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Slav-nsk [51]
3 years ago
9

he ventilating fan of the bathroom of a building has a volume flow rate of 28 L/s and runs continuously. If the density of air i

nside is 1.20 kg/m3, determine the mass of air vented out in one day. The mass of air is kg.
Engineering
1 answer:
stiks02 [169]3 years ago
7 0

Answer:

\Delta m = 2903.04 kg

Explanation:

The mass of vented out in one day is:

\Delta m = \rho_{air}\cdot \dot V \cdot \Delta t\\\Delta m = (1.20 \frac{kg}{m^{3}} )\cdot (28 \frac{L}{s}) \cdot (\frac{1 m^{3}}{1000 L} )\cdot (24 h)\cdot (\frac{3600 s}{1 h} ) \\\Delta m = 2903.04 kg

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Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

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Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

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