Question

Power cycle Heat pump cycle $dot{Q}_H$ $dot{Q}_C$ $dot{W}_{cycle}$ $dot{Q}'_H$ $dot{Q}'_C$ $dot{W}'_{cycle}$ (a) 60 40 20 80 60 20 (b) 120 80 40 100 80 20 Hot reservoir, $T_H = 500 K$ $dot{Q}_H$ $dot{W}_{cycle}$ Power cycle $dot{Q}_C$ $dot{Q}'_H$ $dot{Q}'_C$ Cold reservoir, $T_C = 300 K$ $dot{W}'_{cycle}$ Heat pump

          Power cycle
Heat pump cycle
$dot{Q}_H$
$dot{Q}_C$
$dot{W}_{cycle}$
$dot{Q}'_H$
$dot{Q}'_C$
$dot{W}'_{cycle}$ 
(a)
60
40
20
80
60
20
(b) 120
80
40
100
80
20
Hot reservoir, $T_H = 500 K$
$dot{Q}_H$
$dot{W}_{cycle}$
Power cycle
$dot{Q}_C$
$dot{Q}'_H$
$dot{Q}'_C$
Cold reservoir, $T_C = 300 K$
$dot{W}'_{cycle}$
Heat pump

Power cycle
Heat pump cycle
dotQH
dotQC
dotWcycle
dotQ'H
dotQ'C
dotW'cycle
(a)
60
40
20
80
60
20
(b) 120
80
40
100
80
20
Hot reservoir, TH = 500 K
dotQH
dotWcycle
Power cycle
dotQC
dotQ'H
dotQ'C
Cold reservoir, TC = 300 K
dotW'cycle
Heat pump

Added by Glenn -.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Supreeta N

11/16/2023

Step 1

According to the first law of thermodynamics, the energy input to the system (QH) should be equal to the sum of the work output (W) and the energy rejected from the system (Qc). In this case, QH = 60 kW, Qc = 40 kW, and W = 20 kW. QH = W + Qc 60 kW = 20 kW + 40 Show more…

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Shown in the figure below is a system consisting of a power cycle and a heat pump cycle, each operating between hot and cold reservoirs whose temperatures are 500 K and 300 K, respectively. All energy transfers are positive in the directions of the arrows. The accompanying table provides two sets of steady-state data, in kW. For each set of data, determine if the system is operating in accordance with the first and second laws of thermodynamics. Power cycle: QH = 60 kW, Qc = 40 kW, W = 20 kW Heat pump cycle: QH = 80 kW, Qc = 60 kW, W = 20 kW (a) (b) Hot reservoir, T = 500 K Power cycle: W = 60 kW, Qc = 40 kW Heat pump cycle: QH = 80 kW, W = 20 kW Cold reservoir, Tc = 300 K