Fundamentals Of Engineering Thermodynamics
9th Edition
ISBN: 9781119391388
Author: MORAN, Michael J., SHAPIRO, Howard N., Boettner, Daisie D., Bailey, Margaret B.
Publisher: Wiley,
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Chapter 4, Problem 4.49P
To determine
The temperature change of the water.
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A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a
rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 Ibf/in.?, and 180°F, respectively; at the exit the pressure is 120
Ibf/in.? The pump requires 1/ 15 hp of power input. Water can be modeled as an incompressible substance with constant density of
60.58 lb/ft³ and constant specific heat of 1 Btu/lb · °R.
Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump.
A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a
rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 Ibf/in.², and 180°F, respectively; at the exit the pressure is 120
Ibf/in.2 The pump requires 1/ 15 hp of power input. Water can be modeled as an incompressible substance with constant density of
60.58 Ib/ft3 and constant specific heat of 1 Btu/lb · °R.
Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump.
AT =
i
°R
* Your answer is incorrect.
A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a
rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 lbf/in.², and 180°F, respectively; at the exit the pressure is 90 lbf/in.²
The pump requires 1/15 hp of power input. Water can be modeled as an incompressible substance with constant density of 60.58
lb/ft3 and constant specific heat of 1 Btu/lb. °R.
Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump.
ΔΤ :
= i 0.36
°R
Chapter 4 Solutions
Fundamentals Of Engineering Thermodynamics
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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- A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 lbf/in.2, and 180°F, respectively; at the exit the pressure is 60 lbf/in.2 The pump requires 1/ 35 hp of power input. Water can be modeled as an incompressible substance with constant density of 60.58 lb/ft3 and constant specific heat of 1 Btu/lb · °R. Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump.arrow_forwardA pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 lbf/in.2, and 180°F, respectively; at the exit the pressure is 90 lbf/in.² The pump requires 1/25 hp of power input. Water can be modeled as an incompressible substance with constant density of 60.58 lb/ft3 and constant specific heat of 1 Btu/lb. °R. Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump. AT = °Rarrow_forward1. As shown in the figure below, Refrigerant 134a enters a condenser operating at steady state at 70 lbf/in2, 160 °F and is condensed to saturated liquid at 60 lbf/in on the outside of tubes through which cooling water flows. In passing through the tubes, the cooling water increases in temperature by 20 'F and experiences no significant pressure drop. Cooling water can be modeled as incompressible with v-0.0161 ft'/lb and c = 1 Btu/lb R. The mass flow rate of the refrigerant is 3100 lb/h. Neglecting kinetic and potential energy effects and ignoring heat transfer from the outside of the condenser, determine: (a) The volumetric flow rate of the entering cooling water, in gal/min (b) The rate of heat transfer, in Btu/h, to the cooling water from the condensing refrigerant (5 points) Refrigerant 134a P= 70 in. T= 160 F 3100 heh 7,-7,-20F-20R Reirigerant 134a [P-60 lbin V Saturated liquidarrow_forward
- Current Attempt in Progress A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 Ibf/in.?, and 180°F, respectively; at the exit the pressure is 90 Ibf/in.2 The pump requires 1/25 hp of power input. Water can be modeled as an incompressible substance with constant density of 60.58 Ib/ft and constant specific heat of 1 Btu/lb · °R. Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump. AT = i °Rarrow_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 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.arrow_forward
- An oil pump operating at steady state delivers oil at a rate of 10 Ib/s through a 1-in.-diameter exit pipe. The oil, which can be modeled as incompressible, has a density of 100 lb/ft and experiences a pressure rise from inlet to exit of 40 Ibf/in?. There is no significant elevation difference between inlet and exit, and the inlet kinetic energy is negligible. Heat transfer between the pump and its surroundings is negligible, and there is no significant change in temperature as the oil passes through the pump. Determine the velocity of the oil at the exit of the pump, in ft/s, and the power required for the pump, in hp.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_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
- An oil pump operating at steady state delivers oil at a rate of 11 lb/s through a 1-in.-diameter exit pipe. The oil, which can be modeled as incompressible, has a density of 70 Ib/ft³ and experiences a pressure rise from inlet to exit of 40 Ibf/in?. There is no significant elevation difference between inlet and exit, and the inlet kinetic energy is negligible. Heat transfer between the pump and its surroundings is negligible, and there is no significant change in temperature as the oil passes through the pump. Determine the velocity of the oil at the exit of the pump, in ft/s, and the power required for the pump, in hp.arrow_forwardRefrigerant 134a enters an insulated diffuser as a saturated vapor at 120°F with a velocity of 1400 ft/s. The inlet area is 1.4 in?. At the exit, the pressure is 400 Ibf/in2 and the velocity is negligible. The diffuser operates at steady state and potential energy effects can be neglected. Determine the mass flow rate, in Ib/s, and the exit temperature, in °F.arrow_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_forward
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