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(3) P = 0.0400 T= 28.98 x = 0.913 H= 2342.64 H*= -204.85 M = 103.5 (9) P= 1.0135 T= 22 Sub = 78.0 H = 92.31 H*= -2455.18 M = 5494.5 3 (8) P= 1.0135 T= 12 Sub = 88.0 H = 50.47 H*= -2497.02 M = 5494.5 P[bar] T[C] H[kJ/kg] H*[kJ/kg] M[kg/s] Data for the above Schematic Calculate: a. The heat rejected by the steam in kW. b. Heat rejected in the Cooling water in kW. C. Find LMTD in °C and in °F.

(3) P = 0.0400 T= 28.98 x = 0.913 H= 2342.64 H*= -204.85 M = 103.5 (9) P= 1.0135 T= 22 Sub = 78.0 H = 92.31 H*= -2455.18 M = 5494.5 3 (8) P= 1.0135 T= 12 Sub = 88.0 H = 50.47 H*= -2497.02 M = 5494.5 P[bar] T[C] H[kJ/kg] H*[kJ/kg] M[kg/s] Data for the above Schematic Calculate: a. The heat rejected by the steam in kW. b. Heat rejected in the Cooling water in kW. C. Find LMTD in °C and in °F.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Practice Problem 2: (3) P = 0.0400 T= 28.98 X = 0.913 H = 2342.64 (9) P = 1.0135 T= 22 Sub = 78.0 H = 92.31 -> H*= -204.85 H*= -2455.18 M = 103.5 M = 5494.5 3 (8) P= 1.0135 T= 12 Sub = 88.0 H = 50.47 H*= -2497.02 M = 5494.5 P[bar] T[C] H[kJ/kg] H*[kJ/kg] M[kg/s] Data for the above Schematic Calculate: a. The heat rejected by the steam in kW. b. Heat rejected in the Cooling water in kW. Find LMTD in °C and in °F. d. If U=2500 W/(m2.°C), find the surface area of the pipes. С.
Transcribed Image Text:Practice Problem 2: (3) P = 0.0400 T= 28.98 X = 0.913 H = 2342.64 (9) P = 1.0135 T= 22 Sub = 78.0 H = 92.31 -> H*= -204.85 H*= -2455.18 M = 103.5 M = 5494.5 3 (8) P= 1.0135 T= 12 Sub = 88.0 H = 50.47 H*= -2497.02 M = 5494.5 P[bar] T[C] H[kJ/kg] H*[kJ/kg] M[kg/s] Data for the above Schematic Calculate: a. The heat rejected by the steam in kW. b. Heat rejected in the Cooling water in kW. Find LMTD in °C and in °F. d. If U=2500 W/(m2.°C), find the surface area of the pipes. С.
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