Fundamentals Of Engineering Thermodynamics, 9e
9th Edition
ISBN: 9781119391432
Author: MORAN
Publisher: WILEY
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Chapter 4, Problem 4.69P
To determine
The mass flow rate of air.
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The figure belows shows three components of an air-conditioning system, where T3= 115°F and m˙3= 1.5 lb/s. Refrigerant 134a flows through a throttling valve and a heat exchanger while air flows through a fan and the same heat exchanger. Data for steady-state operation are given on the figure. There is no significant heat transfer between any of the components and the surroundings. Kinetic and potential energy effects are negligible.
Modeling air as an ideal gas with constant cp = 0.240 Btu/lb · °R, determine the mass flow rate of the air, in lb/s.
The figure belows shows three components of an air-conditioning system, where T3 = 115°F and ng = 3 lb/s. Refrigerant 134a
flows through a throttling valve and a heat exchanger while air flows through a fan and the same heat exchanger. Data for steady-state
operation are given on the figure. There is no significant heat transfer between any of the components and the surroundings. Kinetic
and potential energy effects are negligible.
Air
T = 535°R
p= 0.240 Btu/lb-"R
Saturated liquid R-134a
T3, m3
Fan
Wey=-0.2 hp
Throttling
valve
5
www
Saturated vapor
Ps=P4
P.= 60 lbfin.?
T;= 528°R +2
- Heat exchanger
Modeling air as an ideal gas with constant c, = 0.240 Btu/lb - °R, determine the mass flow rate of the air, in Ib/s.
mi =
Ib/s
= 95°F and m3 = 1.5 lb/s. Refrigerant 134a
The figure belows shows three components of an air-conditioning system, where T3
flows through a throttling valve and a heat exchanger while air flows through a fan and the same heat exchanger. Data for steady-
state operation are given on the figure. There is no significant heat transfer between any of the components and the surroundings.
Kinetic and potential energy effects are negligible.
Air
Tj = 535°R
C,= 0.240 Btu/I6•°R
Saturated liquid R-134a
T3, ṁ3
Fan
Wey = -0.2 hp
Throttling
valve
4
Saturated vapor
P5=P4
P4 = 60 lbf/in.2
T = 528°R
-Heat exchanger
Modeling air as an ideal gas with constant c, = 0.240 Btu/lb· °R, determine the mass flow rate of the air, in Ib/s.
i
Ib/s
Chapter 4 Solutions
Fundamentals Of Engineering Thermodynamics, 9e
Ch. 4 - Prob. 4.1ECh. 4 - Prob. 4.2ECh. 4 - Prob. 4.3ECh. 4 - Prob. 4.4ECh. 4 - Prob. 4.5ECh. 4 - Prob. 4.6ECh. 4 - Prob. 4.7ECh. 4 - Prob. 4.8ECh. 4 - Prob. 4.9ECh. 4 - Prob. 4.10E
Ch. 4 - Prob. 4.11ECh. 4 - Prob. 4.12ECh. 4 - Prob. 4.13ECh. 4 - Prob. 4.14ECh. 4 - Prob. 4.15ECh. 4 - Prob. 4.1CUCh. 4 - Prob. 4.2CUCh. 4 - Prob. 4.3CUCh. 4 - Prob. 4.4CUCh. 4 - Prob. 4.5CUCh. 4 - Prob. 4.6CUCh. 4 - Prob. 4.7CUCh. 4 - Prob. 4.8CUCh. 4 - Prob. 4.9CUCh. 4 - Prob. 4.10CUCh. 4 - Prob. 4.11CUCh. 4 - Prob. 4.12CUCh. 4 - Prob. 4.13CUCh. 4 - Prob. 4.14CUCh. 4 - Prob. 4.15CUCh. 4 - Prob. 4.16CUCh. 4 - Prob. 4.17CUCh. 4 - Prob. 4.18CUCh. 4 - Prob. 4.19CUCh. 4 - Prob. 4.20CUCh. 4 - Prob. 4.21CUCh. 4 - Prob. 4.22CUCh. 4 - Prob. 4.23CUCh. 4 - Prob. 4.24CUCh. 4 - Prob. 4.25CUCh. 4 - Prob. 4.26CUCh. 4 - Prob. 4.27CUCh. 4 - Prob. 4.28CUCh. 4 - Prob. 4.29CUCh. 4 - Prob. 4.30CUCh. 4 - Prob. 4.31CUCh. 4 - Prob. 4.32CUCh. 4 - Prob. 4.33CUCh. 4 - Prob. 4.34CUCh. 4 - Prob. 4.35CUCh. 4 - Prob. 4.36CUCh. 4 - Prob. 4.37CUCh. 4 - Prob. 4.38CUCh. 4 - Prob. 4.39CUCh. 4 - Prob. 4.40CUCh. 4 - Prob. 4.41CUCh. 4 - Prob. 4.42CUCh. 4 - Prob. 4.43CUCh. 4 - Prob. 4.44CUCh. 4 - Prob. 4.45CUCh. 4 - Prob. 4.46CUCh. 4 - Prob. 4.47CUCh. 4 - Prob. 4.48CUCh. 4 - Prob. 4.49CUCh. 4 - Prob. 4.50CUCh. 4 - Prob. 4.51CUCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.31PCh. 4 - Prob. 4.32PCh. 4 - Prob. 4.33PCh. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Prob. 4.37PCh. 4 - Prob. 4.38PCh. 4 - Prob. 4.39PCh. 4 - Prob. 4.40PCh. 4 - Prob. 4.41PCh. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - Prob. 4.48PCh. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - Prob. 4.51PCh. 4 - Prob. 4.52PCh. 4 - Prob. 4.53PCh. 4 - Prob. 4.54PCh. 4 - Prob. 4.55PCh. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Prob. 4.58PCh. 4 - Prob. 4.59PCh. 4 - Prob. 4.60PCh. 4 - Prob. 4.61PCh. 4 - Prob. 4.62PCh. 4 - Prob. 4.63PCh. 4 - Prob. 4.64PCh. 4 - Prob. 4.65PCh. 4 - Prob. 4.66PCh. 4 - Prob. 4.67PCh. 4 - Prob. 4.68PCh. 4 - Prob. 4.69PCh. 4 - Prob. 4.70PCh. 4 - Prob. 4.71PCh. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.74PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76PCh. 4 - Prob. 4.77PCh. 4 - Prob. 4.78PCh. 4 - Prob. 4.79PCh. 4 - Prob. 4.80PCh. 4 - Prob. 4.81PCh. 4 - Prob. 4.82PCh. 4 - Prob. 4.83PCh. 4 - Prob. 4.84PCh. 4 - Prob. 4.85PCh. 4 - Prob. 4.86PCh. 4 - Prob. 4.87PCh. 4 - Prob. 4.88P
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- The figure belows shows three components of an air-conditioning system, where T3 = 105°F and m3 = 1.5 lb/s. Refrigerant 134a flows through a throttling valve and a heat exchanger while air flows through a fan and the same heat exchanger. Data for steady-state operation are given on the figure. There is no significant heat transfer between any of the components and the surroundings. Kinetic and potential energy effects are negligible. Air T1 = 535°R C, = 0.240 Btu/lb-°R Saturated liquid R-134a T3, ṁ3 3. Fan = -0.2 hp Throttling valve Saturated vapor P5=P4 P4 = 60 lbf/in.2 T2= 528°R 2 Heat exchanger Modeling air as an ideal gas with constant c, = 0.240 Btu/lb · °R, determine the mass flow rate of the air, in Ib/s. m¡ = i Ib/sarrow_forwardThe figure belows shows three components of an air-conditioning system, where T3 = 95°F and m3 = 3 lb/s. Refrigerant 134a flows through a throttling valve and a heat exchanger while air flows through a fan and the same heat exchanger. Data for steady-state operation are given on the figure. There is no significant heat transfer between any of the components and the surroundings. Kinetic and potential energy effects are negligible. m₁ = Saturated liquid R-134a T3, m3 i Throttling valve lb/s P4 = 60 lbf/in.² Air T₁=535°R Cp = 0.240 Btu/lb-ºR Fan wwww ²+² 2 T₂=528°R Wey=-0.2 hp Modeling air as an ideal gas with constant cp = 0.240 Btu/lb. °R, determine the mass flow rate of the air, i lb/s. Saturated vapor P5 P4 -Heat exchangerarrow_forwardIn an air conditioning system running at steady-state, m ̇ = 0.7 kg/s of refrigerant 3 134a in saturated liquid state at 48◦C flow through a throttling valve reducing its pressure to a value of p4 = 4 bars. The system is shown in Fig. 1. Then the refrigerant flows through the internal side of a heat exchanger exiting at saturated vapor with p5 = p4. Air enters the external side of the heat exchanger at T1 = 300 K and exits at T2 = 295 K moved by a fan ̇ Figure 1: Problem 1 that consumes WCV = 0.15 kW. Determine the mass flow rate of the air, in kg/sarrow_forward
- Separate streams of air and water flow through the compressor and heat exchanger arrangement shown in the figure below, where m, 0.6 kg/s and To = 30°C. Steady-state operating data are provided on the figure. Heat transfer with the surroundings can be neglected, as can all kinetic and potential energy effects. The air is modeled as an ideal gas. 1 Air P₁ = 1 bar T₁ = 300 K m Compressor A P2=3 bar -2 T₂=600 K Determine: WEVA ܒܝ www wwww +6 T6, P6-Ps P₁ = 9 bar T₁=800 K Compressor B 3- P3 P2 T₁=450 K Heat exchanger 5+ 4 Water T5= 20°C Ps= 1 bar (a) the total power for both compressors, in kW. (b) the mass flow rate of the water, in kg/s. WevB =arrow_forwardTwo separate streams of air and steam flow through a compressor system and a heat exchanger, as shown in the figure. The diagram provides steady state operating data for this system. The transfer of heat with the surroundings can be neglected, as well as the effects of kinetic and potential energy. Air can be considered an ideal gas. Decide:a) The total power generated by both compressors: kW b) The mass flow of water: kg/sarrow_forward4.30 Refrigerant 134a enters a heat exchanger operating ai steady state as a superheated vapour at 10 bars. 60°C. where it is cooled and condensed to saturated liquid at 10 bars. The mass flow rate of the refrigerant is 10 kg/min. A separate stream of air enters the heat exchanger at 37°C with a mass flow rate of 80 kg/min. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the exit air temperature, in °C.arrow_forward
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