1)
F2f L D By 1 and 2 we get V ef V ef V (2) M V 2f L ef D ef (3) F is friction factor and is a function of Reynold s number. Re D V µ (4) De.... K 0.292 m and ρ 1.18 RT Inserting the values of De, ρ, µ and V (V is velocity of air in duct), in eq. 4, we get Re 1.105 10 Flow is turbulent and we use f 0.0791 Re. In the equation 3 substitute the values of e, e, L,De,f,V, to get. M80.94 per kg of fluid. Power 80.94 0.5 1.18 0.048 kw Now 1 kw 1.341 hp horse power Horse power of fan 0.064 hp. Exercise 2 A fan draws exhaust gases at 800 from the hood of a furnace as shown below Figure 2: Arrangement of hood to discharge exhaust gases
The exhaust gas flow rate is at 1 atm and 298 k The duct is rectangular cross section 0.2m 0.3m and is joined by an elbow as shown in the figure. Total length of the duct is 90 m. Calculate horse power of the fan from the following data Friction losses due to contraction and expansion are 0.4 and 1 respectively. µ ρ 1.7710 m s at 1073 K. Use f 0.0791Re to calculate friction factor. Equivalent length for elbow L 20. D Hint: Apply mechanical energy balance between plane 1 and 2 and get the following expression g Δ Z F M (5) g Δ Z V 2f L L ef D D ef (6) Substituting the values. Power = 8.8 hp. Exercise 3 Apply mechanical energy balance equation to calculate velocity of gas flowing in a pipe Velocity of gas flowing in a pipe is calculated by measuring the difference between the static pressure and the impac t pressure by the pitot tube at a given point in the flow. The pitot tube consists of two openings: impact and static. Impact opening is directed to receive the impact of the flow and the static opening remains at parallel to the direction of flow. Mechanical energy can be applied at plane 1 which is upstream from the impact point and plane 2 just at the impact point to find the relationship between pressure difference and velocity. Kinetic energy of the gas is converted to pressure at plane 2. At point 1 velocity is known and the pressure is that determined by static opening of pitot tube. At point 2 velocity is zero and pressure is that detected by impact opening. Mechanical energy balance simplifies to V P 0 P (7) V P P V (8)
The frictional losses are taken into account by the discharge coefficient C P which depends on the design of impact and static openings of the pitot tube. Thus eq. 8 is V C P P P (9) Note that the pitot tube measures the pressure at a particular point in the flow and the velocity will also correspond to that point. In order to obtain the complete velocity profile, it is necessary to traverse the pitot tube radially in order to be able to measure the pressure and then to calculate the velocity. The following relations can be used to calculate the average velocity: For laminar flow V V 0.5 0 Re 2100 (10) and in turbulent flow region 10 10. V 0.62 0.04 log D V V µ (11) For gases at low speeds 60 / and isothermal conditions we may use eq. 8 as well. At higher velocity of gases, density of the gas is not constant and Bernouli equation is to be written in a differential form i.e. P P P V (12) For the compressible fluid flowing under adiabatic conditions and assuming ideal gas law, the relation between P and is P constant (13) Where γ is the isentropic exponent of the gas. Its value is 1.3 for monoatomic, 1.4 for diatomic and 1.669 for He and argon. By 12 and 13 we get. V V C P P 1 P (14) P Assignment: 1) It is proposed to install a fan to draw air at rest in a horizontal straight duct of 250mm 350mm cross section. The duct is 60 m long. The air enters the duct at 30 m min measured at 298 K and 755 mm Hg. Calculate the horse power of the fan, if the fan discharges air at 755 mm Hg pressure.
1mmHg 133.2 N m. Friction loss due to contraction and expansion are 0.4 and 1 respectively. K Viscosity air 1.78 10 Figure 1: Horizontal straight duct 2) A fan draws exhaust gases at 800 0 C from the hood of a furnace as shown below Figure 2: Arrangement of hood to discharge exhaust gases The exhaust gas flow rate is at 1 atm and 98 k The duct is rectangular cross section 0.2m 0.3m and is joined by ass elbow as Shown in the figure. Total length of the duct is 90 m. Calculate horse power of the fan from the following data Friction losses due to contraction and expansion are 0.4 and 1 res pectively. µ ρ 1.7710 m s at 1073 K. Use f 0.0791Re to calculate friction factor. Equivalent length for elbow L 20. D