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Tcecarenko [31]
3 years ago
10

Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5

5°C at a rate of 2 kg/s, determine the exit temperature of air. Solve using appropriate software.
Engineering
1 answer:
Furkat [3]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

W = \dot{m}c_p \Delta T

Where

C_p = Specific heat

\dot{m}= Mass flow rate

\Delta T = Change in Temperature

Our values are given as

C_p = 1.005kJ/kgK \rightarrow Specific heat of air

T_1 = 50\°C

\dot{m} = 2kg/s

W = 8kW

From our equation we have that

W = \dot{m}c_p \Delta T

W = \dot{m}c_p (T_2-T_1)

Rearrange to find T_2

T_2 = \frac{W}{\dot{m}c_p}+T_1

Replacing

T_2 = \frac{8}{2*1.005}+(50+273)

T_2 = 326.98K \approx 53.98\°C

Therefore the exit temperature of air is 53.98°C

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2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

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PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
3 years ago
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