At exit, v3-350 fps, y-7 ftlb Find (a) the speed v2 at section 2, and Thermodyna

Question

At exit, v3-350 fps, y-7 ftlb Find (a) the speed v2 at section 2, and Thermodynamics 1: Exam 2 Direction: Use A4 for your solutions. (b) the flow and area at the exit section. 7. If a pump discharges 75 gpm of water whose specific weight is 61.5 lb/ft (g-31.95 fps), find (a) the mass flow rate in lb/min, and (b) total time required to fill a vertical cylinder tank 10 ft in diameter and 12 ft high Problem Solving: Box your final answer. 1. The weight of an object is 50 lb. What is its mass at standard condition? 2. Five masses in a region where the acceleration due to gravity is 30.5 ft/s2 are as follows: m is 500 g of mass; m2 weighs 800 gcm weighs 15 poundals; m4 weighs 3 lbr m is 0.10 slug of mass. What is the total mass expressed (a) in grams, (b) in pounds, and (c) in slugs. 8. The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer, as shown in the figure. Determine the gage pressure of air in the tank if h 0.2 m, h2 -0.3 m and h3 0.46 m. Take the densities of water, oil and mercury to be 1000 kg/m3, 850 kg/m3 and 13,600 kg/m3 3. Two liquids of different densities (density 1 is 1500 kg/m. density 2 is 500 kg/m) are poured together into a 100-L tank, filling it. If the resulting density of the mixture is 800 kg/m find the respective quantities of liquids used. Also, find the weight of the mixture, local g -9.675 mps oil Air 4. The pressure of a boiler is 9.5 kg/cm2. The barometric pressure of the atmosphere is 768 mm of Hg. Find the absolute pressure in the boiler Water Mercury 5. Show that the specific heat of a substance in Btu/(lb)F) is numerically equal to cal (g)(C). 8 6. Two gaseous streams enter a combining tube and leave as a single mixture. These data apply at the entrance section: For one gas, Ar 75 in.2, vi500 fps, y-10 ftlb For other gas, A-50 in., m-16.67 lb/s, p 0.12 lb/ ft

Explanation / Answer

1. Given

weight of the object W = 50 lb

mas m = ?

we know that weight is W = mg

m = W/g

here first converting the weight into Newtons

W = 50 lb = 50*32.174 ft/s2 = 1608.7 ft*lb/s2

now one lb = 0.45359237 kg

m = (1608.7*0.45359237 )(1/9.80665) kg

m = 74.41 kg

2.

Given masses are  

m1 = 500 g

m2 = 800 g

m3 = 15 poundals

m4 = 13 lbf

m5 = 0.10 slugs

we know that 1 pound of mass = 453.592 g ===> 15*453.592 g = 6803.88 g

1 lbf = 4.44822 N ==> m = 4.44822/9.80665 kg = 0.45359 kg = 453.59 g

===> 13 lbf = 453.59*13 g = 5896.67 g

1 slug = 14593.9 g ==> 0.1 slug of mass = 1459.39 g

a) total mass in grams is  

M = 500+800+6803.88 + 5896.67 +1459.39 g

M = 15459.940 g

b) in pounds we know that 1 pound = 453.592 g

so x pounds = 15459.940 g

x = 34.0834

that is total mass in pounds is m = 34.0834 pounds

c)

we know one slug = 14593.9 g

then x slug = 15459.940 g

x = 1.0593426

total mass in slugs is m = 1.059343 slugs

3. Given densities are

d1= 1500 kg/m3

d2= 500 kg/m3

we know that 1 kg/m^3 = 0.001 kg/L

so densities are  

d1 = 1500*0.001 kg/L = 1.5 kg/L

d2 = 500*0.001 kg/L = 0.5 kg/L

resultant density of liquid is d = 800 kg/m^3 = 0.8 kg/L

volume of tank is v = v1+v2 = 100 L ,==> v1 = 100-v2 --------(1)

we know that the density d = m/v ==> m = d*V

here d1 = m1*v1 , d2 = m2*v2 and d = d1+d2 ---(2)

m*v = m1*v1+m2*v2

here m = d*v = 0.8*100 = 80 kg

so m = m1+m2 = 80 kg

m = m1+m2

80 = v1*d1+v2*d2 = (100-v2)d1 +v2*d2

substituting the values of densities  

80 = (100-v2)1.5 +v2*0.5

v2 = 70 L so v1 = 100-v2 = 30 L

masses m1 = d1*v1 = 1.5*30 = 45 kg, m2 = d2*v2 = 0.5*70 = 35 kg

total weight of the mixture is  

m = m1+m2 = 45+35 kg = 80 kg

weight is W = m*g = 80*9.675 N = 774 N

4.

Given boiler pressure is p1 = 9.5 kg/cm^2

barometer pressure p2 = 768 mm of hg

we know that 1 mm of hg = 0.00135951 kg /cm^2

s0 768 mm of hg = 1.0441 kg/cm^2

the absolute pressure is P = P1+p2 = 9.5+1.0441 kg/cm^2 = 10.5441 kg/cm^2

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