PPE - PROBLEM SET # 1
INSTRUCTION: Encircle the letter that corresponds to the correct answer of your choice.
MULTIPLE CHOICE:
1. A vacuum gage connected to a tank reads 30 kpa at a location where the barometric readi
...
PPE - PROBLEM SET # 1
INSTRUCTION: Encircle the letter that corresponds to the correct answer of your choice.
MULTIPLE CHOICE:
1. A vacuum gage connected to a tank reads 30 kpa at a location where the barometric reading is 755 mm Hg.
Determine the absolute pressure in the tank.
A. 70.6 kpaB. 84.23 kpa C. 90.34 kpa D. 98.45 kpa
Pvac= 30 Kpa Patm = 755 mm Hg
Pabs = Patm- Pvac
Patm= ( 755mm Hg x 101.325kPA/ 760 mm Hg) - 30 kpa
Pabs = 70.65 Kpa
2. Determine the pressure exerted on a diver at 30 m below the free surface of the sea. Assume a barometric
pressure of 101 kpa and the specific gravity of sea water is 1.03.
A. 404 kpa B. 410 kpa C. 420 kpa D. 430 kpa
h = 30 m S.G = 1.03 Patm= 101 kpa
S.G= ρfluid/ ρwater
P = ( ρwater)(S.G)(h)+ Patm
= (9.81KN/m3)(1.03)(30m)+101 kpa
P = 404.129kPa
3. A compressed spring above the piston exerts a force of 60 N on the piston. If the atmospheric pressure is 95
kpa, determine the pressure inside the cylinder.
A. 108.20 kpa B. 112.56 kpa C. 123.4 kpa D. 137.40 kpa
4. If the absolute temperature is 300oK, what is the equivalent in oR?
A. 530.34oR B. 534.23oR C. 537.20 oR D. 540.6oR
°K = 300 °R =? °F= 9/5 (°C) +32 °R = °F + 460
°K =°C+273 = 9/5(27°) +32 = 80.6 + 460
°C = 27° °F= 80.6 °R = 540.6
5. If the oF scale thrice the oC scale, what are the value oF and oC?
A. 26.67oC and 50oF B. 26.67oC and 80oF C. 16.67oC and 70oF D. 56.67oC and 60oF
°F= 9/5 (°C) +32 °F = 3°C
°F = 3°C °F = 3(26.67)
°F = 80
3°C = 9/5 (°C) +32
°C = 26.67
6. Water enters the heater at 30oC and leaves at 150oF, what is the temperature difference in oC?
A. 25.55oC B. 35.55oC C. 45.55oC D. 55.55oC
T1 = 30°C T2 = 150°F T = (T2-T1)
°C = 5/9 (°F-32) = (65.55-30)
= 5/9 (150-32) T = 35.55°C
°C = 65.55
7. A 5 kg plastic tank that has a volume of 0.2 m3 is filled with liquid water. Assuming the density of water is
1000 kg/m3, determine the weight of the combine system.
A. 195 kg B. 200 kg C. 205 kg D. 210 kg
m = 5 kg v = 0.2 m3
Ww = mass/volume
mw = (Ww) (v)
mw= (1000 kg/m3) (0.2 m3)
mw = 200kg
mT =mw+m = 200 kg + 5kg
mT= 205kg
8. Determine the mass of the air contained in a room whose dimensions are 15 ft x 20 ft x 20 ft. Assume the
Density of the air is 0.0724 lb/ft3.A. 11.49 slugs B. 13.49 slugs C. 15.49 slugs D. 17.49 slugs
L = 15ft; W = 20ft; H = 20ft ρ = mass/volume
A = (L) (W) (H) m = (ρ) (v)
= (15) (20) (20) = (600ft3) (0.0724 lb/ft3)
A = 600 ft3 = (434.4 lb) (1 slug/32.174 lb)
m = 13.50 slug
9. The water flows in the channel 200 mm x 300 mm at the rate of 2 m/sec. What is the volume flow in
ft3/sec? A. 1.24 ft3/s B. 2.24 ft3/s C. 3.24 ft3/s D. 4.24 ft3/s
L = (200mm) (1m/1000mm) (3.281ft/1m) Vf = AV
= 0.6562ft = (L) (W) (V)
W = (300mm) (1m/1000mm) (3.281ft/1m) = (0.6562ft) (0.9843ft)
(6.562ft/s)
= 0.9843ft Vf = 4.24 ft3/s
V = (2m/s) (3.281ft/1m)
= 6.562ft/s
10. Two gaseous streams are mixed together with one stream contains a diameter of 120 mm and specific
Gravity of 0.86 and speed of 5 m/s and on the other stream is 150 mm and density of 890 kg/m3 at 2 m/s.
Find the mass flow rate at the exit if diameter of exit is 200 mm.
A. 65 kg/s B. 70 kg/s C. 75 kg/s D. 80 kg/s
D1 = (120mm) (1m/1000mm) m1 = (A1) (V1) (SG1 xWw)
= 0.12 m = (π/4) (0.12m) 2 (5m/s) (0.86x 1000kg/m3)
D2 = (150mm) (1m/1000mm) m1 = 48.631 kg/s
= 0.15m m2 = (A2) (V2) (W2)
V1 = 5m/s = (π/4) (0.15m) 2(2m/s) (890 kg/m3)
V2 = 2 m/s m2 = 31.455 kg/s
S.G1 = 0.86 mt = m1 + m2
W2 = 890 kg/m3 = 48.631 + 31.455
mt = 80.0862 kg/s
11. In one section of water flow the pressure is 1500 kpa. If internal energy is 20 KJ/kg find the enthalpy at this
Point. A. 21.50 KJ/kg B. 23.50 KJ/kg C. 25.50 KJ/kg D. 27.50 KJ/kg
P = 1500 kpa; U = 20 KJ/kg
h = u + PV
= 20 KJ/kg + (15000kpa) (1/1000 m3/kg)
h = 21.5 KJ/kg
12. In a constant temperature process at 150oC, heat is transferred with an entropy change of 0.5 KJ/K.
Determine the heat added for the system. A. 201.50 KJ B. 211.50 KJ C. 221.50 KJ D. 231.50 KJ
T = 150°C; S = 0.5 KJ//K
S = Q/T
Q = (S) (T)
= (0.5KJ/K) (150+273)
Q = 211.5 KJ
13. Two hundred kg of water is added to 4000 kg of alcohol with specific gravity of 0.8. Determine the specific
Volume after mixing, m3/kg. A. 0.0124 B. 0.02124 C. 0.00124 D. 0.000124
m1 = 200 kg ρm= m/v V = 1/ρm
m2= 4000 kg ρm= m1+m2/V1+V2 =1/807.962
SG = 0.8 = _______200+4000__________ = 1.24x10-3
ρm= m/v (200kg/1000kg/m3) + (4000kg/800) V = 0.00124
m3/kg
ρm= m1+m2/V1+V2 ρm = 807.962 m3/kg
14. A batch of concrete consisted of 240 lbs fine aggregate, 380 lbs coarse aggregate, 100 lbs, cement, and 5
Gallons water. The specific gravity of the sand and gravel may be taken as 2.65 and that of the cement as
3.10. How much by weight of cement is required to produce one cubic yard?
A. 547.14 lb/yd3 B. 647.14 lb/yd3 C. 747.14 lb/yd3 D. 847.14 lb/yd3
Vol of water = (5 gal) (1ft3/7.481 gal)
= 0.6684ft3
Vol of sand & gravel = (240+380)lb / (2.65) (62.4lb/ft3)
= 3.749 ft3
Vol of cement = 100lb /3.10 (62.4lb/ft3)
= 0.51696ft3
Total Vol = 0.6684+3.749+0.51696
= 4.934 ft3
Weight of cement/ft3 of concrete mixture = 100lb/4.934ft3
= (20.267lb/ft3) (3ft/1yd) 3
Weight of cement/ft3 of concrete mixture= 547.29 lb/yd3
15. A vessel has a pressure of 200 Kpag. The atmospheric pressure is 10 m of water equivalent. Find the
absolute pressure in m of water.
A. 28.39 m of H2O B. 30.39 m of H2O C. 32.39 m of H2O D. 34.39 m of H2O
Pg = 200kpa
Patm = 10 m of H2O
Pabs = Pg + Patm
= (200kpa) (10.33 m H2O/101.325) + 10m
Pabs = 30.39 m H2O
16. Water flows in a pipe at the rate of 10 kg/s. If speed of flow is 10 m/s, find the pipe diameter.
A. 30.23 mm B. 35.68 mm C. 38.39 mm D. 42.39 mm
m = 10 kg/s; v = 10 m/s Q = AV
Q = mv 100N = (π/4) (D) 2 (10m/s)
= (10kg/s) (10m/s) = 3.568m
Q = 100 N D = 35.68 mm
17. R-134a flows in a pipe at 30oC with a specific volume of 0.04434 m3/kg. The internal energy and enthalpy
of R-134a are 250.8 KJ/kg and 273.0 KJ/kg respectively. The pressure of the refrigerant in Mpa is
A. 0.5 B. 0.4 C. 0.3 D. 0.1
T = 30 °C v= 0.04434 m3/kg U = 250.8 KJ/kg h = 273.0 KJ/kg
P=?
h = u+ (p) (v)
273.0 KJ/kg = 250.8KJ/kg + (P) (0.04434m3/kg)
P = (500.67 kpa) (1Mpa/1000kpa)
P = 0.50067 Mpa
18. Water is heated on an electrical range with a power rating of 1.5 KW for a period of 18 minutes. The initial
and final temperatures of the water are 15oC and 85oC and 70% of electrical heat is transferred to the
water. What is the amount of water? A. 1.4 kg B. 3.9 kg C. 5.5 kg D. 9.2 kg
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