An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at 7°C and 105 kPa at a rate of 0.55 m3/s while warm air enters at 34°C and 105 kPa. The air leaves the room at 24°C. The ratio of the mass flow rates of the hot to cold airstreams is 1.6. Using variable specific heats, determine (a) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room. (AMIE Summer 2024, 10 marks)
The gas constant of air is R = 0.287 kPa.m3/kg.K. The enthalpies of air are obtained from air table as
h1 = h @280 K = 280.13 kJ/kg
h2 = h @ 307 K = 307.23 kJ/kg
hroom = h @ 297 K = 297.18 kJ/kg
(a) We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as
Mass balance:
\(\begin{array}{l}{m_m} - {m_{out}} = \Delta {m_{system}} = 0\\\therefore {m_m} = {m_{out}}\\{m_1} + 1.6{m_1} = {m_3} = 2.6{m_1}\,as\,{m_2} = 1.6{m_1}\end{array}\)
Energy balance:
\(\mathop {{E_m} - {E_{out}}}\limits_{\scriptstyle{\rm{rate of net energy transfer}}\atop\scriptstyle{\rm{by heat, work, mass}}} = \mathop {\Delta {E_{system}}}\limits_{\scriptstyle{\rm{rate of change in internal}}\atop\scriptstyle{\rm{kinetic, potential etc}}{\rm{.}}} = 0\)
\(\begin{array}{l}{E_{in}} = {E_{out}}\\{m_1}{h_1} + {m_2}{h_2} = {m_3}{h_3}\end{array}\)
Combining the two gives
\(\begin{array}{l}{m_1}{h_1} + 1.6{m_1}{h_2} = 2.6{m_1}{h_3}\\{h_3} = ({h_1} + 1.6{h_2})/2.6\end{array}\)
Substituting,
h3 = (280.13 + 1.6 307.23)/2.6 = 296.81 kJ/kg
From air table at this enthalpy, the mixture temperature is
T3 = T@h = 296.81 kJ/kg = 296.6 K = 23.6 C
(b) The mass flow rates are determined as follows
\(\begin{array}{l}{\upsilon _1} = \frac{{R{T_1}}}{P} = \frac{{(0.287)(7 + 273)}}{{105}}\\ = 0.7654\,{m^3}/kg\\{m_1} = \frac{{{\upsilon _1}'}}{{{\upsilon _1}}} = \frac{{0.55{m^3}/s}}{{0.7654{m^3}/kg}} = 0.7186kg/s\\{m_2} = 2.6{m_1} = 2.6(0.7186) = 1.868kg/s\end{array}\)
The rate of heat gain of the room is determined from
Qgain =m3 (hroom - h3 ) (1.868 kg/s)(297.18 - 296.81) kJ/kg 0.691 kW
Therefore, the room gains heat at a rate of 0.691 kW.
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A 0.3-L glass of water at 200C is to be cooled with ice to 50C. Determine how much ice needs to be added to the water, in grams, if the ice is at (a) 00C and (b) -200C. Also determine how much water would be needed if the cooling is to be done with cold water at 00C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 00C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L. (AMIE Summer 2024, 10 marks, figures changed)
The density of water is 1 kg/L, and the specific heat of water at room temperature is c = 4.18 kJ/kg0C. The specific heat of ice at about 00C is c = 2.11 kJ/kg0C). The melting temperature and the heat of fusion of ice at 1 atm are 00C and 333.7 kJ/kg,.
\({m_w} = \rho V = 1x0.3 = 0.3\,kg\)
We take the ice and the water as our system, and disregard any heat and mass transfer. This is a reasonable assumption since the time period of the process is very short. Then the energy balance can be written as
\(\mathop {{E_{in}} - {E_{out}}}\limits_{\scriptstyle{\rm{Net energy transfer}}\atop\scriptstyle{\rm{by heat, work \& mass}}} = \mathop {\Delta {E_{system}}}\limits_{\scriptstyle{\rm{Change in internal kinetic}}\atop\scriptstyle{\rm{potential etc}}{\rm{. energies}}} \)
\(\begin{array}{l}0 = \Delta U\\0 = \Delta {U_{ice}} + \Delta {U_{water}}\\{[mc{(0 - {T_1})_{solid}} + m{h_{if}} + mc{({T_2} - 0)_{liquid}}]_{ice}} + {[mc({T_2} - {T_1})]_{water}} = 0\end{array}\)
Noting that T1, ice = 00C and T2 = 50C and substituting gives
\(m[0 + 333.7 + (4.18)(5 - 0)] + (0.3)(4.18)(5 - 20) = 0\)
m = 54.6 g
(b) When T1, ice = -200C instead of 00C, substituting gives
\(\begin{array}{l}m[(2.11)(0 - ( - 20)] + 333.7 + (4.18)(5 - 0)\\ + (0.3)(4.18)(5 - 20) = 0\end{array}\)
m = 0.0487 kg = 48.7 g
Cooling with cold water can be handled the same way. All we need to do is replace the terms for ice by a term for cold water at 0 C:
\(\Delta {U_{coldwater}} + \Delta {U_{water}} = 0\)
\([mc{({T_2} - {T_1})_{coldwater}} + [mc({T_2} - {T_1})] = 0\)
Substituting,
\([{m_{coldwater}}(4.180(5 - 0)] + (0.3)(4.18)(5 - 20) = 0\)
m = 0.9 kg = 900 g
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Nitrogen gas is compressed from 80 kPa and 27°C to 480 kPa by a 10-kW compressor. Determine the mass flow rate of nitrogen through the compressor, assuming the compression process to be (a) isentropic, (b) polytropic with n = 1.3, (c) isothermal, and (d) ideal two-stage polytropic with n = 1.3. (AMIE Summer 2024)
The gas constant of nitrogen is R = 0.297 kJ/kg.K. The specific heat ratio of nitrogen is k = 1.4.
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\({W_{comp,in}} = m\frac{{kR{T_1}}}{{k - 1}}\left\{ {{{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)}^{(k - 1)/k}} - 1} \right\}\)
\(10 = m\frac{{(1.4)(0.297)(300)}}{{1.4 - 1}}\left\{ {{{\left( {\frac{{480}}{{80}}} \right)}^{0.4/1.4}} - 1} \right\}\)
Giving m = 0.048 kg/s
(b) Polytropic compression with n = 1.3
\({W_{comp,in}} = m\frac{{nR{T_1}}}{{n - 1}}\left\{ {{{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)}^{(n - 1)/n}} - 1} \right\}\)
\(10 = m\frac{{(1.3)(0.297)(300)}}{{1.3 - 1}}\left\{ {{{\left( {\frac{{480}}{{80}}} \right)}^{0.3/1.3}}} \right\}\)
m = 0.051 kg/s
(c) Isothermal compression
\({W_{comp,in}} = mRT\ln \frac{{{P_1}}}{{{P_2}}}\)
\(10 = m(0.297)(300)\ln \left( {\frac{{480}}{{80}}} \right)\)
m = 0.063 kg/s
(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each stage is the same, and its value is determined to be
\({P_x} = \sqrt {{P_1}{P_2}} = \sqrt {80x480} = 196\,kPa\)
The compressor work across each stage is also the same, thus total compressor work is twice the compression work for a single stage:
\({W_{comp,in}} = 2m{w_{comp,I}} = 2m\frac{{nR{T_1}}}{{n - 1}}\left\{ {{{\left( {\frac{{{P_x}}}{{{P_1}}}} \right)}^{(n - 1)/n}} - 1} \right\}\)
\(10 = 2m\frac{{1.3x0.297x300}}{{1.3 - 1}}\left\{ {{{\left( {\frac{{196}}{{80}}} \right)}^{0.3/1.3}} - 1} \right\}\)
m = 0.056 kg/s
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Verify the validity of the last Maxwell relation for steam at 250 degree C and 300 kPa. (AMIE Summer 2024, 5 marks)
The last Maxwell relation states that for a simple compressible substance, the change in entropy with pressure at constant temperature is equal to the negative of the change in specific volume with temperature at constant pressure. various values are taken from table (Superheated water).
\({\left( {\frac{{\partial s}}{{\partial P}}} \right)_{T = {{250}^0}C}} \cong - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_{P = 300kPa}}\)
\({\left[ {\frac{{{s_{400kpa}} - {s_{200kpa}}}}{{(400 - 200)kPa}}} \right]_{T = {{250}^0}C}} \cong - {\left[ {\frac{{{V_{{{300}^0}C}} - {V_{{{200}^0}C}}}}{{{{(300 - 200)}^0}C}}} \right]_{P = 300kPa}}\)
\(\frac{{(7.3804 - 7.7100)kJ/kgK}}{{(400 - 200)kPa}} \cong \frac{{(0.87535 - 0.71643){m^3}/kg}}{{{{(300 - 200)}^0}C}}\)
\( - 20.00165{\rm{ }}{m^3}/kg.K{\rm{ }} \cong {\rm{ }} - 20.00159{\rm{ }}{m^3}/kg.K\)
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A fluid is confined in a cylinder by a spring-loaded, frictionless piston so that the pressure in the fluid is a linear function of the volume (p = a + bV). The internal energy of the fluid is given by the following equation
U = 34 + p V
Where U is in kJ, p in kPa, and V in m3. If the fluid changes from an initial state of 170 kPa, \(0.03{m^3}\) to a final state of 400 kPa, \(0.06{m^3}\), with no work other than that done on the piston, find the direction and magnitude of the work and heat transfer. (AMIE, Summer 2023)
\(\begin{array}{l}{U_2} - {U_1} = 5({p_2}{V_2} - {p_1}{V_1})\\ = 5(400x0.06 - 170x0.03)\\ = 94.5\,kJ\end{array}\)
Now,
p = a + bV
170 = a + b x 0.03
400 = a + b x 0.06
Solving the above two equations, we get
a = -60
b = 7666.67
Therefore
\(\begin{array}{l}W = \int_1^2 {pdV = \int_1^2 {(a + bV)dV} } \\ = a({V_2} - {V_1}) + b\frac{{{V_2}^2 - {V_1}^2}}{2}\\ = \left( {{V_2} - {V_1}} \right)\left[ {a + \frac{b}{2}({V_1} + {V_2})} \right]\\ = (0.06 - 0.03)\left[ { - 60 + \frac{{7666.67}}{2}(0.06 + 0.03)} \right]\end{array}\)
= 8.55 J
\(\begin{array}{l}Q = W + {U_2} - {U_1}\\ = 8.55 + 94.5 = 103.05\,kJ\end{array}\)
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A
turbine is supplied with steam at a gauge pressure of 1.4 MPa The
steam, after expansion in the turbine, flows into a condenser maintained
at a vacuum of 710 mm of Hg. The barometric pressure is 772 mm of Hg.
Express the inlet and exhaust steam pressure in pascal (absolute). Take
density of mercury as \(13600\,kg/{m^3}\) and acceleration due to
gravity as \(9.81\,m/{s^2}\).
(AMIE Summer 2023, 10 marks)
Given data
Gauge pressure = 1.4 MPa = \(1.4x{10^6}\,Pa\)
Vacuum pressure = 710 mm of Hg ;
Barometric pressure = 772 mm of Hg;
p = \(13600\,kg/{m^3}\)
g = \(9.81\,m/{s^2}\)
Atmospheric pressure
\(\rho gh = 13600x9.81x0.772 = 103x{10^3}\,N/{m^2}\)
Inlet steam pressure = Gauge pressure + Atmospheric pressure
\(1.4x{10^6} + 103x{10^3} = 1.503x{10^6}\,N/{m^2}\,or\,Pa\)
= 1.503 MPa
Condenser pressure = Barometric pressure - Vacuum pressure
= 772 - 710 = 62 mm of Hg = 62 x 133.3 = 8265 \(N/{m^2}\) or Pa
= 8.265 kPa
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