1.1 If certain gasoline weighs 7,000 N/m, what are the values of its density, specific volume, and specific gravity relative to water at 15°C? Use Appendix 3 Solution: Given that, Specific weight of gasoline =7000 N/m³ We have to find, Density, ρ =? At 15°C Specific volume, V =? At 15°C Specific Gravity, S=? At 15°C Density, ρ =999.1 kg/m³at 15°C & Specific Volume, V =1/ρ V =1/999.1 V =0.001 m³/kg & Specific gravity, S =гs/гw S=7000/9.810×1000 (As гw=9810 N/m³ at 15°C for water) S=0.714
1.2 A certain gas weighs 16 N / m ³ at a certain temperature and pressure. What are the values of its density, specific volume, and specific gravity relative to air weighing 12 N / m ³? Solution: Given that, Specific Weight of gas, г gas=16 N/m³ Specific Weight of air, гair =12 N/m³ We have to find, Density of gas relative to air, ρ =? Specific Volume of gas relative to air, V=? Specific Gravity of gas relative to air, S=? a) As, г=ρg ρ= г/g ρ=16/9.81 ρ=1.63 Kg/m³ b) As, V=1/ ρ
V=1/1.63=0.613 m³/kg c) As, S= г gas/ гair S=16/12 S=1.33
1.3 The specific gravity of glycerin is 1.26 .compute its density and specific weight. Solution: Given that, Specific Gravity, S =1.26 We have to find, Density, ρ =? Specific Gravity, гs/гw =? As, Specific Gravity, S = гs/гw гs= гw ×s гs =9810×1.26 (As гw=9810 N/m³ at 15°C for water) гs =12360.6 N/m³ As, г=ρg In our case, гs =ρg ρ= гs /g ρ=12360.6/9.81 ρ=1260 kg/m³
1.4 If the specific weight of a liquid is 8.0 kN / m ³, what is its density? Solution: Given that, Specific Weight, гs =8 KN/m³ = 8000 N/m³ We have to find, Density, ρ =? As, г=ρg ρ= гs /g ρ=8000/9.81 ρ=815.5 kg/m³
1.5 If the specific volume of a gas is 0.72 m ³/ kg, what is its specific weight in N / m ³? Solution: Given that, Specific Volume, V =0.72 m³/kg We have to find, Specific Weight, гs =? As, V=1/ρ ρ=1/V ρ=1/0.72 ρ=1.38 kg/m³ As, г=ρg г=1.38×9.81 г=13.54 N/m³ 1.6 Initially when 1000.00 m L of water at 10 ˚ C are poured into a glass cylinder the depth of the water column is 100.00 cm .The water and its container are heated to 80 ˚ C .Assuming no evaporation, what then will be the depth of the water column if the coefficient of thermal expansion for the glass is 3.6 x 10 ^ -6 mm/ mm per ˚ C? Solution: Given that, Volume of water, V=1000mL =0.001 m³ Specific Weight, г1 at 10°C =9840 N/m³ Specific Weight, г2 at 80°C =9530 N/m³ Height of Cylinder, h =100cm=1m As, Weight =Specific weight×Volume W=гV (г1V1)at 10°C=(г1V1)at 80°C V2=г1V1/г2
V2=9804×0.001/9530 V2=0.001028 m³ At 10°C, Volume of Cylinder, V=πr²h r²=v1/πh=0.001/π×1 r=0.01784m Now at 80°C, Volume of Cylinder, V=πr²h h=v1/πr² h=.001028/π×(0.01784)² h=1.028m=1028mm 1.7 Water in a hydraulic press is subjected to a pressure of 100 MN / m ², abs at 20 ˚ C. If the initial pressure is 100 kN / m ², abs, what will be the percentage decrease in specific volume? Use table 1.1. Solution: Given that, Initial Pressure, P 1=100 KN/m² Final Pressure, P 2 =100 MN/m² We have to find, Percentage decrease in specific Volume, -∆V/V=? Now, Bulk Modulus at 100 KN/m² & 20°C, E v1 =2130 MN/m² Bulk Modulus at 100 MN/m² & 20°C, E v2 =2730 MN/m² Mean of Bulk Modulus, Ev=(Ev1+Ev2)/2 Ev =(2130+2730)/2 Ev=2430 MN/m² As, -∆V/V=∆P/Ev -∆V/V=(P2-P1)/Ev -∆V/V=(100-0.1)×100/2430 -∆V/V=4.11% 1.8 At a depth of 8 km in the ocean the pressure is 81.8 Mpa .Assume specific weight at the surface is 10.05 KN / m ³ and that the average volume modulus is 2.34 * 10 ^ 9 N / m ² for that pressure range. (a)
What will be the change is specific volume between that at the surface and at that depth? (b) What will be the specific volume at that depth? (c) What will be the specific weight at that depth? Solution: Given that, Depth of ocean=8Km Pressure at 8Km, P2=81.8 MPa Pressure at top surface of ocean, P1=0Pa Specific weight at surface of ocean,гs=10.05 kN/m³=10050 N/m³ Bulk modulus,Ev=2.34 GN/m² We have to find, a) Change in specific volume,∆V=? b) Specific Volume at 8km,V2=? c) Specific Weight at 8km,гw=? a) As, V1=1/ρ=g/г=9.81/10050 V1=0.000976 m³/Kg As, -∆V/V=∆P/Ev ∆V=-∆P×V/Ev ∆V=-(81800000-0)×0.000976/2340000000 ∆V=-0.00003412 m³/Kg b) As, ∆V=V2-V1 V2=∆V+V1 V2=0.00003412+0.000976 V2=0.000942 m³/Kg c) As, г=ρg & ρ=1/V Then, гw=g/V2 гw=9.81/0.000941 гw=10425.08 N/m³
1.9 (a) what is the percentage change in the specific volume in Prob. 1.8? (b)What is the percentage change in the specific weight Prob.1.8? Solution: a) We have to find Percentage change in Specific Volume, As -∆V/V=∆P/Ev -∆V/V=(81800000×100)/2340000000 -∆V/V=3.5% b) We have to find change in Percentage change in Specific Weight, As, Percentage change in г = (гw-гs) ×100/гw Percentage change in г = (10425.08-10050)×100/10425.08 Percentage change in г =3.6%
1.10 To two significant figures what is the bulk modulus of water in MN/m ² at 50˚C under a pressure of 30 MN/m ²? Use table 1.1. Solution: Given that, Bulk Modulus, Ev=? At 50°C & P=30 MN/m² Now using table 1.1, Bulk modulus,Ev=2410 MN/m² 1.11 Approximately what pressure in N/m ² must be applied to water to reduce its volume 2 percent? Solution: Given that, Percentage decrease in water, -∆V/V=2%=0.02 Bulk Modulus for Cold water, Ev= 2200 MN/m² As -∆V/V=∆P/Ev ∆P=-∆V×Ev/V ∆P=0.02×2200 ∆P=44 MN/m² 1.12 A vessel contains 85 L of water at 10˚C and
atmospheric pressure .If it is heated to 70˚C what will be the percentage change in volume? What weight of water must be removed to maintain the volume at the original value? Use Fig. 1.1. Or Appendix 3 Solution: Given that, Volume of vessel at 10°C,V1 =85L=0.085 m³ Specific Weight at 10°C ,г1 =9804 N/m³ Specific Weight at 70°C ,г2=9589 N/m³ We have to find, Percentage change in Volume, ∆V/V1=? Weight of water required to be removed, W=? As, Weight =Specific weight×Volume W=гV (г1V1)at 10°C=(г2V2)at 70°C V2=г1V1/г2 V2=(9804×0.085)/9589 V2=0.087 m³ As, ∆V=V2-V1 ∆V=0.087-0.085=0.002m³ ∆V/V1=(0.002×100)/0.085=2.3% increase Weight of Original Volume of water,W2=г2×V1 W2=9589×0.085 W2=815.065 N Similarly, W1=г1×V1 W1=9804×0.085 W1=833.34 N As, ∆W=W1-W2 ∆W=833.34 - 815.065 ∆W=18.28 N (This weight should be removed from water) 1.13 A cylinder tank (diameter = 10 m and depth = 5.00 m) contains water at 20˚C and is brimful. If the water is heated to 50˚C, how much water will spill over the edge of the tank? Use Appendix 3. Solution:
Given that, Diameter of Cylinder,D =10 m Depth of Cylinder,d =5 m TemperatureT1=20°C Temperature after HeatingT2=30°C We have to find, Change in Volume, ∆V =? As, Volume of Cylinder, V1=πD²h/4 V1=π×(10)²×5/4 V1=392.7m³ Now From Appendix 3, Specific Weight at 20°C, гw1=9789 N/m³ Specific Weight at 50°C, гw2=9689 N/m³ Consider water at 20°C, As, W=г×V In our case, W=гw1×V1 W=(9789)(392.7) W=3.84 MN Consider water at 50°C, As, W=г×V In our case, V2=W/гw2 V2=3840000/9689 V2=396.75 m³ Now, Change in Volume, ∆V=V2-V1 ∆V= 396.75 - 392.7 ∆V= 4.05 m³ 1.14 A closed heavy steel chamber is filled with water at 10 ˚ C and at atmospheric pressure. If the temperature of the water and the chamber is raised to 30 ˚ C, what will be the new pressure of the water? The coefficient of thermal expansion of the steel is 1.7 x 10-5 per ˚ C. assume the chamber is unaffected
by the water pressure .Use fig.1.1. Solution: Given That, Initial temperature of Water in Steel chamber,T1 = 10°C Temperature After Heating,T2 = 30°C+273=300k We have to find, Pressure after Heating, P=? As we know that, г=gP/RT Or P=гRT/g--------- (i) & R=Ř/molecular mass For water molecular mass is 18 R=8314/18 R=462 Nm/Kg k Specific Gravity at 30°C,г=9764 N/m³ Putting required values in equation (i) P=гRT/g P=9764×462×303/9.81 P=139.32 MPa
1.16 If Natural gas has a specific gravity of 0.6 relative to air at 1000 mbar, abs and 15 ˚ C, what is its specific weight and specific volume at that same pressure and temperature .What is the value of R for the gas? Solution: Given that, Specific Gravity of Natural gas, S=0.6 Pressure of air, P=1000 mbar, abs=100000 Pa Temperature, T=15°C+273=288K We have to find,
a) Specific Weight of gas, г gas =? b) Specific Volume of gas, V gas =? c) Value of R=? a) As, г=gP/RT then, гair=(9.81)(100000)/(287)(288) гair=11.87 N/m³
(As value of R=287 NM/KgK for air)
As, S= г gas/ гair Or гgas=S г air г gas = (0.6)(11.87) г gas = 7.12 N/m³ b) As, г=ρg or г=g/V or V=g/г V= 9.81/7.12 V=1.3776 m/Kg c) As, г gas =gp/RT R=gp/ г gasT R=(9.81)(100000)/(7.12)(288) R=478.4 Nm/Kgk
1.28 A closed heavy steel chamber is filled with water at 10 ˚ C and at atmospheric pressure. If the temperature of the water and the chamber is raised to 30 ˚ C, what will be the new pressure of the water? The coefficient of thermal expansion of the steel is 1.7 x 10-5 per ˚ C. assume the chamber is unaffected by the water pressure .Use fig.1.1. Solution:
a)
viscosity of water at 20 ˚ C = 1.002 * 1000 N-s / m² Viscosity of water at 90 ˚ C = 0.315 * 1000 N-s / m² Ratio = viscosity of water at 20 ˚ C / viscosity of water at 90 ˚ C = 1.002 x 1000/ 0.315 x 1000 = 3.181
(b) Viscosity of crude oil at s= 0.925 from figure 1.2 at 15 ˚ C = 0.11 N-s / m² Viscosity of gasoline at s= 0.680 from figure 1.2 at 15 ˚ C = 0.00032 N-s / m² Ratio = viscosity of crude oil / viscosity of gasoline = 0.11/ 0.00032 = 343.75 (c) Viscosity of SAE 30 western oil at 150 ˚ C = 0.03 N-s / m² Viscosity of SAE 30 eastern oil at 150 ˚ C = 0.05 Ns / m² Viscosity of SAE 30 western oil at 25 ˚ C = 2.5 N-s / m² Viscosity of SAE 30 eastern oil at 25 ˚ C = 2 N-s / m² Ratio= (µ of SAE 30 western oil at 150 ˚ C - µ of SAE 30 eastern oil at 150 ˚ C) / (µ of SAE 30 western oil at 25 ˚ C - µ of SAE 30 eastern oil at 25 ˚ C) = (0.03 – 0.05) / (2.5 – 2) = 0.04 1.29 At 15 ˚ C what is the kinematic viscosity of the gasoline in Fig.1.3, the specific gravity of which is 0.680? Solution: Given that, T=15˚C Specific gravity = 0.680 From fig. 1.3, Kinematic viscosity = 4.5 x 10-7 m²/ s
1.30 To what temperature must the fuel oil with the higher specific gravity in Fig. 1.3 be heated in order that its kinematic viscosity may be reduced to three times that of water 4 ˚ C? Solution: kinematic viscosity of water at 4˚C = 1.527 x 10-6 m²/ s kinematic viscosity of fuel oil = kinematic viscosity of water at 4˚C / 3 = (1.527 x 10-6) / 3 = 5 x 10-7 m²/ s Now we see the figure 1.3 at the given viscosity of oil and see the temperature which comes to be approximately 375 ˚ C 1.31 The absolute viscosity of a certain gas is 0.0107 cP while its kinematic VISCOSITY IS 164 cSt, both measure at 1.013 mbar, abs and 95 ˚ C .Calculate its approximate molecular weight, and suggest what gas it may be? Solution: Viscosity of gas = 1.07 x 10-5 N-s/ m² kinematic viscosity = 164 cSt = 164 x 10 -6 m² / s Pressure = 1.013 mbar = 1.013 x 105 m / m² Temperature = 95˚C Density = Viscosity / kinematic = (1.07 x 10-5)/ (164 x 10-6) = 0.5944 kg / m³ We know that, г=gP/RT R = Pressure / (Density x Temperature) = 101300/ (368 x 0.5944) = 4631.08 N-m/ kg K Now, R=Ř/molecular mass Molecular mass x R = 8312 Molecular mass = 8312 / 4631.08 = approx. 2 Hence as the molecular mass comes to be 2 the gas in hydrogen
1.32 Compare the ratio of the absolute viscosities of air and water at 20 ˚ C with that of their kinematic viscosities at the same temperature and at 100 kN. Solution: Pressure = 100000 kN abs We have to find the ratio between, (Viscosity of air at 20 ˚ C/ viscosity of water at 20 ˚ C): (kinematic viscosity of air at 20 ˚ C / kinematic viscosity of water at 20 ˚ C) Seeing the values from Appendix 3, (1.002 x 10-3 / 1.81 x 10 -5): (1.003 x 10-6 / 1.50 x 10-5) 833: 1 1.33 A flat plate 30 cm x 50 cm slides on oil ( µ = 0.8 N-s / m ²) over a large plane surface .What force is required to drag the plate at 2 m/s ,if the separating oil film is 0.4 mm thick ? Solution: Given that, Plate = 30 cm x 50 cm Area of the plate = 30 x 50 = 1500 cm² = 0.15 m² µ = 0.8 N-s/ m² y = 4 mm= 0.004 m v = 2 m/sec As, Force= (A. µ. V / y) Substituting the values, F= (15 x 0.8 x 2) / 0.0004 = 600 N 1.34 A space of 25 mm width between two large plane surfaces is filled with SAE 30 western lubricating oil at 25 ˚ C. (a) What force is required to drag a very thin plate of 0.35 m ² area between the
surfaces at a speed of 0.1 m/s if this plate is equally spaced between the two surfaces? (b)If it is at a distance of 8.5 mm from one surface? Solution: Given that, Y= 25 mm SAE western oil T= 50 ˚ C A= 0.35 m² V= 0.1 m /sec a) As the plates are equally spaced hence , y1= y2 = 25 / 2 = 12.5 mm µ = 6 x 10 -2 N-s/ m² F= (A. µ. V / y) Substituting the values, F= 0.35 x 6 x 10-2 x 0.1 x (1000/ y1 + 1000/ y2) = 0.336 N b) F= (A. µ. V / y) = 0.35 x 0.006 x 0.1 x (1/8.5 + 1/ 16.5) = 0.374 N 1.35 A hydraulic lift of the type commonly used for greasing automobiles consists of a 254 mm-diam ram which slides in a 254.15 mm-diam cylinder, the annular space being filled with oil having a kinematic viscosity of 3.7 x 10-4 m ²/ s and specific gravity of 0.85. If the rate of travel of the ram is 0.15 m/s, find the frictional resistance when 3m of the ram is engaged in the cylinder. Solution: D1 = 254mm = 0.254m , r1 = 0.127 m D2 = 254.15mm = 0.25415m , r2 = 0.127075m Kinematics viscosity v = 3.7 x 10 – 4 m 2/s du = 0.15 m/s Specific gravity S = 0.85 L = 3m Frictional Resistance = ? We know that
V=u/p U = v /p -------- 1 Where P = Pw x S = 1000 x 0.85=850 So, 1 becomes U = 3.7 x 10 – 4 x 850= 0.314 m 2/s Also we know that, F = u x du /dy x A Where u = 0.314 m2/s A = 3.142 x D x L A= 3.142 x 0.254075 x 3 A= 2.395 m2 dy = r2 – r1 dy= 7.5 x 10 –5
----------------- 2
So, 2 becomes F = 0.314 ( 0.15 / 7.5 x 10-5) x 2.395 F=1504.06 KN
1. 41 In the figure, oil of viscosity µ fills the small gap of thickness Y. Determine an expression for the torque T require to rotate the truncated cone at constant speed ŵ. Neglect fluid stress exerted on the circular bottom. Solution: As, U= ωr
Then, For small gap Y, du/dy=U/Y=ωr/Y
As, =זµdu/dy=µωr/Y---------(a) dA=2πrds=2πrdy/cos ά From equ (a) df= זdA =(µωr/Y)( 2πrdy/cos ά) dT=rdF=2πµωr3dy/Y cos ά r = y tan ά dT=(2πµωtan3α y3/Ycos α)dy T= (2πµωtan3α /Ycos α) ∫ y3dy T=(2πµωtan3α /4Ycos α)[(a+b)4 – a4} a+b a
1.42 Distilled water at 10˚C stands in a glass tube of 8.0 mm diameter at a height if 25.0 mm. What is the true static height? Solution: Given that, T = 10 oC D = 8 mm Hr = 25mm Static height, Hs =? We know that H = 4ơcosθ/(rD)
-------- 1 Surface Tension (ơ) at 10o C = 0.0742 N/m (From Appendix # 3) rw at 10oC =9.804 KN/m3 θ for pure water = 0o So 1 becomes H = 4(0.0742)cos0o / (9.804 x 1000 x 0.008) H= 3.784 x 10 –3 m H= 3.784 mm So, Hs = Hr – H Hs = 21.216 mm.