Concept explainers
(a)
Interpretation:
Draw and label of the given stream of batch process with the expression for molar and mass flow rates for benzene in terms of the total molar flow rate of the stream.
Concept introduction:
Molar flow rate is explained as the number of moles that passes through an area in per unit time.
Similarly, mass flow rate is that particular mass of any content which passes through an area in per unit time.
(b)
Interpretation:
Draw and label of the given stream with the expression for the mass of nitrogen(kg) in terms of total moles(n mol) of this mixture.
Concept introduction:
Mass flow rate is that particular mass of any content which passes through an area in per unit time.
(c)
Interpretation:
Draw and label of the given stream with the expression for the molar flow rate of ethane (
Concept introduction:
Mass flow rate is that particular mass of any content which passes through an area in per unit time.
(d)
Interpretation:
Draw and label of the given stream with the expression for the molar flow rate of O2 and for the mole fractions of H2 O and O2 in the gas in terms of
Concept introduction:
Molar flow rate is explained as the number of moles that passes through an area in per unit time.
Mole fraction is the fraction of a particular content present in total mixture mole.
(e)
Interpretation:
Draw and label of the given stream with the expression for the gram-moles of N2 O4 in terms of n (mol mixture) and
Concept introduction:
Molar flow rate is explained as the number of moles that passes through an area in per unit time.
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ELEM.PRIN.OF CHEMICAL PROC.-W/ACCESS
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- step by steparrow_forwardThe power out of an adiabatic steam turbine is 5 MW and the steam enters turbine at 2 MPa and velocity of 50 m/s, specific enthalpy (h) of 3248 kJ/kg. The elevation of the inlet is 10 m higher than at the datum. The vapor mixture exits at 15 kPa and a velocity of 180 m/s, specific enthalpy (h) of 2361.01 kJ/kg. The elevation of the exit is 6 m higher than at the datum. Let g = 9.81 m/s². Assuming the ideal gas model and R = 0.462 KJ/(kg.K). The steam specific heat ratio is 1.283. Calculate:arrow_forwardThe power out of an adiabatic steam turbine is 5 MW and the steam enters turbine at 2 MPa and velocity of 50 m/s, specific enthalpy (h) of 3248 kJ/kg. The elevation of the inlet is 10 m higher than at the datum. The vapor mixture exits at 15 kPa and a velocity of 180 m/s, specific enthalpy (h) of 2361.01 kJ/kg. The elevation of the exit is 6 m higher than at the datum. Let g = 9.81 m/s². Assuming the ideal gas model and R = 0.462 KJ/(kg.K). The steam specific heat ratio is 1.283. Calculate:arrow_forward
- O Consider a 0.8 m high and 0.5 m wide window with thickness of 8 mm and thermal conductivity of k = 0.78 W/m °C. For dry day, the temperature of outdoor is -10 °C and the inner room temperature is 20°C. Take the heat transfer coefficient on the inner and outer surface of the window to be h₁ = 10 W/m² °C and h₂ = 40 W/m² °C which includes the effects of insulation. Determine:arrow_forwardCalculate the mass flow rate of the steam. Determine Cp and C₁ of steam.arrow_forwardstep by step pleasearrow_forward
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