ES202 – Fluid & Thermal Systems – Winter 2007-2008 Homework Answer and Hints Updated 9:00 am – 14 February 2008
Set	Problem	Answer	Hint
4	2-40	110 psia	Apply conservation of linear momentum to the pistons
	2-62	3.39 kPa
	2-69	562.5 kgm^3
5	2-96	12.0 kPa; 2.0 m
	10-16	30,900 lbf	Use hydrostatic force concepts to find force of water on gate and then use this information along with cons. of linear momentum to analyze the gate.
6	10-34	1050 N
	10-45	9.64 x 10^8 N; 17.1 m; 8.35 x 10^8 N	This is the total force (sum of the force due to the atmospheric pressure AND the water). The answer would be different if you just used the force due to the water alone. If you were attempting to study the dam to see if the water will move the dam, the forces due to atmospheric pressure would just cancel out. HOWEVER, if there was a very small leak under the dam, the hydrostatic force pushing up vertically on the dam would increase the possibility of tipping the dam.
7	12-27	2.43 m/s	Apply Bernoulli between the free surface of the gasoline and the leak opening in the tank wall.
	12-35	0.52 in	Apply Bernoulli (correctly) along any single streamline between section 1 and section 2. Section 1 and section 2 are located directly over the opening of the manometer. Recognizing that the pressure variation across parallel streamlines is hydrostatic.
8	12-41	6.24 cm	Apply Bernoulli (correctly) along any single streamline between section 1 and section 2. Section 1 and section 2 are located directly over the opening of the manometer. Recognizing that the pressure variation across parallel streamlines is hydrostatic.
	12-42	4.48 ft^3/s	Apply Bernoulli (correctly) along any single streamline between section 1 and section 2. Section 1 and section 2 are located directly over the opening of the manometer. Recognizing that the pressure variation across parallel streamlines is hydrostatic.
	12-46	33.8 m/s
9	12-61	68.7 kPa	Apply mechanical energy balance between two locations where information is known so that you have one equation and one unknown.
	12-64	0.204 m, 40 W	The useful pumping power is just the pumping power that would be required to overcome the losses in the fitting. Physically this is the size of a pump required to circulate a fluid at the specified rate in a closed loop that consists of only the fitting and a pump.
	12-56	2.4 W; 6.9 cm; 40 Pa	Fan-motor efficiency is the ratio of the (Power into the air from the fan)/(Electrical power into the fan motor). Apply mechanical energy balance.
11	12-65	55 kW	Mechanical energy balance.
11	12-69	50.9 m/s; 201 kW	Mechanical energy balance.
12	12-63	Pump 1150 kW; Turbine 530 kW	Mechanical energy balance
12	14-38	69.9 kPa; 4.40 kW	Friction factor relation. Is the flow laminar OR turbulent? It makes a difference
13	14-42	14.2 lbf/ft^2; 0.37 W	Friction factor relation. Is the flow laminar OR turbulent? It makes a difference.
13	14-45	124 Pa	Friction factor relation. Is the flow laminar OR turbulent? It makes a difference.
14	14-61	320 kPa; 27 kPa	These differ from the answers in the textbook because the author forgot to include the kinetic energy correction factor alpha in the KL value for the expansion loss.
14	14-79	hL = 65.8 m; P1 = 734 kPa	Write the MEB between the free surfaces of the two tanks, but recognize that the pressures at the free surfaces are not equal.
15	14-87	217 W	Write MEB between the pump inlet and the pump inlet. You are left with an expression for w_pump in terms of the gh_loss.
15	14-75	0.00595 m^3/s	If you write the MEB between 1 and 2, you end up with a single equation that ONLY depends on the velocity in one of the pipes ---- that means if I gave you the velocity you could solve for everything else in the equation. HOWEVER, only one unique value of the velocity will satisfy the equation! So this requires an iterative solution like EXAMPLE 14-5 in the textbook. Think of moving everything to one side of the equal sign … 0 = F(V) where there is only one V value that will make this true --- V is the root of the equation. So you can plot (solve) for F(V) versus V and see when F(V)=0.
16	14-80	2.96 kW	Write the MEB between the free surface of the oil in the reservoir and the oil in the tanker. Then evaluate the major and minor losses to find the pump work assuming no losses in the pump. Finally calculate the actual shaft work into the pump using the pump efficiency.
16	14-86	0.87 ft^3/s	The inlet to your system is somewhere in the middle of the room upstream of the fan – the fan is inside the dryer. To find the power you solve the problem with no piping attached to the dryer. Then you assume the power is unchanged and solve for the new flow rate when the piping is attached.
21	5-35	153 min	Closed system energy balance. Water is not an incompressible substance or ideal gas.
21	6-43	1.31 kg/s	Steady-state energy balance and mass balance. R-134a is not an incompressible substance or ideal gas.
21	6-55	182 Btu/s	Steady-state energy and mass balance. Water is not an incompressible substance or ideal gas.
22	6-49	Approx. 12 kg/s & 1.6 m^2	Conservation of mass and ideal gas relation
22	6-89	Approx. 103 kW and 133^oC
22	8-168	Approx. 0.17 kg/s and 0.33 kW/K	Water as a real substance. Conservation of energy and mass, and entropy accounting.
23	8-134	Approx. 72 Btu/lbm	Air as ideal gas and definition of turbine efficiency
23	8-135	Approx. 59^oC and 1.7 kW	R-134a as real substance and definition of compressor efficiency.
24	23-89	bwr = 50.1%, η=29.6%	Assume ideal gas and as requested in the problem use the ideal gas equations with constant-cp values evaluated at room temperature. THUS you are assuming ideal gas but NOT using the ideal gas tables for h, s^o, etc.
24	23-93	659 kW, BWR = 0.625, η=0.319	Assume ideal gas and use the ideal gas tables. This problem has a WRINKLE that makes it slightly more difficult. First you must recognize that the turbine exhaust is a 100 kPa. Second you will discover that T3 is unknown which means you can’t find the heat transfer for the combustor or the work for the turbine directly. To find T3 you have to look at the turbine performance. With T4 and turbine efficiency known you have sufficient information to find the inlet temperature. Here is how ---- … Using turbine efficiency à h3 – h4 = ηturbine (h3 – h4s) where there are TWO unknowns: T3 and T4s … Using what we know about how state 4s and state 3 are related: s4s-s3 = 0 = s^o4s – s^o3 – R [ln(P4s/P3)] which also has two unknowns: T3 and T4s. Solution requires solving these two equations BUT this requires an “iterative solution”: (1) guess T3 (or T4s) and then solve for T4s (or T3) using both equations; (2) compare the value you get from both equations and adjust your guess so that you get the same values from both equations. An alternative approach is to guess T3 and then calculate an efficiency; when your efficiency matches the given one you know your finished. How you proceed is immaterial and usually based on what things are easiest to start guessing, e.g. T4s < T4.
25	23-130/131	130 Turbine – 699.3 kJ/kg Boiler – 2177.8 kJ/kg 31.4% 131 Turbine 0.877 Cycle 0.274	Use the “steam” tables for water (real substance) and work your way around the cycle. Problem 130 is the ideal cycle. Problem 131 includes a inefficient turbine. Problem 131 is just a minor modification of 130. DO NOT rework the problem from scratch use what you’ve already found from 130.
25	23-136	38.9%; 36.0 kg/s; 8.4^oC	Use the “steam” tables for water (real substance). Need to find the net work per unit mass for the cycle before you can find the mass flow rate of water in the cycle. Also treat the cooling water in part (c) as an incompressible substance with a constant c value (just what we did in ES201). If necessary you may neglect the pressure drop in the cooling water loop.
27	15-27	231 N	CD value comes from Table 15-2. Density calculated using ideal gas equation. Notice that under these conditions the drag does not depend, at least directly, on the fluid viscosity. Because shear force acting on the circular disk does not have a component in the direction of the drag force, we would say that the drag force is purely pressure drag for this geometry.
27	15-65	54.4 N	CD for a cylinder in cross flow is found in Figure 15-34 as a function of ReD. If you had not been given the velocity but instead had been given the drag force, you would have three equations and three unknowns that must be solved simultaneously: Force balance on the cylinder (CD and V), Figure 15-34 (CD and ReD) and the definition of Reynolds number (ReD and V).
27	15-68	0.136 m/s; 42.9 min	Solution depends on applying conservation of linear momentum to find the steady-state (or terminal velocity). The drag force can be calculated using the Stokes law formula directly or by using the drag coefficient CD = 24/ReD for Stokes flow (very low Reynolds number). After you complete your solution you should check the ReD for the flow to see if it satisfies the limits for Stokes flow. If not then you must use Figure 15-34 that gives CD vs ReD for the complete range of Reynolds numbers.
29	15-47	5.87 lbf	Check the plate Reynolds number, ReL, first to see whether the flow on the plate is laminar or combined laminar-turbulent flow. Then apply the appropriate average skin-friction-coefficient equation to find Cf. Once you know the average skin-friction coefficient you can calculate the drag force.
29	25-52	15.9 N	Same approach as for the problem above.
30	15-88	CD = 0.0235 CL = 0.17	Density of standard air at an altitude of 4000 m is 0.819 kg/m^3. (This table was left out of the book. Value based on standard temperature and pressure at 4000 m in the standard atmosphere). Drag force at cruising conditions found from the basic relation between power, force, and velocity. Once the drag force is known then use the definition of the drag coefficient to calculate a value. Same goes for the lift coefficient once you know the weight and cruising speed.
30	15-89	3^o	Figure 15-43

Updated 5 December 2007 by D. E. Richards