1. Water flows between two points 1, 2. The volumetric flow rate is 20 litres/min. Point 2 is 50 m higher than point 1. The pipe internal diameters are 0.5 cm at point 1 and 1 cm at point 2. The pressure at point 2 is 1 atm. Calculate the pressure at point 1.
a) 4.6 bar
b) 5.6 bar
c) 4.1 bar
d) 5.1 bar
Explanation: ΔP/ρ + Δv2/2 + gΔh + F = W
ΔP = P2 – P1 (Pa)
Δv2 = v22 – v12
Δh = h2 – h1 (m)
F = frictional energy loss (mechanical energy loss to system) (J/kg)
W = work done on system by pump (J/kg)
ρ = 1000 kg/m3
Volumetric flow is 20/ (1000.60) m3/s
= 0.000333 m3/s
v1 = 0.000333/(π(0.0025)2) = 16.97 m/s
v2 = 0.000333/ (π(0.005)2) = 4.24 m/s
(101325 – P1)/1000 + [(4.24)2 – (16.97)2]/2 + 9.81.50 = 0
P1 = 456825 Pa (4.6 bar).
2. Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4m × 5m × 6m at 100 kPa and 25° C.
a) 1.16 kg/m3, 0.00116, 140 kg
b) 1.17 kg/m3, 0.00117, 140 kg
c) 1.15 kg/m3, 0.00115, 140 kg
d) 1.14 kg/m3, 0.00114, 140 kg
Explanation: At specified conditions, air can be treated as ideal gas.
The gas constant of air is R = 0.287 k Pa-m3/Kg K
The density of the air is determined from the ideal-gas relation, P = ρ R T to be
ρ = P/RT = (100 kPa)/((0.287 k Pa-m3/Kg K) (25+273.15)K) = 1.17 kg/ m3
Then the specific gravity of the air becomes
SG = (ρ)/ρH2O = (1.17 kg/m3)/(1000 kg/m3) = 0.00117
Finally, the volume and mass of the air in the room are
V = (4m) (5m) (6m) = 120 m3
m = ρV = (1.17 kg/m3) (120 m3) = 140 kg.
3. Heat transferred to raise or lower the temperature of a material is called Specific heat.
a) True
b) False
Explanation: Heat transferred to raise or lower the temperature of a material is called sensible heat; change in the enthalpy of a system due to variation in temperature is called sensible heat change, whereas the term specific heat refers to heat capacity expressed on a per-unit-mass basis.
4. The symbol “Û” refers to?
a) Amount
b) Rate of enthalpy
c) Molar flow rate
d) Specific internal energy
Explanation: The symbol “Û” refers to Specific internal energy in a close system in Enthalpy change in non-reactive processes whereas in an open- system symbol “Ĥ” refers to specific enthalpy.
5. Calculate how much energy (kJ) is required to heat a bottle containing 300 mL of liquid water from room temperature (20 °C) to its normal boiling point (100 °C), while still remaining a liquid.
a) 104.4 kJ
b) 100.4 kJ
c) 140.4 kJ
d) 104.5 kJ
Explanation: Enthalpy (“Water”, 20, 000,”l”)
= 6.032 kJ/mol \((∫_{20}^{100} C_{pH_2O} dT)\)
= \( (\frac{6.032 kJ}{mol}) (\frac{1g}{ml}) (\frac{mol}{18.02 g}) (\frac{300 ml}{1})\) = 100.4 kJ
6. What term is used for Temperature measured by Thermometer?
a) Wet- bulb temperature
b) Dry- bulb temperature
c) Dew point temperature
d) Normal temperature
Explanation: The dry-bulb temperature (DBT) is the temperature of air measured by a thermometer freely exposed to the air but shielded from radiation and moisture. DBT is the temperature that is usually thought of as air temperature, and it is the true thermodynamic temperature.
7. Estimate the following properties of humid air at 41°C and 10% relative humidity:
Absolute humidity = 0.0048 kg H2O/ kg DA
Wet-bulb temperature = 19°C
Humid volume = 0.895 M3/ kg DA
Dew point = 3°C
Specific enthalpy = 54.2 – 0.7 = 53.5 kJ/ kg DA
What is the amount of water in 150 m3 of air at these conditions?
a) 0.605 kg H2O
b) 0.705 kg H2O
c) 0.805 kg H2O
d) 0.905 kg H2O
Explanation: \((\frac{150 m^3\,humid \,air}{1}) (\frac{kg \,DA}{0.895 \,M^3}) (\frac{0.0048 \,kg \,H_2O}{kg \,DA})\) = 0.805 kg H2O
8. What is the enthalpy of 130 g formic acid at 70°C and 1 atm relative to 25°C and 1 atm?
Cp for formic acid in the temperature range of interest is 0.524 cal g-1 °C-1.
a) 4.68 kcal
b) 3.06 kcal
c) 2.06 kcal
d) 2.68 kcal
Explanation: ΔH = ( 130 g) (0.524 cal g-1 °C-1) (70-25)°C
ΔH = 3065.4 cal or
ΔH = 3.06 kcal
Relative to H=0 at 25°C the enthalpy of formic acid at 70°C is 3.06 kcal.
9. Processes for phase change of vapour to liquid is called?
a) Vaporization
b) Fusion
c) Sublimation
d) Condensation
Explanation: Condensation is the change of the physical state of matter from gas phase into liquid phase, and is the reverse of evaporation.
10. 50 g benzaldehyde vapour is condensed at 179°C, What is the enthalpy of the liquid relative to the vapour?
Given: The molecular weight of benzaldehyde is 106.12, the normal boiling point is 179.0°C and the standard heat of vaporization is 38.40 kJ gmol-1. For condensation the latent heat is – 38.40 kJ gmol-1.
a) -18.09 kJ
b) -17.09 kJ
c) – 18.06 kJ
d) – 17.06 kJ
Explanation: ΔH = 50 g (- 38.40 kJ gmol-1). |(1 gmol)/(106.12 g)| = -18.09 kJ
Therefore, the enthalpy of 50 g benzaldehyde liquid relative to the vapour at 179°C is – 18.09 kJ. As heat is released during condensation, the enthalpy of the liquid is lower than the vapour.