PRINT COMPANION ENGINEER THERMO
9th Edition
ISBN: 9781119778011
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 = 2 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 m˙4= 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.
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.
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.
Chapter 4 Solutions
PRINT COMPANION ENGINEER THERMO
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 = 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_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_forward
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