FUND OF ENG THERMODYN-WILEYPLUS NEXT GEN
9th Edition
ISBN: 9781119840589
Author: MORAN
Publisher: WILEY
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Chapter 4, Problem 4.67P
a.
To determine
The exit temperature of the refrigerant from the valve.
b.
To determine
The mass flow rate of refrigerant in kg/s at the exit.
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The figure below provides steady-state data for a throttling valve in series with a heat exchanger. Saturated liquid Refrigerant 134a
enters the valve at a pressure of 9 bar and is throttled to a pressure of p2 = 1 bar. The refrigerant then enters the heat exchanger,
exiting at a temperature of 10°C with no significant decrease in pressure. In a separate stream, liquid water at 1 bar enters the heat
exchanger at a temperature of 25°C with a mass flow rate of m4 = 4 kg/s and exits at 1 bar as liquid at a temperature of 15°C. Stray
heat transfer and kinetic and potential energy effects can be ignored.
-Heat exchanger
1
2
3
wwww
P3 = P2
T3 = 10°C
Saturated
liquid R-134a
at p, = 9 bar
Valve
P2
5
T5 = 15°C
P5 = P4
Water
T = 25°C
P4 = 1 bar
Determine:
(a) the temperature, in °C, of the refrigerant at the exit of the valve.
(b) the mass flow rate of the refrigerant, in kg/s.
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.
Q.34
Chapter 4 Solutions
FUND OF ENG THERMODYN-WILEYPLUS NEXT GEN
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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- 4.30 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_forwardT-8arrow_forwardSeparate 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_forward
- = 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/sarrow_forwardThe 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/sarrow_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
- Steady-state operating data are provided for a compressor and heat exchanger in the figure below. The power input to the compressor is 50 kW. As shown in the figure, nitrogen (N2) flows through the compressor and heat exchanger with mass flow rate of 0.25 kg/s. The nitrogen is modeled as an ideal gas. A separate cooling stream of helium, modeled as an ideal gas with k=1.67, also flows through the heat exchanger. Stray heat transfer and kinetic and potential energy effects are negligible. Find: a) Enthalpy change of Nitrogen from inlet to the compressor and exit from Heat exchanger, ( ℎ1-ℎ3) in kJ/kg, b) Enthalpy change of Helium from inlet to and exit from Heat exchanger, (ℎ5-ℎ4) in kJ/kg, c) Mass flow rate of the helium in kg/s.arrow_forwardSeparate streams of steam and air flow through the turbine and heat exchanger arrangement shown in the figure below, where air enters location 5 at a rate of 1000 kg/min. The left turbine (Turbine 1) is able to produce 12,000 kW of power. 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. W2 = ? Turbine Turbine 2 P3 = 10 bar T3 = ? T2 = 400°C_ P2= 10 bar T = 240°C P4 = 1 bar Steam in P1 = 20 bar +6 T = 600°C www T5 = 1500 K -5 Ps = 1.35 bar Heat exchanger V T = 1200 K P6 = 1 bar Air in Determine: T3, in °C. • the mass flow rate of steam at 1, in kg/s. • the power output of the second turbine, in kW. • the magnitude of heat transfer between the steam and air, in kW. • the direction of the heat transfer (i.e., to the steam or from the steam).arrow_forwardThe figure belows shows three components of an air-conditioning system, where 105°F and 4.5 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.arrow_forward
- The 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_forwardThe 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_forwardA small nuclear reactor is cooled by passing liquid sodium liquid sodium out of the reactor at 2 bar and 400 ° C. It is cooled to 320 ° C by passing through a heat exchanger before returning to the reactor. In the heat exchanger heat is transferred from the sodium to the water, which enters the exchanger at 100 bar and 49 ° C and exits at the same pressure as saturated steam. The mass flow of sodium is 10,000 kg / h, its specific heat is constant and is 1.25 kJ / kg "C and the pressure drop is negligible. Determine (a) the mass flow in kg / h of evaporated water. in the heat exchanger. and (b) the heat flux transferred between the two fluids in kJ / h Neglect the variations of kinetic and potential energy through it.arrow_forward
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